Pearson Edexcel GCSE Maths · Topic P
Combined events, sample spaces, trees, conditional probability and risk. Every question below is the real Pearson question as it is printed on the paper, with its mark scheme. Then practise with the original printable pack.
Tier
Total recorded: 0 of 105 marks
28 original questions written in the style of the real papers. Attempt one on paper, then reveal the mark scheme and award yourself the marks. Marks you award are saved in this browser against that question.
1
120 people were asked whether they own a bicycle. 75 of them were adults and the rest were children. Of the adults, 45 own a bicycle. Of the children, 27 own a bicycle.
(a)
(1)
(b)
(2)
(c)
(1)
(Total for Question 1 is 4 marks)
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2
A bag contains 5 red counters and 3 blue counters. A counter is taken at random, its colour recorded, and then it is put back in the bag. A second counter is then taken at random.
(a)
(1)
(b)
(2)
(Total for Question 2 is 3 marks)
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3
A spinner can land on red, blue, green or white. The table shows the probabilities of some of the colours. P(red) = 0.4, P(blue) = 0.25, P(green) = 0.15.
(a)
(2)
(b)
(2)
(Total for Question 3 is 4 marks)
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4
Two fair ordinary six-sided dice are rolled. The scores on the two dice are added together to give a total.
(a)
(2)
(b)
(2)
(Total for Question 4 is 4 marks)
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5
A biased coin is flipped 200 times. It lands on heads 130 times.
(a)
(2)
(b)
(1)
(c)
(1)
(Total for Question 5 is 4 marks)
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6
A café offers a set lunch. A customer chooses one sandwich (cheese, ham or tuna) and one drink (tea or coffee).
(a)
(2)
(b)
(1)
(Total for Question 6 is 3 marks)
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7
90 students travel to school by bus or on foot. 54 of the students are in Year 10 and the rest are in Year 11. Of the Year 10 students, 21 walk to school. Of the Year 11 students, 18 walk to school.
(a)
(1)
(b)
(2)
(Total for Question 7 is 3 marks)
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8
160 households are surveyed. 92 households own a car. Of the households that own a car, 48 also own a bicycle. Of the households that do not own a car, 34 own a bicycle.
(a)
(2)
(b)
(2)
(Total for Question 8 is 4 marks)
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9
A fair spinner has 8 equal sections. Three sections are blue. The spinner is spun 240 times.
(2)
(Total for Question 9 is 2 marks)
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10
A spinner has three equal sections labelled A, B and C. It is claimed that the spinner is fair. In 300 spins it lands on A 126 times, on B 93 times and on C 81 times.
(a)
(1)
(b)
(2)
(Total for Question 10 is 3 marks)
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11
A tennis player gets 52 of her first serves in during 80 attempts.
(a)
(2)
(b)
(1)
(Total for Question 11 is 3 marks)
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12
A biased spinner can land on red or blue. Leila spins it 30 times and gets red 18 times. Marcus spins the same spinner 400 times and gets red 264 times.
(a)
(1)
(b)
(1)
(c)
(2)
(d)
(1)
(Total for Question 12 is 5 marks)
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13
A spinner can land on A, B, C or D. P(A) = 0.18, P(B) = 0.27 and P(C) = 0.31.
(a)
(2)
(b)
(1)
(Total for Question 13 is 3 marks)
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14
A bag contains only red, yellow and green counters. P(red) = 2x, P(yellow) = x + 0.1 and P(green) = 0.3.
(a)
(3)
(b)
(1)
(Total for Question 14 is 4 marks)
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15
The same coin is tossed in two experiments. In Experiment A the coin is tossed 20 times. In Experiment B the coin is tossed 500 times.
(2)
(Total for Question 15 is 2 marks)
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16
A spinner lands on red 14 times in its first 40 spins. It is then spun 360 more times. Across all 400 spins it lands on red 134 times.
(a)
(1)
(b)
(2)
(c)
(2)
(Total for Question 16 is 5 marks)
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17
Nadia chooses one hat and one scarf. The hats are black and grey. The scarves are red, blue and green.
(a)
(2)
(b)
(1)
(Total for Question 17 is 3 marks)
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18
A code is made by choosing one digit from 1, 2, 3 and 4, followed by one letter from A, B and C.
(a)
(2)
(b)
(2)
(Total for Question 18 is 4 marks)
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19
A three-digit number is made using each of the digits 2, 4 and 7 exactly once.
(a)
(2)
(b)
(2)
(Total for Question 19 is 4 marks)
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20
A fair coin is tossed and a fair spinner numbered 1, 2 and 3 is spun.
(a)
(1)
(b)
(2)
(Total for Question 20 is 3 marks)
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21
Two fair four-sided spinners are each numbered 1, 2, 3 and 4. Both spinners are spun.
(a)
(2)
(b)
(2)
(Total for Question 21 is 4 marks)
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22
Spinner A is fair and numbered 1, 2, 3 and 4. Spinner B is fair and numbered 1, 2, 3, 4 and 5. Each spinner is spun once.
(a)
(2)
(b)
(2)
(c)
(1)
(Total for Question 22 is 5 marks)
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23
A box contains 4 cards with vowels on them and 6 cards with consonants on them. A card is chosen at random, replaced, and then a second card is chosen.
(3)
(Total for Question 23 is 3 marks)
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24
A bag contains 5 red counters, 3 green counters and 2 blue counters. Two counters are taken at random without replacement.
(a)
(2)
(b)
(3)
(Total for Question 24 is 5 marks)
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25
A bag contains n red counters and 4 blue counters. Two counters are taken at random without replacement. The probability that both counters are blue is 1/6.
(5)
(Total for Question 25 is 5 marks)
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26
120 students are surveyed. 68 students learn a musical instrument, 45 students play a school sport, and 30 students do both.
(a)
(2)
(b)
(2)
(Total for Question 26 is 4 marks)
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27
In a group of 80 students, 46 study French, 38 study Spanish and 22 study both languages.
(a)
(2)
(b)
(3)
(Total for Question 27 is 5 marks)
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28
A factory makes 150 items on the morning shift and 250 items on the evening shift. Nine morning-shift items and 10 evening-shift items are defective. One of the 400 items is chosen at random.
(a)
(2)
(b)
(2)
(Total for Question 28 is 4 marks)
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155 questions shown as printed, 9 reference-only · newest series first
Question and mark-scheme extracts © Pearson Education Ltd. Reproduced for educational use; the original paper is linked on every question. Contact hello@oscarsongtutoring.co.uk for removal.
Edexcel 1MA1 · November 2024 · Paper 1F · Q11 · 5 marks
5 marks
Examiner report
In part (a) of this question learners were asked to use the information given to complete the two-way table about the type of film liked by adults and children. There were a good proportion of learners who could complete this fully correctly using the given information and working out the missing values. Where incorrect answers were seen these generally had the given information correctly completed in the table but had errors in working with these numbers in order to determine the missing values. Some learners were able to complete some of the given values correctly but made errors in placing some values possibly due to the need to identify both the appropriate row and appropriate column to input the values in. In part (b) of the question learners were asked to give the probability that a person chosen at random from the 500 was an adult who said that they liked action films the best. A majority of learners were able to give the required probability. There were a significant proportion of incorrect answers also seen, common incorrect answers were (the probability of an adult chosen at random saying that they liked action films the best), or 100/150, which was awarded 1 mark for the correct numerator, or completely incorrect fractions using a variety of different numbers from the table. There were a minority of learners who had a correct answer, but made errors when trying to simplify or convert their probability fraction (in these cases subsequent working was ignored) or gave their probability as a ratio which is not an acceptable form but which does gain method marks. Learners should be reminded that probabilities should be given as a fraction, decimal or percentage.
500 people were asked which film type they liked best: comedy, thriller or action. 280 were adults. 100 adults preferred action. 80 children preferred thriller. In total, 150 preferred action and 200 preferred comedy.
Edexcel 1MA1 · November 2024 · Paper 1F · Q19 · 4 marks
4 marks
Examiner report
In part (a) of this question learners were asked to work with the probabilities provided to find an estimate for the number of times the dice would land on 2. Learners found this to be a challenging question. Whilst a minority were able to find the required estimate, often by first finding the missing probabilities and then using these to make the estimation, the majority were only able to find the sum of the unknown probabilities or could not make any progress with the question. A common error was to add the given probabilities and divide by 2. The most common mistake was to give an answer of 0.15. In part (b) of the question learners were asked how the answer to (a) would be affected if Kasim’s assumption was wrong. Correct answers often referred to the answer to (a) being larger or there being more 2’s. The majority of learners were not able to give a correct response to this question, referencing the probabilities changing, stating that the answer would change but not explaining how or saying that the answer would not change.
For a biased six-sided die, P(3)=0.10, P(4)=0.30, P(5)=0.05 and P(6)=0.25. Karim initially assumes P(1)=P(2) and rolls the die 500 times.
Edexcel 1MA1 · November 2024 · Paper 1H · Q2 · 4 marks
4 marks
Examiner report
In part (a) most learners made a correct start by finding the sum of the four probabilities in the table. Many went on to gain the first process mark, usually for subtracting 0.7 from 1 to find the sum of the two unknown probabilities although a few learners did so by multiplying 0.7 by 350 to find an estimate for the number of times the dice lands on 3, 4, 5 or 6. A good proportion of learners went on to divide 0.3 by 2 to find the probability of the dice landing on 2 but instead of completing the process many gave 0.15 as the final answer and gained no more marks. Some divided 500 by 0.15. Those that did go on to show a complete process gained the second process mark. Arithmetic errors made when working out 0.15 × 500 meant that the accuracy mark was sometimes lost. After finding the sum of the probabilities in the table some learners divided 0.7 by 2 instead of subtracting it from 1. They gained no marks. Some missed the point about the dice being biased and used 500 ÷ 6 to work out the number of times the dice lands on each number. In part (b) a good number of learners were able to give a correct explanation that the answer to part (a) will be greater or that the number of 2s will increase. Explanations that were not acceptable often stated that the answer to part (a) would change but failed to mention how it would change or stated that the probability would increase. Many of those with an answer of 0.15 in part (a) concentrated on probability rather than the number of expected outcomes. Since part (a) asked for an estimate for the number of times the dice will land on 2, answers in (b) relating to an increase in the probability did not gain any marks.
For a biased six-sided die, P(3)=0.10, P(4)=0.30, P(5)=0.05 and P(6)=0.25. Karim initially assumes P(1)=P(2) and rolls the die 500 times.
Edexcel 1MA1 · November 2024 · Paper 1H · Q19 · 4 marks
4 marks
Examiner report
This proved to be a very challenging question with relatively few learners able to work out the probability of player A winning the chess tournament. Some learners were able to make a start to the process and scored the first mark for finding at least one correct product. Often this product was 0.6 × 0.5, the probability of A winning against both B and C, and this was often followed by 0.6 × 0.3, the probability of A winning against both B and D. Most were unable to go on to show a complete process and it was common to see the sum of these two products, 0.48, given as the final answer. Those that multiplied their product by a correct probability, e.g. 0.6 × 0.5 × 0.2, earned the second mark and some were able to go on and show a complete process. The final accuracy mark was sometimes lost because of arithmetic errors in the multiplication of decimals. A common incorrect answer was 0.6 × 0.5 × 0.3, obtained by multiplying the probabilities of player A winning against each of the other three players. This incorrect triple product gained no marks. Tree diagrams were often attempted but in many cases these did not help learners to solve the problem because of the unfamiliar nature of this question. Many did not know how to combine the probabilities and often added them rather than multiplying.
Chess semi-finals are A v B and C v D; winners meet. P(A beats B)=0.6, P(A beats C)=0.5, P(A beats D)=0.3, P(C beats D)=0.2.
Edexcel 1MA1 · November 2024 · Paper 2F · Q7 · 5 marks
5 marks
Examiner report
There were occasions in this question where learners confused the statistical terms mean, mode, median and range, resulting in an incorrect method being used in either part (a) or more commonly in part (b). Part (a) was answered well, with many learners aware of the need to sum the ages and divide the total by 8. There were a few arithmetic errors when adding, but if the full method was shown, learners could still gain 1 mark. A variety of incorrect answers were seen, and very often these seemed to be statistics other than the mean. The most common errors seen were 10 (the mode), 11.5 (the median) and 6 (the range), all of which gained no marks. Part (b) also had many correct answers from working out the range. Sometimes either the maximum or minimum was incorrectly chosen, and as in part (a) we also saw some incorrect statistics, most often the mean. Less success was evident in part (c), with a larger than expected number of learners not being able to show the probability correctly on the scale provided. The most common incorrect response was to misinterpret the divisions shown on the scale and mark at 1 8 rather than 1 4 .
Eight ages are 14,10,10,13,15,9,15,10.
Edexcel 1MA1 · November 2024 · Paper 2F · Q16 · 3 marks
3 marks
Examiner report
This question required learners to show working to support their choice and many showed a valid start to the problem by finding one suitable probability or proportion to gain the first mark and credit was often given for responses using ratios of 10 : 30 or 7 : 18. Comparable probabilities in a common format such as percentages, decimals, or written as fractions with common denominators or even common numerators were required for the second mark. However, whilst an initial probability or proportion was commonly seen, many learners made no attempt to convert to a common comparable format so gained no further credit. Of those who recognised the need to compare in a common format, many gained full marks for providing accurate comparable figures and a correct decision. For those working with ratio, many were unable to find two ratios with either the same left- hand side or the same right-hand side in order to make a comparison, with some then attempting to convert to fractions unsuccessfully.
Box A has 10 red and 30 green sweets. Box B has 7 red and 18 green sweets.
Edexcel 1MA1 · November 2024 · Paper 2F · Q23 · 2 marks
2 marks
Examiner report
As with many questions requiring a written explanation about errors made in the way information is presented, this question appears to have been challenging for many learners. To gain the marks, learners needed to give clear statements about both the incorrectly placed 17 and the omission of 1 from the diagram. General comments such as “add the missing odd numbers” were incomplete and gained no credit. Many also gave ambiguous comments about “removing the 17 from the ‘A’ circle”, which suggested that they thought the overlap was not part of Set A and was considered to be a contradiction. When comments were not fully clear, credit could still be awarded if the diagram contained evidence to support the statements, for example in cases where learners had clearly indicated which 17 they wanted to remove or had added the number 1 in the right place.
The universal set E is the odd numbers between 0 and 20. Tom's Venn diagram contains: A-only {3, 9, 15, 17}; A ∩ B {17}; B-only {5, 7, 11, 13, 19}; outside both circles { }.
Edexcel 1MA1 · November 2024 · Paper 2H · Q4 · 2 marks
2 marks
Examiner report
The first of the written response questions and one that did cause problems. In many cases, learners understood the errors but a lack of clarity in their response let them down. The first mark was for realising and articulating that 17 should only occur in the intersection and not in the set ' A B . Many simply said that 17 needed to be removed from A, which did not gain the credit as it was too ambiguous (17 obviously is in set A). The second mark was for realising that the number 1 was missing from the diagram and stating where it should be. Many failed to gain the credit as they said things such as ‘include the other odd numbers’ and needed to specify the 1. Also, those that said ‘1 needs to go outside’ did not gain the mark as this could refer to outside of the rectangle.
The universal set E is the odd numbers between 0 and 20. Tom's Venn diagram contains: A-only {3, 9, 15, 17}; A ∩ B {17}; B-only {5, 7, 11, 13, 19}; outside both circles { }.
Edexcel 1MA1 · November 2024 · Paper 2H · Q22 · 5 marks
5 marks
Examiner report
The final question was a probability problem in which learners had to generate their own expressions without any scaffolding provided in the question. Many tried to work simply with numerical probabilities, and this scored zero. To gain the first mark a correct second probability was needed, and this did need to include algebra. The second mark was for a correct product, either for both red, both yellow, or one of each. The third mark was for forming a correct equation. Those who gained the first two, often failed to gain this mark as they only worked with one product and not two. The final two marks came from manipulating the algebra into an equation to solve (either quadratic or linear), and then the correct answer. Summary Based on the performance on this paper, learners/centres should: Spend time learning how to describe the various regions in Venn diagrams clearly, ideally using set notation. Practise articulating their mathematics in verbal discussions to aid them in written responses. Learn the shapes of common graphs. In particular remembering quadratic graphs are curves and not a series of line segments. With transformations take care to read the question carefully so they know which shape is the object and which is the image. Also, that if asked for a single transformation giving two will result in zero marks. Continue to work on annotating working clearly, especially in extended problem questions, to ensure all marks can be awarded. Practise further working with iterative formulae. Whenever possible use accurate figures without prematurely rounding or truncating in calculations and only round at the final stage. Practise working with multiple negative signs when working with algebra.
A box has only red/yellow counters; 3/5 are red. Two are drawn without replacement; probability same colour is 41/80.
Edexcel 1MA1 · November 2024 · Paper 3F · Q20 · 4 marks
4 marks
Examiner report
Part (a) saw some success with a good number of learners gaining 2 marks for correctly completing the tree diagram. Of those that didn’t, many gained 1 mark for correctly filling out the branch for Game 1, but made errors on Game 2 branches such as mixing up the order of 0.3 and 0.7. In part (b) many did not understand the need to multiply the 2 probabilities and instead added or simply gave their answer as 0.7. If a learner did not have correct probabilities in (a) they could still gain 2 marks on the follow through in (b) and this was seen on occasions.
Dan wins each card game with probability 0.3. He plays two independent games.
Edexcel 1MA1 · November 2024 · Paper 3H · Q4 · 4 marks
4 marks
Examiner report
A standard question for which most learners gained a good number of marks. Part (a) was well answered although there were a small number of learners who didn’t correctly place the probabilities for Game 2. Of those that gained 0 marks for part (b) it was usually because the probabilities were added instead of being multiplied. A significant number of candidates gained no marks for part (b) as although they calculated 0.7×0.7 they spoilt their method by multiplying or dividing this by 2 or by subtracting it from 1 or adding to another product. All of these resulted in 0 marks being awarded as M1 was for 0.7×0.7 as a standalone product. A number of learners scored 2 marks through the follow through mark from their diagram.
Dan wins each independent game with probability 0.3; plays twice.
Edexcel 1MA1 · June 2024 · Paper 1F · Q13 · 3 marks
3 marks
Examiner report
In (a) slightly less than half of candidates were able to find the required probability. Where correct answers were seen these were most commonly given as 100%. Common incorrect answers were likelihood statements such as ‘very likely’ or ‘certain’ or attempts to calculate a probability using the values of the £10 and £20 notes such as 30 or 40. In part (b) candidates needed to work out the value of the 1p and 2p coins in the bag, use this to identify the number of 1p and 2p coins and give the probability of taking a 1p coin from the bag. Only a minority of candidates were able to reach the correct probability. Many more were able to gain 1 mark which was often due to finding the number of 1p coins and 2p coins. Some candidates were able to identify that there were 20p worth of 1p coins and 20p worth of 2p coins, but did not go on to find the number of 1p and 2p coins. Common incorrect answers included 2 which appeared to be from an incorrect assumption that there were equal numbers of 1p coins and 2p coins in the bag rather than the values being equal or 40 or 40. If given as a percentage candidates should note they need to include the % sign. Some candidates incorrectly gave the probability answer in a ratio form.
There are only £10 notes and £20 notes in a wallet.
Edexcel 1MA1 · June 2024 · Paper 1F · Q23 · 3 marks
3 marks
Examiner report
The first part of this question was not well answered. Only a minority of candidates could write down the numbers that were in set P’ of the Venn diagram. The most common incorrect answer was to give the elements of set P suggesting that candidates were not familiar with the notation for ‘not’. In part (b) of this question candidates were asked to find the probability that a number chosen at random from the universal set ξ was in the set 𝑃∪𝑄. There were only a minority of candidates who could give a fully correct answer. A greater proportion of candidates were able to gain partial marks for answer of the form 𝑎 9 with 1 ≤𝑎≤8 having correctly identified that there were 9 values in the universal set. 9 was a common response with candidates seemingly confusing union and intersection of sets. A smaller number of candidates were able to identify the values in the set 𝑃∪𝑄 but did not know what to do in order to find the required probability, often adding the values in the set 𝑃∪𝑄 and the values in the universal set and using these as the numerator and denominator of the probability that they gave as the answer. Common incorrect answers included 0, 4 and 4 together with likelihood words and giving the answer as a ratio.
In a Venn diagram, P only contains 12 and 15; P ∩ Q contains 18; Q only contains 10 and 14; and 11, 13, 16 and 17 lie outside both circles.
Edexcel 1MA1 · June 2024 · Paper 1H · Q4 · 3 marks
3 marks
Examiner report
In part (a) almost half of the students correctly identified the numbers that are in set P. The most common errors were to list the three numbers in set P or the two numbers, 12 and 15, that are only in set P. Some students wrote down the correct six numbers but also included 18 and gained no mark. In part (b) many students scored one mark for writing a probability with the correct denominator of 9. A correct numerator was seen less frequently. Some students identified the members of the set P Q and scored one mark but then added the five numbers. It was evident that many students were unable to identify the region P Q and the most common mistake was to give the probability that the number is in set P Q . Another incorrect method was finding P(P) × P(Q).
In a Venn diagram, P only contains 12 and 15; P ∩ Q contains 18; Q only contains 10 and 14; and 11, 13, 16 and 17 lie outside both circles.
Edexcel 1MA1 · June 2024 · Paper 1H · Q16 · 2 marks
2 marks
Examiner report
A good number of students gained the first mark for giving the probability that the first sweet is orange as n n + . Often this probability was seen on a tree diagram. About half of these students were able to give the probability that the second sweet is orange as n n − and then multiply the two probabilities together. Just n n n n n n − − = + + was sufficient for both marks to be awarded but it was pleasing to see some students include the intermediate step ( 1) ( 1) n n n n − + . Many students gave clear concise responses which often involved cancelling the n. However, there were many who overcomplicated the problem by multiplying the fractions, then factorising the numerator and denominator before finally cancelling the common factor. In some cases, ( 1) ( 1) n n n n − + was incorrectly expanding to give n n − + before writing the correct final answer. If any incorrect algebra was shown after a correct product then the accuracy mark could not be awarded. A common error was to give the second probability as n n − + . Some students did not know what to do with the probabilities, often adding them rather than multiplying. Others hoped that a random mixture of algebraic fractions similar to that given in the question could somehow be manipulated to miraculously come up with the correct final answer. This approach did not earn any credit.
There are n orange sweets and 1 white sweet in a bag. Saira takes and eats one sweet at random, then takes and eats another without replacement. Show that the probability she eats two orange sweets is (n - 1)/(n + 1).
Edexcel 1MA1 · June 2024 · Paper 2F · Q11 · 1 mark
1 mark
Examiner report
Providing a valid reason with a decision was required in this written response question, with most students being awarded the mark for giving very clear and concise responses. A very wide range of explanations were seen and accepted, as shown in the mark scheme, and it was pleasing to see very few instances of contradictory reasons also being given compared to similar past questions. Where the mark was not given, the student had often argued that the statement was correct. Other responses that scored 0 involved statements that did not clearly indicate a decision, or statements that included incomplete statements such as 2 had a 90° angle without relating the angle to probability.
A spinner has three unequal sectors labelled 1, 2 and 3. Majid says the probability of landing on 2 is 1/3 because there are three sectors. Is Majid correct? Give a reason.
Edexcel 1MA1 · June 2024 · Paper 2F · Q23 · 4 marks
4 marks
Examiner report
This question was a good discriminator for this cohort, with the full range of different possible scores seen. Almost all students attempted the question and very few blank responses were seen. Many were able to gain the first mark, usually by working out 1 − 0.16 and thereby showing some understanding of probability and some gained the first mark for working correctly with ratio to find the total number of discs in the bag, but this was seen less frequently. A score of 2 marks was also often seen, usually where students got as far as a probability of 0.46 but then did not know how to go further, or for working with probability and ratio separately but not knowing how to combine both skills correctly to make further progress through the problem. Of the students who understood how to approach the problem and showed a complete process, many rounded their numbers prematurely, often reaching answers like 69, 70.5 or 71 and scored all but the accuracy mark. Centres should remind students to work with accurate values fully until the end of the question and avoid premature rounding or truncating.
There are only red discs, blue discs and yellow discs in a bag.
Edexcel 1MA1 · June 2024 · Paper 2H · Q4 · 4 marks
4 marks
Examiner report
This question was the first main challenge of the paper, and it was just that for a good number. The most common successful approach was to use the number and probability of yellow to find the total number of discs, followed by dividing by the ratio to get the number of red (and often blue) discs. It was then quite straight forward to work out the number of red discs. Others did find success by working with the probability and ratio first to find the probability of red, and then with the total to get the number of red. Some lost marks after working with ratio to gain the number of red and blue but didn’t clearly identify the value for red and lost the final mark. A significant number of students prematurely rounded or truncated their probability value, often working with 0.46 or 0.47 which resulted in incorrect values of 69 or 71 for the answer. Where it was clear from working that correct processes were being followed it was still possible for all process marks to be gained.
There are only red discs, blue discs and yellow discs in a bag.
Edexcel 1MA1 · June 2024 · Paper 2H · Q18 · 5 marks
5 marks
Examiner report
Students were awarded at least one mark for a correct probability for a second or third counter, and these were often seen in probability trees. Students were able to gain the second mark or the third mark or both for a correct product for either 2 reds and 1 green or for 3 reds, with fewer showing the need to add the probabilities rather than multiplying. The problem normally came in the final step for a full process as students often missed one or two of the possible ways of getting 2 reds and a green, and so their process was incomplete. There was an alternative approach seen which involved looking at how many possible combinations of 3 counters there are, and how many combinations involved no yellow counter. These two values were then combined to find a probability. Another alternative was to notice that any possible selection of the three counters from the reds and greens was valid, and since only one green was available, this led quickly to the correct answer. There was also credit given for the students who dealt with the counters being replaced but did all the correct products and added. This was in the form of a Special Case B2.
There are only 4 red counters, 3 yellow counters and 1 green counter in a bag.
Edexcel 1MA1 · June 2024 · Paper 3F · Q27 · 4 marks
4 marks
Examiner report
Part (a) saw mixed results with 0, 1 and 2 marks awarded regularly. A good number of students were able to complete the probability tree diagram correctly and gain 2 marks. Of those that gained 1 mark, completing coin B’s branches incorrectly was the most common error. In part (b) few students gained 2 marks for a correct answer. Of those that didn’t, most scored 0 marks and there was a huge variety of incorrect methods seen, such as 0.6 + 0.55 or showing the correct product but doing further incorrect work such as adding to another product. It should be noted that M1 could be gained for 0.6 × 0.55 only, so any additional work lost the method mark.
Tim has two biased coins, coin A and coin B.
Edexcel 1MA1 · June 2024 · Paper 3H · Q6 · 4 marks
4 marks
Examiner report
As would be expected on the higher paper, the completion of the tree diagram was done very well, with a majority of students scoring both marks in part (a). A few students seemed unsure of the probabilities for coin B seemingly thinking it was a conditional probability. In part (b) students scored one of the two marks for writing but calculating incorrectly 0.6×0.55; it was surprising how many students gained only this one mark given that they had access to a calculator. Of those that gained 0 marks for part (b) it was usually because the probabilities were added instead of being multiplied. A significant number of students gained no marks for part (b) as although they calculated 0.6×0.55 they spoilt their method by multiplying or dividing this by 2 or by subtracting it from 1 or adding to another product. All of these resulted in 0 marks being awarded as M1 was for only 0.6×0.55. Students need to be aware that workings seen in part a) ie. 0.33 on the end of the correct path was superseded by new workings in part b) so it is important to show correct processes in the relevant workings area.
Tim has two biased coins, coin A and coin B.
Edexcel 1MA1 · November 2023 · Paper 1F · Q9 · 3 marks
3 marks
Examiner report
For part (i) and (ii) the majority of students answered the questions correctly to gain 2 marks. Of those that didn’t, giving the answer in an incorrect notation, such as ratio, was often the issue; students should be reminded that correct notation for probabilities are fractions, decimals or percentages. Some students gave an answer of 8 22 for (ii), e.g. the probability for ‘pink’ instead of ‘not pink’. In (iii) the majority of students gained B1 for a correct answer with 0 and 0 given in equal measure. Incorrect answers generally involved incorrect notation such as 0:22 or worded answers, e.g. impossible.
A bag contains 22 counters: 5 blue, 9 red and 8 pink. One counter is chosen at random.
Edexcel 1MA1 · November 2023 · Paper 1H · Q18 · 4 marks
4 marks
Examiner report
A pleasing number students made a good start to this probability question, using the two given probabilities to work out the probability that spinner B lands on red, and scored the first mark. Almost half of these students then went on to show a complete process and the majority of those that did show a complete process were able to complete the arithmetic and give a correct final answer. Some students gained the second mark for writing at least one correct product but were unable to show a complete process. Probability tree diagrams were very common and helped some students to formulate their solutions. A common error was to use 1 24 as the probability that spinner B lands on red. Students doing this gained no marks.
Spinner A and spinner B are each spun once. The probability that spinner A lands on red is 1/4. The probability that both spinner A and spinner B land on red is 1/24
Edexcel 1MA1 · November 2023 · Paper 2F · Q11 · 2 marks
2 marks
Examiner report
Both part (a) and part (b) required students to select an appropriate description of the likelihood of an event occurring. A correct selection in part (b) occurred more frequently than in part (a), with over three quarters of the cohort correctly identifying evens for part (b).
A fair ordinary six-sided die is rolled once.
Edexcel 1MA1 · November 2023 · Paper 2H · Q10 · 3 marks
3 marks
Examiner report
The probability tree question is a familiar one and most were able to gain at least one mark in part (a) for at least two correct probabilities, with many completing the tree correctly. In part (b) however there was less success. A number gained a single mark for one correct product. The issue it seems was that many didn’t really understand the meaning of ‘at least’ and worked out the probability of only passing one test. This resulted in many gaining no further credit as their method was incomplete.
Shakir has to complete two tests. He can either pass or fail each test. The probability that he will pass the first test is 0.87. If he passes the first test, the probability he will pass the second test is 0.94. If he fails the first test, the probability he will pass the second test is 0.73
Edexcel 1MA1 · November 2023 · Paper 2H · Q21 · 4 marks
4 marks
Examiner report
This question proved too hard for most candidates, which is to be expected for a question at this level. The conditional aspect of the probabilities was normally missed, and very few were able to gain the first mark for finding the number of gold medals. Failure to gain this first mark meant no marks were awarded at all. A small number were awarded 1 mark for a correct expression for the number of golds. Those who went one step further and formed the correct equation typically completed the problem and gained all 4 marks, Summary Based on the performance on this paper, students should: • Remember correct money notation when working with problems involving currency. • Practise further with solving equations that involve negative coefficients. • Spend more time working with bearings and be encouraged to form diagrams to support. • Ensure they read questions carefully so, like in the iteration questions, they complete the question as it is asked. • Ensure they are familiar with the formula sheet as it will be available in Summer.
Vicky has a collection of medals. A Venn diagram shows the numbers of medals in her collection, where E = {all medals}, A = {English medals} and B = {gold medals}. The region in A only contains 7x − 11 medals, the region in both A and B contains x medals, the region in B only contains 5x − 3 medals, and the region outside both A and B contains 34 medals. Vicky is going to take at random a medal from her collection. Given that the medal is gold, the probability that the medal is English is 2/11
Edexcel 1MA1 · November 2023 · Paper 3F · Q15 · 5 marks
5 marks
Examiner report
It was quite common for candidates to score full marks on the frequency tree. Where this did not always happen the majority of candidates scored 1 mark placing the 64 and 36 correctly. Most candidates scored well on part (b), recognising which values from the frequency diagram to use. Some candidates wrote their probability out of 100 or as a ratio but this was less common. A mistake seen a few times was when the candidates had missed the 'given' part of the question and included the UK holidays for families and couples which scored them zero.
A travel agent sold 100 holidays, either UK or abroad. There were 64 family holidays and 36 couple holidays. Of the 18 holidays abroad, 11 were sold to couples.
Edexcel 1MA1 · November 2023 · Paper 3F · Q21 · 2 marks
2 marks
Examiner report
This question was mostly answered well as 200 × 0.4 = 80. The common incorrect answer was 200 ÷ 0.4 = 500.
A biased spinner has P(A)=0.4, P(B)=0.21, P(C)=0.32 and P(D)=0.07. It is spun 200 times.
Edexcel 1MA1 · November 2023 · Paper 3H · Q2 · 2 marks
2 marks
Examiner report
This question discriminated well between students working at the lower grades. Most students were successful and gained the 2 marks available. However, lower attaining students often divided 200 by 0.4 instead of multiplying. Other responses included cases where 0.4 × 200 was seen in working but then answers such as 200, 0.8%, 80% and 0.8 appeared on the answer line.
A biased spinner can land on A, on B, on C or on D. The table shows the probabilities that when the spinner is spun it will land on each letter. Letter: A B C D Probability: 0.4 0.21 0.32 0.07 Luka will spin the spinner 200 times.
Edexcel 1MA1 · November 2023 · Paper 3H · Q16 · 3 marks
3 marks
Examiner report
Where lower attaining students attempted this question their answers were often restricted to listing the possible numbers for each digit of the passcode without writing down the number of options for each digit. These students could not be awarded any credit. Having said this, the question discriminated well between higher attaining students with each of the possible marks often being awarded. Only a small proportion of students scored full marks for a correct probability. A significant number of students used fractions with denominator 9 in their calculations. They did not appear to appreciate the importance of “Zia tells her friend Amber that...” The most common error leading to the loss of the final mark was for students to include 1 in their list of prime numbers which led to an answer of
Zia has to set a 4-digit security passcode on her phone. Each digit of the passcode is a number from 1 to 9. She can use each number more than once. Zia tells her friend Amber that: the first digit is a cube number; the second digit is a prime number; the third digit is greater than 6; the fourth digit is an odd number. One possible 4-digit passcode is 1 3 8 3. Amber is going to have one attempt at guessing Zia's passcode.
Edexcel 1MA1 · June 2023 · Paper 1F · Q24 · 5 marks
5 marks
Examiner report
The completion of a fully correct Venn diagram in part (a) was rare. Many did not recognise “1” as a square number and many ignored the remaining numbers 2, 6, 8 and 10 from the universal set. Often students duplicated values in more than one section. In part (b), it was very clear that only a very few students understood the complement notation preferring instead to just give the probability that the number chosen was in set B. Many students gave their probability as a decimal without previously writing it as a fraction out of 10. 0.7 the correct answer of course gained full marks but an answer of say 0.3 alone begs the question, where has it come from, 10 ? or 20 ? etc.
E = {1,2,3,4,5,6,7,8,9,10}, A is the set of odd numbers, and B is the set of square numbers.
Edexcel 1MA1 · June 2023 · Paper 1F · Q31 · 2 marks
2 marks
Examiner report
Few solutions were fully correct in this question. 0.2 + 0.3 (= 0.5) was the most common error. Those correctly trying to find the product often gave an answer of 0.6 instead of 0.06 Often a lack of showing working out meant that 0.6 didn’t gain any credit. Students need to be encouraged to show their calculations for this reason. Many students highlighted the correct probabilities but did not go any further. Summary On the evidence of performance on this paper, students need to: ● take more care when reading the questions. ● take care when carrying out arithmetical operations and check their working to avoid careless errors. Also, they need to check that they have answered the question e.g. Q10 and Q14c. ● write clearly so that correct values quoted are not altered in subsequent working. ● show clear methods of working particularly in basic calculations such as finding fractions or percentages of quantities. ● set their working out clearly and systematically, crossing out working that is being replaced. ● write the decision when asked for e.g. Yes or No and keep comments short and simple to avoid contradiction. Explanations should be brief, but comprehensive (concise). ● estimate the demand of a question by looking at how many marks a question is worth (Q28 for example) in allocating time.
On Monday, Shayla works at home with probability 0.7 and at the office with probability 0.3. On Friday, independently, she works at home with probability 0.2 and at the office with probability 0.8.
Edexcel 1MA1 · June 2023 · Paper 1H · Q12 · 3 marks
3 marks
Examiner report
After gaining the first mark for writing at least one correct product many students were able to go on and show a complete method by finding the sum of the three relevant products. Some students chose to subtract the probability of Martha winning both games from 1. The majority of those that showed a full method were able to complete the arithmetic and give a correct final answer. A very common error was failing to include the probability of Martha losing both games and finding the sum of only two products suggesting that the students missed or misunderstood the words 'at least' in the question. Some students did not know what to do with the probabilities, often adding them rather than multiplying.
Martha plays a game twice. The probability that she wins the first game is 5/8. The probability that she wins the second game is 2/9, whether or not she wins the first game.
Edexcel 1MA1 · June 2023 · Paper 2F · Q6 · 2 marks
2 marks
Examiner report
Whilst most of the cohort could answer both parts correctly, students were slightly more successful in part (b).
Shari rolls a fair ordinary six-sided die once.
Edexcel 1MA1 · June 2023 · Paper 2F · Q15 · 3 marks
3 marks
Examiner report
This question required students to compare the likelihood of two outcomes, showing all working, and performance was disappointing. Whilst many students could express at least one probability, very few could show one outcome was more likely than the other and chose to state a conclusion that was unsupported. Comparing the probabilities as equivalent fractions with a common denominator, decimals, percentages or even a ratio was acceptable. A lot of students listed all the numbers on a dice, and highlighted the ones which were less than three, or listed all the numbers on the spinner and highlighted those which were more than five, but did not write this as a probability and therefore gained no marks. These students then went on to conclude that the spinner giving a number greater than 5 was more likely because there were more available options. Some who did write it as a probability often included 3 on the dice and/or 5 on the spinner, suggesting that there was a misunderstanding of ‘less than 3’ and ‘more than 5’. Of those who attempted to convert to a common format, many made the error of writing 6 or as 0.3 or 30%
Charlie throws a fair six-sided die and spins a fair eight-sided spinner numbered 1 to 8.
Edexcel 1MA1 · June 2023 · Paper 2F · Q19 · 5 marks
5 marks
Examiner report
The majority of students gained marks on this familiar frequency tree question, with part (a) being well attempted and many gaining at least one mark. Of those who were not awarded full marks, the vast majority were able to correctly place at least one of the given values, generally for placing 150 correctly, and then could calculate at least 1 or 2 of the missing values. However, many students placed 110 incorrectly as the number of students who have a bicycle and a car, simply reading this value from the question and not realising this was the total number of people with a bicycle and making use of the final piece of information given. Careful reading should be encouraged for this type of question and it was pleasing to see students checking that the values they had placed totalled 240 and correcting as necessary if 240 was not reached. Part (b) required students to use their frequency tree to find a percentage, with a large proportion of students who were successful with the frequency tree, correctly working out the required percentage. Whilst many identified the correct value to use as a numerator, a small number of students calculated this as a percentage of the total people rather than the number of people who had a car and therefore gained no marks. The most common errors from incorrect frequency trees led to 110 and 150 being used in part (b) but could be followed through for both marks if used correctly. Other students simply stated the values from their frequency tree or as a fraction rather than converting them to a percentage or gave a subtraction e.g. 150 – 45 or even 100 – 45.
At a factory, 240 people work in total; 150 have a car; 110 have a bicycle; and 65 bicycle owners do not have a car.
Edexcel 1MA1 · June 2023 · Paper 2H · Q10 · 4 marks
4 marks
Examiner report
A significant number of candidates tried to overcomplicate part (a) often carrying out complex calculations, when the answer was much simpler. A number of candidates worked out adding the result of the first throw of the dice to the table for the second probability. These candidates were given full credit for this response. Part (b) was the second explanation question, and again clarity was lacking for some. It was common to see responses such as “the bigger the value the more accurate”, or “it must be bigger” without any reference to 60 or part (a) and in these cases no mark was awarded.
A biased dice is thrown 60 times. The table shows information about the number that the dice lands on each time. Number on dice: 1, 2, 3, 4, 5, 6 — Frequency: 12, 7, 8, 9, 9, 15. Gethin throws the dice twice.
Edexcel 1MA1 · June 2023 · Paper 2H · Q22 · 2 marks
2 marks
Examiner report
Whilst only being 2 marks, this probability problem was probably found to be the hardest on the paper for students. Very few were able to gain any credit. To gain a single mark they had to either write an expression for getting all blue, or all red for n trials. It was this element that challenged students, with many giving examples for 2 or possibly 3 years, but not the generalisation that the mark scheme demanded, we did on occasion see elements such as but without the brackets, no credit could be given. A fully correct expression was needed for both marks, typically given as but also seen as which is a correct simplified form.
There are only blue pens and red pens in a box. The number of blue pens is four times the number of red pens. Rita takes at random one pen from the box. She records the colour of the pen and then replaces it in the box. Rita does this n times, where n ≥ 2.
Edexcel 1MA1 · June 2023 · Paper 3F · Q7 · 3 marks
3 marks
Examiner report
Part (a) of this question was not particularly well answered with many struggling to choose between “likely” and “evens” (with many crossings out and changes of mind seen). Even when the correct probability of 2 4 was seen candidates tended to go from this to ‘likely’ as their answer. In part (b) an answer of ‘certain’ was commonly seen. Many candidates showed misunderstanding of the 'less than 4’ element of the question by stating ‘impossible’, probably as there was no 4 on the spinner. These were by far the most common responses. Some gave an incorrect answer of 'likely'. In part (c) there were many correct answers of 0.6 Not all candidates showed working, but this was not required to gain marks. Candidates mostly stated their answer as a decimal, with very few giving the fraction or percentage equivalent.
A spinner has four equal sectors labelled 1, 2, 2 and 3. Ralph rolls a biased die for which P(5)=0.4.
Edexcel 1MA1 · June 2023 · Paper 3H · Q24 · 5 marks
5 marks
Examiner report
This question also discriminated well between high attaining students. Most students who made a good attempt at the question recognised the non-replacement nature of the problem and were able to make a start by forming expressions for probabilities in terms of x and/or y. A good proportion of these students were also able to use the ratio given in the question to make a substitution in order to get their expressions in terms of x only or y only. However, a significant number of students used “y = 4x” instead of the correct “y = 3x”. Invariably and sensibly, the overwhelming majority of students opted for getting their expressions in terms of x only. Those students who attempted substituting 3x for y into their probabilities before combining their products had more success than those who substituted later as the algebra became more complex. A significant number of students obtained correct expressions for the probability of Freda taking two pink counters and the probability of her taking two blue counters but gave the probability of her taking two green counters as 2𝑥−5 3𝑥 × 2𝑥−4 3𝑥−1 instead of 2𝑥−5 3𝑥 × 2𝑥−6 3𝑥−1 . This had clearly arisen from an error with signs earlier on in a student’s working. It was encouraging to see a good number of fully correct solutions from students working at the highest level. Summary Based on their performance on this paper, students are offered the following advice: • carry out a check to make sure you have not miscopied any numbers given in a question before you complete the question. • check that you have not miscopied algebraic expressions from a previous line in your working. • practise solving equations involving algebraic fractions. • compare the medians and either the ranges or the interquartile ranges in context when asked to compare two distributions represented by box plots. • ensure you can write and manipulate algebraic expressions accurately particularly with regard to the use of brackets and the signs of terms. • practise solving real life problems involving inverse proportionality. • carry out common sense checks where possible.
There is a total of y counters in a box. There are x pink counters and 5 blue counters in the box. The rest of the counters are green. x : y = 1 : 3. Freda takes at random two counters from the box. Find, in terms of x, an expression for the probability that Freda takes two counters of the same colour. Give your answer as a fraction in the form (ax^2 + bx + c)/(dx^2 + ex) where a, b, c, d and e are integers.
Edexcel 1MA1 · November 2022 · Paper 1F · Q7 · 3 marks
3 marks
Examiner report
Part (a) was answered well, the vast majority clearly understanding the concept of mode. In part (b) also, understanding of probability scales and probability in general was good. In (ii), some simply added a cross to the probability scale, failing to quote the actual numerical probability; this gained no marks. Some students are still using terms such as unlikely in terms of describing probability.
Here is a list of 8 letters. B C A A A A B A
Edexcel 1MA1 · November 2022 · Paper 1F · Q17 · 5 marks
5 marks
Examiner report
Again, understanding of probability was good and many students gained the full 2 marks in part (a). One mark was awarded for sight of 100 – 30 (= 70) or a correct value for the probability of taking a blue counter. 4 was a common incorrect answer by some of the less able students, assuming there was an equal number of coloured counters. Students should be reminded that expressing a probability as a ratio is not an acceptable form for this answer. In part (b), the correct answer of 45 was the modal answer, with incorrect methods following no particular pattern. A common incorrect attempt was also to try to share 30 in the ratio 2 : 3. Part (c) followed through student’s answers in (b). Explanations were usually acceptable, such that 25 counters could not be shared equally between red and yellow. It should be noted that just saying ‘25 can’t be divided by 2’ gained no credit. Also, students should be aware that any calculations given as evidence in such questions need to be accurate.
There are only blue, green, red and yellow counters in a bag. There are 30 blue counters and 100 counters in total. Ashin takes a counter at random.
Edexcel 1MA1 · November 2022 · Paper 1F · Q23 · 2 marks
2 marks
Examiner report
This question was not answered well at all. Sampling techniques were not well understood at this level. Many either multiplied 40 000 by 80 or divided by 15. Attempts to divide 40 000 by 80 often introduced arithmetic errors, some just left 500 as their answer and those that knew to multiply by 15 struggled to do so correctly. Some students tried to work with percentages which rarely gained any credit eg 15%. Students need to be encouraged to consider the suitability of their answers as some offered answers which were greater than 40 000, the total number of cars produced.
A car factory asks 80 people which model they are most likely to buy. The results are: A, 23 people; B, 15; C, 30; D, 12. The factory is going to make 40 000 cars next year. Work out how many model B cars it should make.
Edexcel 1MA1 · November 2022 · Paper 1H · Q10 · 4 marks
4 marks
Examiner report
Part (a) was answered quite poorly. Those who used the table to find that the probability of getting a 5 is 10 gained the method mark but many students ignored the fact that the spinner is biased and gave the probability of getting a 5 as 1 . Using 10 to work out an estimate of scoring 5 both times proved to be beyond many students. A common incorrect method was + 10 = 1 and often 10 or 1 was given as the final answer. In part (b) many of the students who gained the method mark for identifying the proportion of 1s as 6 were able to write this as 15%. Some students showed a method to write 6 as a percentage, e.g. 6 × 100, but could not complete the calculation whereas others did not know how to write it as a percentage. Dividing 40 by 6 was a common error. There were many students who did not identify the proportion of 1s as 6 .
Lina spins a biased 5-sided spinner 40 times. Her results are: score 1, frequency 6; score 2, frequency 8; score 3, frequency 9; score 4, frequency 7; score 5, frequency 10.
Edexcel 1MA1 · November 2022 · Paper 1H · Q20 · 3 marks
3 marks
Examiner report
Many of the higher attaining students made a good start to this probability question, showing understanding that the first card is not replaced. They often wrote at least one correct product of two fractions with denominators 11 and 10 respectively and gained the first mark. Only a small proportion of students showed a complete process and considered all six possible ways of taking cards of different colours. A common mistake was to use only three of the six ways. The majority of those that did show a complete process were able to complete the arithmetic and give a correct final answer. Some students approached the problem by finding the probability that Alfie takes two cards of the same colour and then subtracted the result from 1. A surprisingly large number of students thought that the first card is replaced before the second one is taken and used a denominator of 121. Those giving an answer of 62 121 were awarded 1 mark. For some students a tree diagram was often all they managed; they did not know what to do with the probabilities. Some added rather than multiplied the probabilities.
Alfie has 11 cards. He has 3 blue cards 7 green cards and 1 white card. Alfie takes at random 2 of these cards. Work out the probability that he takes cards of different colours.
Edexcel 1MA1 · November 2022 · Paper 3F · Q26 · 4 marks
4 marks
Examiner report
In the first part of this question, there was a clear split between those candidates who understood the concepts of a tree diagram and were able to insert the correct probabilities and those that seemed to not know that probabilities needed to sum to 1 and often just repeated values in the tree that they could already see. Full marks and zero marks were the most common, although some students were able to identify the 0.7 but not the other two probabilities. In part (b) most candidates attempted some form of calculation although 0.3 + 0.35 = 0.65 was the most common approach and led to zero marks. Very few candidates seemed to know that they needed to multiply the probabilities.
One weekend the Keddie family is going to do a sports quiz and a music quiz. The probability that the family will win the sports quiz is 0.3 The probability that the family will win the music quiz is 0.35
Edexcel 1MA1 · November 2022 · Paper 3H · Q6 · 4 marks
4 marks
Examiner report
In part (a), the great majority of candidates were able to complete the probability tree diagram successfully to score 2 marks. Some candidates gave 2 different probabilities for the probability of a candidate not winning the music quiz when both should have had the same value, 0.65. Part (b) was also answered well and and a majority of candidates scored 2 marks for a fully correct answer. The most common error seen was the addition rather than multiplication of 0.3 and 0.35. Some candidates gave more than one combined outcome and, as a consequence, could not be awarded any marks for this part of the question.
One weekend the Keddie family is going to do a sports quiz and a music quiz. The probability that the family will win the sports quiz is 0.3 The probability that the family will win the music quiz is 0.35
Edexcel 1MA1 · November 2022 · Paper 3H · Q19 · 3 marks
3 marks
Examiner report
A small proportion of lower attaining candidates merely added the three numbers 0.4, 7 and 8 given in the question and tried to work with this. In general, however, the question was quite well answered. When it was not completed successfully, examiners could usually award some credit to a candidate for showing a correct process to find the probability that a counter was red or green, (1 − 0.4 or 0.6). Many candidates went on to find the correct probability for picking a green counter (0.32), thus showing an understanding of how to apply the ratio, but were then unable to make the final link to multiply this by the number of trials (50).
There are only blue counters, red counters and green counters in a box. The probability that a counter taken at random from the box will be blue is 0.4 The ratio of the number of red counters to the number of green counters is 7 : 8 Sameena takes at random a counter from the box. She records its colour and puts the counter back in the box. Sameena does this a total of 50 times. Work out an estimate for the number of times she takes a green counter.
Edexcel 1MA1 · June 2022 · Paper 1F · Q13 · 2 marks
2 marks
Examiner report
Both parts of this question were answered well. In part (a), the usual error was seen using incorrect probability notation. With there being just one mark for this question, answers such as 4 : 15, 4 in 15, 4 out of 15 or similar gained no credit. In part (b), often the correct answer of 0.7 was written as 70% or 10 , both alternatives being perfectly acceptable. Some thought that as there were only two colours of counters in the bag then they must both have the same probability of being chosen, giving 0.3 as their answer.
Probability questions.
Edexcel 1MA1 · June 2022 · Paper 1F · Q17 · 5 marks
5 marks
Examiner report
It was very pleasing to see so many students correctly relating the information given to the diagram before them. Very few errors were made, usually through carelessness when they were. In part (b), the most common answers were either the correct answer of 72 for two marks or for one mark where students just considered the adults and not the whole population. Giving the answer as a ratio 12:72 gained one mark A significant number of students insisted on trying to give their answer as a decimal. If their initial fraction was seen, any attempt at converting to a decimal was ignored. However, some students only offered a decimal answer and this was virtually always incorrect.
72 people did a test. 32 were adults; 20 adults passed. Six children failed.
Edexcel 1MA1 · June 2022 · Paper 1H · Q16 · 4 marks
4 marks
Examiner report
Overall, this question was answered very poorly with a large number of students unable to find an appropriate strategy to work out the probability of passing the practical test. Probability tree diagrams were very common but these were of little use to many students as the probability of passing only one of the two parts, 0.36, was often placed incorrectly on branches of the tree diagrams. Some students wrote down a relevant product, often this was 0.75 × x, and gained the first mark but many were not able to go on and make any further progress. Statements such as 0.75 × x = 0.36 were common because many students did not recognise that there are two ways of passing only one of the two parts. Some of the students who did consider the two ways wrote 0.75 × x + 0.25 × x = 0.36 and gained no more marks. It was also common to see 0.75 × x + 0.25 × y = 0.36 which gained the second mark but this did not necessarily get translated into an equation in one variable and so no further progress was made. Many of those that did show a correct equation such as 0.75(1 – x) + 0.25x = 0.36 were able to complete the process and give a correct final answer. A small number of students who got as far as forming and solving a correct equation in one variable were confused about what their variable represented and lost the accuracy mark.
A first-aid test has independent theory and practical parts. P(pass theory)=0.75 and P(pass exactly one part)=0.36.
Edexcel 1MA1 · June 2022 · Paper 2F · Q20 · 4 marks
4 marks
Examiner report
Part (a) saw a pleasing number of successful attempts for a late question on the paper with almost a third of the cohort scoring both marks. Many students were able to place 0.87 correctly, with the 0.94 proving more difficult for others. A commonly seen misconception involved trying to sum the probabilities to make 0.13 or 0.87 in the second stage instead of summing to 1. Centres should remind students of this difference between probability tree diagrams and frequency tree diagrams. Part (b) proved to be more of a challenge for this cohort, with many adding the probabilities instead of multiplying. There were some students who could not identify the required probabilities. This question was not well answered.
Probability tree: P(work train late)=0.13; P(home train late)=0.06 on either branch.
Edexcel 1MA1 · June 2022 · Paper 2H · Q16 · 6 marks
6 marks
Examiner report
This Venn diagram question allowed most students to gain at least some credit. However, this example was challenging so that only a minority of students were able to score all 4 marks. The region most often incorrect was B only due to the fact it was the region that required the most working out. 0 was often omitted which cost students a mark unless they had the other 7 sections correct. Part (b) required students to pull values from the diagram. Many students scored 1 mark for a correct numerator or denominator (more commonly numerator). There was a follow through on this question, which meant that students could gain credit for pulling the correct values from their incorrect Venn diagram. Many students did not understand the ‘given uses A’ and had 100 as the denominator.
Of 100 customers, 16 use none of A,B,C; 4 use all three; AB-only=8; 14 use B and C in total; A total=62; C total=20 and every C-user also uses A or B.
Edexcel 1MA1 · June 2022 · Paper 3F · Q10 · 4 marks
4 marks
Examiner report
This question was well done with most students making an attempt at all parts. Students were largely confident with part (a) but struggled more with part (b) with many putting a cross at 5 8 . Part (c) was mostly accurate, with correct notation in fraction, decimal or percentage form, although some only gained one mark by using incorrect notation such as 5:8 or 5 out of 8. Some did not relate the question to the list at the top of the page and gave an answer of 2, presumably from the probability of choosing an even number from all numbers.
Choose one at random from 2,2,3,5,6,6,8,9.
Edexcel 1MA1 · June 2022 · Paper 3H · Q21 · 4 marks
4 marks
Examiner report
Higher attaining students usually made a good start on this question on probability and identified the non-replacement nature of this problem. They usually presented at least one correct product of three fractions with denominators 9, 8 and 7 respectively. Only a small proportion of students gave all four possible ways of getting an even sum with the resultant correct answer, 21. Students who drew a tree diagram were generally more successful at finding all 4 correct combinations. Very few students approached the problem by finding the probability that the result was an odd number then subtracted from 1. Those students who did do this were rewarded appropriately.
From cards numbered 1–9, three distinct cards are selected at random.
Edexcel 1MA1 · November 2021 · Paper 1F · Q21 · 3 marks
3 marks
Examiner report
Many students were able to gain at least one mark, and often more in this question. The universal set was often incomplete with the omission of the numbers 4, 8, 10 and 16. Some students lost marks by duplicating entries in each region, so it was not uncommon to see, for example, the intersection containing, 6, 6, 18 and 18 or 6 (18) in all three sections.
E is the set of even numbers less than 19. A = {6, 12, 18}. B = {2, 6, 14, 18}.
Edexcel 1MA1 · November 2021 · Paper 1H · Q2 · 3 marks
3 marks
Examiner report
Those students who started by writing 6 and 18 in the intersection generally went on to answer this question quite well and often placed all five numbers correctly inside the circles to gain the first two marks. Some students, however, wrote 6 and 18 in more than one region or wrote just 18 in the intersection with 6 written in two different regions. The outside region, ) ( B A , proved to be more problematic. It was common to see either no numbers at all in this region or many of the numbers that had been placed inside the circles. Those who did attempt to put the rest of the even numbers in this region sometimes failed to include all four numbers.
E is the even numbers less than 19. A={6,12,18}. B={2,6,14,18}.
Edexcel 1MA1 · November 2021 · Paper 1H · Q16 · 4 marks
4 marks
Examiner report
Probability tree diagrams were very common and those students that attempted to draw a tree diagram often achieved the first mark for use of 7 , 7 or 7 , showing understanding that the first counter was not replaced. Those who wrote at least one correct product gained the second mark but often these students were not able to go on and show a complete process. Sometimes this was because they did not consider all three ways that Jude could take exactly one red counter and sometimes it was because they used incorrect probabilities in their products. The majority of those that did show a complete process were able to complete the arithmetic and give a correct final answer. For some students the tree diagram was often all they managed; they did not know what to do with the probabilities. Some added rather than multiplied the probabilities.
A bag contains only 3 red and 5 yellow counters. Jude takes 3 counters at random without replacement.
Edexcel 1MA1 · November 2021 · Paper 2F · Q16 · 5 marks
5 marks
Examiner report
This question required students to interpret a pie chart, firstly to find the number of cars being represented in a sector followed by calculating a probability, with many finding both parts of the question challenging. A range of approaches using either knowledge of the number of degrees in a pie chart or using proportionality could gain credit, with the most successful approach appearing to be to use an appropriate method to find the number of degrees per car or per 10 cars. However, many deducted the number of degrees difference for the two sectors from the number of black cars arriving at the incorrect answer of 125 or dividing by 10 rather than 9 when working with proportionality. Very few identified that the whole pie chart represented 540 cars and used this to work out the number represented by 80 degrees. Students were required to combine knowledge of probability with interpretation of the pie chart in part (b). Whilst many gained a mark for using 50 as the numerator in a fraction, the majority were unable to also give a correct denominator. Common incorrect denominators included 100, presumably using percentage, or 360 from using the total number of degrees in a circle. A small minority of students are still expressing a probability as a ratio rather than a fraction, decimal or percentage. Centres are encouraged to promote the use of fractions for probability in this type of problem.
A pie chart for car colours has a 90° black sector and an 80° white sector. There are 135 black cars. There are 50 grey cars.
Edexcel 1MA1 · November 2021 · Paper 2F · Q17 · 5 marks
5 marks
Examiner report
It was common to award 2 marks or full marks for this question. Students often correctly placed 38 or 15 for the first mark and therefore went on to calculate or place 22 or 7 for the second mark. Many students then failed to attempt to find 60% of 60 but when they did they often went on to correctly complete the rest of the question. The most common error was to find 60% of the wrong value (usually 38). Some students gained a third mark by simply showing that the total number of texts was 36 but not using it in the diagram, this skill in working with percentages was enough for the third mark. Some responses had the correct numbers misplaced in the tree diagram and students should be encouraged to check their answers thoroughly. These students did gain some marks for working but could not gain communication and accuracy marks.
60 people are asked whether they prefer texting or emailing. 38 are women and the rest are men. 15 men prefer email. 60% of all the people prefer text.
Edexcel 1MA1 · November 2021 · Paper 2H · Q20 · 4 marks
4 marks
Examiner report
This question proved a real challenge due to the fact it involved 3 days, rather than the 2 that students are more familiar with. Also, the lack of scaffolding, which is to be expected at this grade, challenged students. Many attempted to draw a probability tree, and that alone was too difficult. A first mark could be gained for a correct product for consecutive days, typically seen for Monday and Tuesday, such as 0.7 × 0.8, but could equally be for Tuesday and Wednesday such as 0.2 × 0.6. Two marks were awarded for a correct triple probability and, when seen, it was normally for RM, RT, RW or NRM, RT, RW. In fact it was the probabilities when it didn’t rain on Tuesday that students found the most challenging. Awarding 3 or 4 marks was very rare.
If it rains, next-day rain probability is 0.8; if dry, next-day rain probability is 0.6. Monday rain probability is 0.7.
Edexcel 1MA1 · November 2021 · Paper 3F · Q26 · 4 marks
4 marks
Examiner report
In part (a) many students were aware that probabilities should sum to 1 and it was not uncommon to therefore see two figures presented as their answer which added to 0.8; 0.6 and 0.2 were common. Splitting this so that one number was 0.2 more than the other proved impossible for many. Part (b) was even less well answered, indeed left unanswered by many. A significant number tried to find the number of counters for each colour but were unsuccessful, not knowing to simply divide by 0.15 Some students tried 18 ÷ 0.15 but then went on with other incorrect methods. 0.15 × 18 = 2.7 was a popular answer. Build up approaches were again evident, always unsuccessful.
A bag contains only red, blue, green and pink counters. P(red)=0.05 and P(blue)=0.15. P(green) is 0.2 more than P(pink). There are 18 blue counters.
Edexcel 1MA1 · November 2021 · Paper 3H · Q6 · 4 marks
4 marks
Examiner report
In part (a), nearly all students used the result that the probabilities in the table should add to one and so scored the mark for showing that the probability of taking a green or pink counter is 1 ̶ (0.05 + 0.15). Fewer students were able to go on to complete the table successfully, many giving two probabilities which were not the correct 0.5 and 0.3 but added to 0.8. Part (b) was answered well and and a majority of students scored 2 marks for a fully correct answer which they usually obtained by dividing 18 by 0.15. Few students used a method which involved finding the number of counters of each colour then adding them.
A bag contains only red, blue, green and pink counters. P(red)=0.05 and P(blue)=0.15. P(green) is 0.2 more than P(pink). There are 18 blue counters.
Edexcel 1MA1 · November 2021 · Paper 3H · Q20 · 2 marks
2 marks
Examiner report
This question was often not attempted but it also proved to be a good discriminator between the highest attaining students. Of these students, a good number were able to gain one mark for ( 2) 𝑛 , the probability of getting no heads (or all tails) in n throws or of getting no tails (or all heads) in n throws. Only the best students were able to get the fully correct expression 1 − ( 2) 𝑛 − ( 2) 𝑛 or equivalent for their final answer.
Pat throws a fair coin n times.
Edexcel 1MA1 · November 2020 · Paper 1F · Q6 · 2 marks
2 marks
A fair 8-sided spinner has three sectors labelled A, one labelled B and four labelled C.
Edexcel 1MA1 · November 2020 · Paper 1H · Q14 · 3 marks
3 marks
Sally plays two games against Martin. In each game, Sally could win, draw or lose. In each game they play, the probability that Sally will win against Martin is 0.3 the probability that Sally will draw against Martin is 0.1 Work out the probability that Sally will win exactly one of the two games against Martin.
Edexcel 1MA1 · November 2020 · Paper 2F · Q22 · 5 marks
5 marks
E = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} A = {even numbers} B = {factors of 10}
Edexcel 1MA1 · November 2020 · Paper 2H · Q2 · 5 marks
5 marks
E = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} A = {even numbers} B = {factors of 10}
Edexcel 1MA1 · November 2020 · Paper 2H · Q20 · 8 marks
8 marks
A bag contains n red sweets and 8 yellow sweets. Sajid takes one sweet at random and eats it.
Edexcel 1MA1 · November 2020 · Paper 3F · Q16 · 3 marks
3 marks
A pack contains 5 blue, 2 green and 4 red cards. Separately, a biased die has probability 0.3 of landing on six.
Edexcel 1MA1 · November 2020 · Paper 3F · Q21 · 2 marks
2 marks
Malik will toss a fair coin 50 times. Separately, Paula tosses another coin 10 times and Simon tosses the same coin 100 times to investigate whether it is biased.
Edexcel 1MA1 · November 2020 · Paper 3F · Q27 · 4 marks
4 marks
Each fair 3-sided spinner has one side labelled 2 and two sides not labelled 2. Amanda spins each spinner once.
Edexcel 1MA1 · November 2020 · Paper 3H · Q5 · 4 marks
4 marks
Each fair 3-sided spinner has one side labelled 2 and two sides not labelled 2. Amanda spins each spinner once.
Edexcel 1MA1 · November 2020 · Paper 3H · Q8 · 3 marks
3 marks
There are some counters in a bag. The counters are blue or green or red or yellow. The table shows the probabilities that a counter taken at random from the bag will be blue or will be green. Colour blue green red yellow 0.32 0.20 The probability that a counter taken at random from the bag will be red is five times the probability that the counter will be yellow. There are 300 counters in the bag. Work out the number of yellow counters in the bag.
Edexcel 1MA1 · November 2019 · Paper 1F · Q11 · 2 marks
2 marks
Examiner report
This question was a good discriminator. More able students scored two marks for a complete list without any repeats. About a third of students scored one mark for listing at least 4 correct outcomes but were not able to provide a fully correct answer. A total of 6 combinations was commonly seen. Some students listed the possibilities of how many heads and how many tails could occur without making it clear the order of the heads and tails within these possibilities, so for example, stating 1 heads and 2 tails without differentiating between HTT, THT and TTH. Of the students who could not be awarded any marks, a good proportion of them gave outcomes for only 2 throws.
Mason throws a coin 3 times. The outcome of each throw is either Heads or Tails. List all the possible outcomes of the 3 throws.
Edexcel 1MA1 · November 2019 · Paper 1F · Q15 · 4 marks
4 marks
Examiner report
Most students completed the frequency tree accurately and scored three marks for their response to part (a). The errors seen were usually caused by arithmetic mistakes. Students should be encouraged to write down their calculations in the working space as those students who did this were usually awarded the first and second marks even where their entries in the diagram were incorrect. In part (b) ଶଽ ସହ was the most common incorrect answer seen. Students who wrote this had clearly not considered carefully the statement in the question that “one of the 120 people is chosen at random”.
120 people were at a hockey match. Each person was asked if they wanted to stand or to sit to watch the match. 75 of the people were female 29 of the males wanted to stand 30 of the people wanted to sit
Edexcel 1MA1 · November 2019 · Paper 1H · Q10 · 6 marks
6 marks
Examiner report
This question was poorly answered by this cohort of students. Any marks gained were usually in part (a) for correct plotting of the given values. However, it was common to see the points plotted in parts of the intervals other than at the ends. Often seen were plots at time values of 30, 40, 50, 60 and 70 minutes; this gained no credit and prevented any success in part (c). Many students drew histograms. A lack of understanding of quartiles and interquartile range was evident in part (b), 30 – 10 = 20 was a common error. In part (c), correct readings from time = 50 and 90 were often found but then either just stated or added together gaining no credit. Many students struggled with the scale on the horizontal axis and often misread the required values.
Cumulative frequencies for puzzle-completion time are: up to 40 min: 5; up to 60: 25; up to 80: 35; up to 100: 38; up to 120: 40.
Edexcel 1MA1 · November 2019 · Paper 2F · Q9 · 6 marks
6 marks
Examiner report
In part (a) the majority of students knew that the median was the middle number even if they tried to find the median without ordering the list. Those who gave the answer as 8.5 (middle of the unordered list) gained one mark. Most students did order the numbers with many then able to give the correct answer. Some attempts at ordering the list omitted one of the numbers, often one of the two 4s. A common error was to identify the two middle numbers as 4 and 6 and give an answer of 4 or 6 or 4, 6 or carry out an incorrect calculation such as 6 – 4 with the two numbers. Some students confused the averages and worked out the mean or the mode. A few found the range. Part (b) was answered very well. Those who did not give the correct probability often gained one mark for a fraction with either the correct numerator or the correct denominator. A few students used incorrect notation for probability such as 2 : 6 and were awarded the method mark only. A common incorrect answer was 1/2. There were relatively few students who gave a likelihood rather than a probability as the answer. In part (c) many students were able to use the information about the mode to identify one of the hidden numbers as 3 and gain the first mark. They then often went on to give the correct answer. Some students showed 3 + 3 + 8 + 5 + 6 = 25 and 25 ÷ 5 = 5 in the working space but then gave 3, 5 as the answer and gained one mark only. It was common to see 3 on the answer line with a number other than 6 and no working out shown. Some students gained the first mark for working with the mean but did not use the mode to complete the solution.
Here is a list of six numbers: 6, 4, 8, 9, 4, 3. Aisha picks at random one of these numbers. Clara has five cards showing 3, ?, 8, 5, ?. The mode of Clara's five numbers is 3 and their mean is 5.
Edexcel 1MA1 · November 2019 · Paper 2H · Q16 · 3 marks
3 marks
Examiner report
Using a tree diagram was the most successful method seen which often led to at least one mark. A significant number of students chose to try to list all possibilities. This approach was largely unsuccessful, either because they failed to list various possibilities in the process, or because their work was not sufficiently organised to produce the necessary lists to enable them to see where they were going. A few used an incorrect replacement process.
Marek has cards numbered 1 to 9. He selects two without replacement and multiplies their numbers.
Edexcel 1MA1 · November 2019 · Paper 3F · Q16 · 4 marks
4 marks
Examiner report
The fundamental part of this question was the first step, where students had to equate the sum of the probabilities to 1. Without using this, students were unable to gain credit for later processes. Once students had found that the sum of the probabilities for blue and green is 0.35, they had to divide this value in the ratio 3 : 4. This was done with varying degrees of success, many dividing by 3 and 4 separately rather than dividing by (3 + 4). Students were able to work in any suitable form (fractions, decimals and percentages) and those who worked in percentages were able to gain process marks with a missing % symbol. However, it did need to be present to gain the final accuracy mark.
A bag contains only red, blue, green, yellow counters. P(red)=0.4, P(yellow)=0.25, and blue:green counts=3:4.
Edexcel 1MA1 · November 2019 · Paper 3F · Q20 · 4 marks
4 marks
Examiner report
The Venn diagram question was answered really well by many students and the award of three or four marks was common. Generally speaking a good proportion of students were able to score at least three marks, normally for a correctly labelled diagram with at least 2 regions correct. When mistakes were made, it was often from values in more than one region, or quite commonly, for the region (A ∪ B)’ to be either incomplete, or containing all the values.
Universal set U={1,...,13}; A={multiples of 3}; B={even numbers}.
Edexcel 1MA1 · November 2019 · Paper 3H · Q11 · 3 marks
3 marks
Probability tree: Monday late 0.07/not 0.93; if Monday late, Tuesday late 0.02/not 0.98; if Monday not late, Tuesday late 0.11/not 0.89.
Edexcel 1MA1 · November 2019 · Paper 3H · Q21 · 2 marks
2 marks
For a biased coin, P(4 heads in 4 throws)=16/81.
Edexcel 1MA1 · June 2019 · Paper 1F · Q17 · 2 marks
2 marks
Examiner report
This is another question where an explanation was required and students continue to find this difficult. Very few students got both parts correct. For part (i) many did not appreciate the fact that more trials meant a better estimate, instead believing that more tails gave a better estimate. Some responses wanted the probabilities to be as close to a whole number as possible, and incorrectly chose Stuart for that reason. Other common errors were saying that Maxine's were better because they gave a closer answer to 50% or 0.5 or even and thus that she was not using a biased coin. A significant number said that Stuart’s was better because he had thrown fewer. For part (ii) the most common incorrect answers were 65% by simply calculating 70% (=7/10) and 60% (30/50) and then finding midpoint or just stating 0.5. Some of the students calculated separate probabilities without combining them. An answer of 37 was also common, showing the failure to write a probability in a correct format, a few ratios were seen.
Stuart gets 7 tails in 10 throws; Maxine gets 30 tails in 50 throws of same biased coin. Prasha will throw once.
Edexcel 1MA1 · June 2019 · Paper 1F · Q22 · 4 marks
4 marks
Examiner report
Part (a) was well answered. The majority of students had three probabilities summing to 1 in the table, earning at least one out of the two marks available. The most common incorrect answer was to give any two numbers that add to 0.8 or 0.8 twice. Part (b) was less successful than part (a). Many students initially made the connection that 0.2 was equivalent to 12 cubes, therefore the 0.4 was equivalent to 24 cubes but did not always add correctly. Quite a few students related the number of cubes to the probability incorrectly, by assuming that if a probability of 0.2 is equivalent to 12 cubes, a probability of 0.4 must be equivalent to 14 cubes, giving the most common incorrect solution of 12+14+14.
Box contains only blue,red,yellow cubes. P(blue)=0.2; red count equals yellow count; there are 12 blue cubes.
Edexcel 1MA1 · June 2019 · Paper 1H · Q1 · 4 marks
4 marks
Examiner report
Part (a) was answered extremely well. Most of the students who failed to complete the table correctly gave two probabilities that sum to 0.8, e.g. 0.6 and 0.2, and gained one mark. The most common approach used in part (b) was to find that 0.4 was equivalent to 24 cubes and then work out 12 + 24 + 24. Some students chose to work out 12 × 5 or, less frequently, 12 ÷ 0.2. Marks were sometimes lost through arithmetic errors, on occasion even simple adding up went wrong. Students who had given incorrect answers of 0.6 and 0.2 in part (a) were often able to work out the total number of cubes as 60 in part (b).
A box contains only blue, red and yellow cubes. P(blue)=0.2; the number of red cubes equals the number of yellow cubes; there are 12 blue cubes.
Edexcel 1MA1 · June 2019 · Paper 1H · Q22 · 5 marks
5 marks
Examiner report
This proved to be a challenging question. Many students failed to find a successful strategy and gained no marks. Working out was often messy and difficult for examiners to follow. Students attempting an algebraic approach often started by forming two equations such as g/(r + g) = 3/7 and (g + 3)/(r + g + 5) = 6/13. To solve these equations students needed to deal with the fractions and eliminate one of the variables and a successful outcome proved to be beyond many. Students who formed the equations g/r = 3/4 and (g + 3)/(r + 2) = 6/7 had a slightly easier route to a solution. Some students formed only one equation, using either 3/7 or 6/13, and scored one mark only. A different approach used by some students was to form an equation such as (3x + 3)/(7x + 5) = 6/13. These students were generally more successful. Some of the students that attempted an algebraic approach were not able to derive the necessary equation(s). Answers of 12 red counters and 9 green counters did not always come from an algebraic approach. Some students realised that the number of counters in the bag at the start must be a multiple of 7 and the number of counters in the bag at the end must be a multiple of 13. Since 5 counters had been added they knew that they were looking for a multiple of 13 that is 5 more than a multiple of 7 and quickly came up with 26 and 21. These students usually went on to score full marks. Many of the unsuccessful attempts were based on using fractions with a denominator of 91 (from 7 × 13). Since 3/7 = 39/91 and 4/7 = 52/91 a common incorrect answer was 52 red counters and 39 green counters. There were many attempts at representing the information in tree diagrams. These were generally unhelpful. It seemed that most students wanted to jump straight to probability tree diagrams and had no other strategies to use. Summary Based on their performance on this paper, students should: practise working out estimates by rounding numbers and develop an understanding of the purpose of rounding so that they can choose appropriate rounded values have a greater awareness of the graphical solution of simultaneous equations practise writing clear concise explanations when using the median and measures of spread to compare two sets of data learn what is required in a proof as opposed to giving specific numerical examples memorise the necessary exact trig values or learn how to work them out
A bag initially has r red counters and g green counters only. P(green)=3/7. The counter is replaced, then 2 red and 3 green counters are added; now P(green)=6/13.
Edexcel 1MA1 · June 2019 · Paper 2F · Q16 · 5 marks
5 marks
Examiner report
In part (a), it was rare to see both parts answered correctly. B was often identified as the least likely coin to land on Heads but A and D were very common choices for the most likely. Many students correctly said that Julie was wrong in part (b) but their reasoning varied widely. Simply saying that coin C was biased was insufficient since this was given in the question. It needed to be clear that the student was referring to the probability of the ଵ ଷ given in the table. Students using decimal or fraction equivalents for ଵ ଷ often made errors in the conversion; 30% was a common error. A number of students referred to even chances or 50-50 chances or ‘odds’ rather than using the correct probability language. Part (c) was often answered well. Many students tried to convert 0.033 to a fraction or a percentage usually making an error in their subsequent working. Many did not understand the use of the word “estimate” in the context of probability and believed they were required to round off the numbers before calculation. Students who found the correct answer of 132 and then rounded to say 130 were not penalised, provided 132 was seen. A number of students were not explicit in the use of 0.033 and so any rounding made (eg. 0.03) without first writing 0.033 was given no credit.
Four biased coins have P(heads): A=0.33, B=0.033, C=1/3, D=30%. Julie claims coin C is equally likely to land heads or tails. Coin B will be thrown 4000 times.
Edexcel 1MA1 · June 2019 · Paper 2H · Q10 · 5 marks
5 marks
Examiner report
As would be expected on the higher paper, the completion of the tree diagram was done very well, with a majority of students scoring both marks in part (a). In part (b) many students scored the first mark for a suitable product, but then lost the remaining marks as their method was incomplete. This was usually because they dealt with the two possibilities of being late on exactly one day, but missed the words “at least” in the question, and failed to consider being late on both days. For those who gained the second mark, the third almost always followed.
Mary's train is late on any day with probability 0.15. A two-stage probability tree covers Thursday and Friday, treated with the same daily probabilities.
Edexcel 1MA1 · June 2019 · Paper 3F · Q24 · 5 marks
5 marks
Examiner report
Many students were able to complete the regions within sets A and B correctly but often included all 1 to 9 numbers in the outer region. A significant minority of students incorrectly wrote their numbers on the circumference of the circles. Part (b) was usually well done and did not rely on getting all the marks in part (a). Some students did not understand the set notation for the intersection.
The universal set is E = {1,2,3,4,5,6,7,8,9}. A = {1,5,6,8,9} and B = {2,6,9}.
Edexcel 1MA1 · June 2019 · Paper 3H · Q1 · 5 marks
5 marks
Examiner report
This question provided a good start to the paper. Nearly all students gained some credit for their answers with most students scoring at least four of the five marks available. Part (a) of this question requiring students to complete the Venn diagram was generally well done though some students did make careless errors when placing the numbers within set A and/or set B. A significant number of students wrote all of the numbers 1 to 9 in the region( )' A B . Part (b) of the question was answered quite well and many students benefited from the mark scheme which allowed a follow through from part (a). It was not unusual for students to gain part marks in part (a) followed by full marks in part (b). Some students reverted to using the sets listed rather than their Venn diagram to answer part (b). A common error was for students to give 6 9 as their answer to this part presumably because students had confused intersection with union.
The universal set is E = {1, 2, 3, 4, 5, 6, 7, 8, 9}. A = {1, 5, 6, 8, 9} and B = {2, 6, 9}.
Edexcel 1MA1 · November 2018 · Paper 1F · Q17 · 5 marks
5 marks
Examiner report
Part (a) of this question was answered well by many students who produced a complete and accurate frequency tree. Nearly all students gained at least one mark in this part of the question. More usually they scored at least 2 marks. The correct answer to part (b), 61, was given by a reasonable proportion of students and many other students could use the frequencies from their diagram in the right way to find a probability which was correct for their diagram. In cases where students earned only 1 mark in part (b), it was often due to an incorrect denominator of either 48 (the number of women) or 80 (the total number of people). This question was a very good discriminator with every mark from 1 to 5 being awarded to a good number of students. Very few students scored no marks for their responses, some of whom gave an improper fraction as their answer to part (b).
There are 80 people. Of these, 48 are women. In total, 61 people like coffee. Eight of the men do not like coffee.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2018 · Paper 1F · Q27 · 4 marks
4 marks
Examiner report
Most students were able to score some marks for their responses to this question. The tree diagram was often successfully completed and where it was not, students were often able to place a correct probability for Amina taking a green ball from the first bag together with a correct probability for obtaining a green ball from the second bag following a red ball being taken from the first bag. The incorrect probabilities 19 and 19 were often seen placed for the probabilities of taking a red or green ball respectively from the second bag following a green ball having been taken from the first bag. Part (b) of the question was answered correctly by only a small percentage of students. By far the most common answer given in part (b) was obtained from the spurious method of adding 10 and 9.
Bag 1 contains balls for which P(red) = 3/10 and P(green) = 7/10. Bag 2 contains balls for which P(red) = 5/9 and P(green) = 4/9. One ball is chosen from each bag.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2018 · Paper 1H · Q22 · 6 marks
6 marks
Examiner report
Again, very few students were able to seriously attempt to answer this question. Many found the “12” a distraction and insisted upon using it to develop numerical probabilities. Of the very few making a correct start it was rare to see a solution beyond the product of two correct probabilities. Summary Based on their performance on this paper, students should: • Read questions carefully and ensure that their answers address the requirement of the question. • Avoid introducing a value for pi when pi is part of the answer. • Practise working out estimates by rounding numbers and develop an understanding of the purpose of rounding so that they can choose appropriate rounded values. • Practise substituting negative values into formulae and work with calculations involving negative numbers. • When using trigonometry, students should be encouraged to quote trigonometric equations before any attempting a calculation.
There are x green pens and x + 3 blue pens, with more than 12 pens in total. Two pens are taken at random without replacement. The probability that they have the same colour is 27/55.
Pearson does not publish a direct link to this question paper.
Edexcel 1MA1 · November 2018 · Paper 2F · Q12 · 2 marks
2 marks
Examiner report
After working out 49 – 20 = 29 to find the number of blue counters most students gave the correct answer of 29/49. There were a surprising number of answers of 29/40 which scored one mark for the numerator of 29. However, an answer of 29/20 scored no marks because the numerator is larger than the denominator. Incorrect answers were often the result of using 29 incorrectly, e.g. writing the answer as 20/29 or as 1/29, or not using it at all. It was pleasing to note that there were relatively few students who gave a likelihood rather than a probability as the answer.
There are 49 counters in a bag. 20 of the counters are red. The rest of the counters are blue. One of the counters is taken at random. Find the probability that the counter is blue.
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Edexcel 1MA1 · November 2018 · Paper 2F · Q14 · 4 marks
4 marks
Examiner report
Responses to part (a) demonstrated a wide variation of knowledge regarding the use of probability language in an appropriate way. Whilst many students were able to explain clearly that the probability for each outcome is the same or one sixth, some contradicted their correct statement by using incorrect probability terminology such as “even chance”. A small number of students failed to state a decision with their explanation and a larger than expected number of students gave an incorrect decision of “yes” alongside an explanation based upon 3 being half 6 and assuming that the probability related to this multiplication fact. Very few students were successful in part (b). The correct decision of “no” was made by many students but in most responses it was supported by an incorrect explanation. The most common mistake was to state that the probability should be 2/12 because there are 12 outcomes with two chances of getting a 6. A large number of students explained that the probability will be 1/6 because the probability of getting a 6 is 1/6 on each throw. There were other students who made the wrong decision and stated that Andy is correct because 1/6 + 1/6 = 2/6. In part (c) the majority of students listed the 12 correct combinations. These were written in a variety of ways with words and with abbreviations. It was pleasing that many of the lists were systematic and written in a logical order. Those that did not use such a logical approach were more likely to miss out or repeat combinations. Failure to gain at least one mark was usually due to students failing to appreciate that the outcomes should be pairs – a common incorrect answer was H, T, 1, 2, 3, 4, 5, 6.
Victoria throws an ordinary fair 6-sided dice once. She says, “The probability of getting a 3 is half the probability of getting a 6”
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Edexcel 1MA1 · November 2018 · Paper 2F · Q20 · 6 marks
6 marks
Examiner report
Many students scored 3 of the 4 marks in part (a) for correctly placing the eight numbers in sets A, B and C. Common mistakes included writing 20 and/or 8 in more than one region and writing three 8s in the intersection of all three sets. The outside region, ) ( C B A , proved to be much more problematic. It was very common to see either no numbers at all in this region or duplicates of numbers that had already been placed inside the circles or all numbers listed. It should be emphasised to students that each number in the universal set should appear just once in a Venn diagram. In part (b) many students scored one mark for the correct denominator of 12 or for a denominator (usually 8) that followed through correctly from their Venn diagram. Common incorrect denominators were 25 and 11. A correct numerator was seen less frequently and it was evident that many students were unable to identify the region B A .
E = {even numbers between 1 and 25} A = {2, 8, 10, 14} B = {6, 8, 20} C = {8, 18, 20, 22}
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Edexcel 1MA1 · November 2018 · Paper 2H · Q1 · 6 marks
6 marks
Examiner report
Many students made a good attempt at part (a), usually gaining 3 marks. Common errors included repeating the 8 in both regions of the intersection of A and B, and failing to put the numbers 4, 12, 16 and 24 outside the circles. Students need to be aware that all the numbers in the Universal set need to appear somewhere in the Venn diagram. Part (b) was not as well answered, even though a follow through was allowed from their (sometimes incorrect) Venn diagram. The main error was to write the number 8 as the numerator rather than 1. The number used as the denominator did not always match the total of the numbers that they had presented in part (a).
E = {even numbers between 1 and 25} A = {2, 8, 10, 14} B = {6, 8, 20} C = {8, 18, 20, 22}
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Edexcel 1MA1 · November 2018 · Paper 2H · Q16 · 4 marks
4 marks
Examiner report
In part (a) students approached this question in many different ways. Many did 1 – 0.65 = 0.35 and thought this was the answer. More did 0.65 0.35 = 0.2275 and left it there. Those who drew a tree diagram usually saw that there were two options and doubled to get the correct answer. In part (b) students had more success with many getting 42. The most common method was 78 ÷ 0.65 arriving at 120. Some used 78 0.35 but then some did not seem to know how to proceed.
There are only red counters and blue counters in a bag. Joe takes at random a counter from the bag. The probability that the counter is red is 0.65 Joe puts the counter back into the bag. Mary takes at random a counter from the bag. She puts the counter back into the bag.
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Edexcel 1MA1 · November 2018 · Paper 3F · Q6 · 2 marks
2 marks
Examiner report
Almost every response in part (a) had the cross placed correctly at 2. In part (b) the correct answers of 6 or 3 were very regularly seen. A small number of students gave an incorrect response of 2 but this was rare.
A fair six-sided die is numbered 1 to 6.
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Edexcel 1MA1 · November 2018 · Paper 3F · Q14 · 4 marks
4 marks
Examiner report
This probability was generally answered very well. In part (a) most students got the correct answer. However, there were a significant number of students who put the correct value of 0.3 in the table, but then had a different incorrect value on the answer line. In this case the answer line takes precedence and so these students lost the mark. Part (b) was answered well again with the only answer seen other than correct one was typically 3. In part (c) a good number of students were able to answer this common question. Those who went wrong either incorrectly interpreted the table and found 4 of 60, or divided by 0.2 rather than multiplied.
A four-sided spinner is numbered 1, 2, 3 and 4. The probability table gives P(1) = 0.2, P(3) = 0.4 and P(4) = 0.1.
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Edexcel 1MA1 · June 2018 · Paper 1F · Q7 · 5 marks
5 marks
Examiner report
Parts (a) and (b) were generally correctly answered although errors were more prevalent in part (b), ‘C’ and ‘A’ respectively being the most common errors. In part (c), it was pleasing to see many students successfully gaining full marks for a clear and correct structured solution. It was enough to show without the need for any statement of ‘proof’. Virtually all students gained at least one mark for determining that there were 6 green counters. Sight of 8/12 gained 2 marks for many students. Some students calculated 2/3 of 12 and showed the answer to be 8. Occasionally students did not seem to understand what was required for a ‘show that’ question of this nature.
A probability scale places D at 1 and B between 0 and 1/2. A bag has 12 counters: 3 red, 1 blue, 2 yellow and 6 green.
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Edexcel 1MA1 · June 2018 · Paper 1F · Q18 · 3 marks
3 marks
Examiner report
A great many students clearly did not understand or know the concept of union and intersection, although the majority were able to pick up one mark for correctly listing either set A or set B either numerically or in a Venn diagram and identifying it as such. Many students got the union and intersection the wrong way around. In part (a), a common error was to simply relist the two sets thereby repeating the common elements of 15 and 25. There were several alternative acceptable answers to part (b) and those showing some understanding of intersection were usually successful.
A is the set of multiples of 5 between 14 and 26. B is the set of odd numbers between 14 and 26.
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Edexcel 1MA1 · June 2018 · Paper 1H · Q16 · 3 marks
3 marks
Examiner report
Once students had identified a suitable strategy to use they were often able to work out the probability that Sam takes a red counter, though there were many errors in arithmetic. Successful solutions often started with students working out that the probability of taking a red counter or a blue counter is 0.8 and then finding 3/20 of 0.8. Division of 0.8 by 20 often lead to arithmetic errors. Some students worked out that if 20 red counters and blue counters represent 80% of the total number of counters then the total number of counters must be 25, often using an argument such as “80% = 20, 20% = 5”. Those that attempted to divide 20 by 0.8 often got into difficulties. A common error was for students to work out that the total number of counters is 24, e.g. 80% = 20, 20% of 20 = 4, 20 + 4 = 24. Some students assumed that there are 100 counters altogether and were able to work out that there are 12 red counters. Many students failed to find a successful strategy and in these cases the working out was often messy and difficult for examiners to follow. There were quite a few attempts at representing the information in tree diagrams and in ratios. These were generally unhelpful.
There are only red, blue and purple counters in a bag. The ratio red:blue is 3:17. The probability of taking a purple counter is 0.2.
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Edexcel 1MA1 · June 2018 · Paper 2F · Q13 · 2 marks
2 marks
Examiner report
Many correct answers were seen with the most common correct answer being , sometimes simplified to . Expressing the answer as a percentage was also seen, i.e. x 100 = 3.4% and 100 ̶ 3.4 = 96.6%. The correct decimal answers were rare. Most students started well and often found 350 ̶ 12 =338 but then were confused as to what to do with 338. Some ignored the value altogether and chose to write as their answer on the answer line, which then scored no marks. The most common incorrect answers were to give the probability that she won the raffle as or , or to calculate 350 ÷ 12 = 29.16.. or to or try to work with percentages with either 350 ÷ 12 = 29.16% or 100 – 29.16 = 70.84%.
Laura owns 12 of 350 raffle tickets for one prize.
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Edexcel 1MA1 · June 2018 · Paper 2F · Q14 · 3 marks
3 marks
Examiner report
This question was well attempted, with many gaining full marks for correctly interpreting the question and showing the ability to calculate the missing values and place them in the frequency tree correctly. Of those who were not awarded full marks, the vast majority were able to correctly place at least one of the given values, and then able to calculate at least 1 or 2 of the missing values. However, these were often located in the incorrect place in the frequency tree, showing a lack of understanding of what those missing values actually represented or an inability to re-read the question and check where to put the individual answers. The most common numerical mistakes tended to be made on the last branch for males, for example a common incorrect answer was to divide 22 by 2. 23 and 18 were the most common ‘calculated value’ seen. Students preferred not to show any working out and most of the time the answers in the frequency tree were not backed up by calculations, provided the values given were correct this wasn’t a problem. A minority of students completed the frequency tree with probabilities rather than frequencies.
There are 45 workers: 22 male, 23 female. Of 34 right-handed workers, 16 are female.
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Edexcel 1MA1 · June 2018 · Paper 2H · Q8 · 4 marks
4 marks
Examiner report
Again, this question was answered well in general, with most students gaining at least three marks. It was evident however, that some students made their lives hard by using very inefficient methods. These tended to be students who didn’t use a structure such as a two-way table or frequency tree. Having produced a correct diagram, it was disappointing to see a number of students give an incorrect fraction for their answer and thus lose the final mark; most commonly seen was ଷ . Although a frequency tree or two-way table is the most efficient method of solution, students should be encouraged to write their subtractions in the working space as a single arithmetic slip leads to a lot of wrong numbers and clear working is needed in these circumstances in order to award method marks.
There are 60 people: 38 male, 22 female. Britain has 32 including 11 female; Italy has 8 males; Spain has 12 people.
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Edexcel 1MA1 · June 2018 · Paper 2H · Q15 · 5 marks
5 marks
Examiner report
This part was answered well with two marks frequently gained. Errors included getting the probabilities on the last set of branches the wrong way round and making an arithmetic error when subtracting 0.45 from 1. When students chose to work with products but some added probabilities from the tree diagram instead. Those who found products typically went on to get all three marks although arithmetic errors were again in evidence. It was disappointing to see some students give a final answer that was greater than 1.
A team has P(win Saturday)=0.45. If it wins Saturday, P(win Sunday)=0.67; otherwise P(win Sunday)=0.35.
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Edexcel 1MA1 · June 2018 · Paper 3F · Q14 · 5 marks
5 marks
Examiner report
The majority of students knew what a stem and leaf diagram was in part (a) and most of those who had done an earlier unordered version completed their ordered diagram successfully to gain 2 marks. There were cases of an omitted number in the leaves; a simple check that there were 20 pieces of data would have been beneficial for these students. A key was not always present, but when offered was usually correct. In part (b) many students were able to identify 6/20 (or equivalent) from their data, though some students miscounted the number of students failing, with 7 seen quite often. Others that lost marks were comparing those that had passed instead of those that have failed. One mark was achieved by either identifying the correct number of fails (6) or by knowing a quarter of 20 was 5. Some students worked on the misconception that 71 was the percentage who passed rather than the value of the pass mark. Also, a lot of students commented on the denominator being 4, when there were 20 pieces of data, therefore not understanding the probability. It was common for students to make observations which did not make for a comparison. Many responses were not specific enough making vague statements about more or less than a quarter passed without evidence.
The marks are 76, 82, 84, 69, 80, 64, 70, 81, 75, 91, 87, 67, 80, 70, 94, 76, 81, 69, 71 and 77. The pass mark is 71. Omar claims that the probability a randomly chosen student failed is 1/4.
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Edexcel 1MA1 · June 2018 · Paper 3F · Q22 · 2 marks
2 marks
Examiner report
Most students recognised that the probabilities for the first throw did not add up to 1. Many lost out on a second mark because they were not clear in their explanation as to which of the second throw branches had an error. Very few students annotated the tree diagram, but those who did generally got at least one mark. The misconception that the dice had been thrown 3 times was stated but only rarely.
A biased die has P(4)=0.65 and is thrown twice. Amir's tree shows first-throw branches P(4)=0.65 and P(not 4)=0.25. After a first 4 it shows second-throw branches P(4)=0.35 and P(not 4)=0.65. After a first not 4 it shows P(4)=0.65 and P(not 4)=0.35.
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Edexcel 1MA1 · June 2018 · Paper 3F · Q24 · 5 marks
5 marks
Examiner report
Many responses in part (a) were able to demonstrate the understanding that the four probabilities summed to one. When shared in the given ratio, the second mark was awarded for the correct probability for red or white, though some gave 0.15 and 0.15 or 0.1 and 0.2 rather than the correct 0.2 and 0.1. The other main method seen for this item was to calculate the total number of counters in the bag by taking a proportional approach to the 18 blue counters. Those responses that achieved the correct red probability then sometimes either failed to achieve the full correct method and no further marks were awarded or went on to achieve the correct total of 8. It was surprising to find answers as fractions or decimals (usually based on assuming that 18 was the total number of counters). In part (b) many responses focussed on the fact that two halves make a whole. Most common was to comment on the splitting of an odd total and incorrectly stated that this was not possible. The best responses were able to articulate the connection to the context of this problem and that halving an odd number of marbles would result in splitting a marble – therefore there would need to be an even number of marbles to begin with.
A bag has red, white, blue and yellow counters. P(blue)=0.45, P(yellow)=0.25, 18 are blue, and P(red)=2P(white).
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Edexcel 1MA1 · June 2018 · Paper 3H · Q4 · 2 marks
2 marks
Examiner report
This question was generally well answered and descriptions of the things wrong with the tree diagram were generally clearly written. Nearly all students were able to identify the fact that the two probabilities for the first throw of the dice did not sum to one and make a valid comment on this. Most students were also able to see that two of the probabilities for the second throw were the wrong way round but some of the explanations were a little too vague for the award of the mark. Examiners noted that those students who annotated the tree diagram usually scored this mark.
A biased die has P(4)=0.65 and is thrown twice. Amir's tree shows first-throw branches P(4)=0.65 and P(not 4)=0.25. After a first 4 it shows second-throw branches P(4)=0.35 and P(not 4)=0.65. After a first not 4 it shows P(4)=0.65 and P(not 4)=0.35.
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Edexcel 1MA1 · June 2018 · Paper 3H · Q6 · 5 marks
5 marks
Examiner report
This question was a good discrminator. In part (a) nearly all students realised that they needed to subtact 0.45 and 0.25 from 1 to find the total of the probabilities of taking a red counter or a white counter. They could often also work out the separate probabilities, 0.2 and 0.1. However, only more able students could find the total number of counters in the bag then use the 0.2 to find the number of red counters. Instead many students worked out 0.2 × 18, 18 being the number of blue counters in the bag, so 3.6 (sometimes rounded to 4) was a commonly seen incorrect answer. Other common incorrect approaches involved reversing the probabilities for the numbers of red and white counters or working out the probability for a red counter as 0.15 (0.30 ÷ 2). Very few students used an algebraic approach. In part (b) there were many clear explanations seen but there were also many incomplete or unclear statements seen. For example “you cannot half an odd number” was a commonly seen response which just fell short of an acceptable response. Students needed to use the fact that 0.5 multiplied by an odd number will not give an integer in some way to score the mark in this part of the question. Answers including statements equivalent to “you cannot have half a marble” usually scored the mark available.
A bag has red, white, blue, yellow counters. P(blue)=0.45, P(yellow)=0.25, blue count 18, P(red)=2P(white).
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Edexcel 1MA1 · June 2018 · Paper 3H · Q20 · 5 marks
5 marks
Examiner report
There were relatively few fully complete and correct solutions to this question due to most students either making an error when calculating the number of students who only speak Spanish or failing to appreciate the non-replacement nature of the probability calculation. Having said that, many students were able to make a good start on the question and they drew an appropriate Venn diagram to help them. Students using a different approach were seldom successful. Students using a Venn diagram were generally able to place some elements in the Venn diagram correctly though many students interpreted the statement “7 speak German and Spanish” as “7 speak German and Spanish but not French”. This meant that further calculations would inevitably result in incorrect values placed in the diagram. Where a value 0 is found, in this case the number of students who speak only German, students are advised to write this on the Venn diagram rather than leaving a blank space. Students who did not get a correct value for the number of students who only speak Spanish could still access a mark for a correct probability calculation using their value.
Of 50 people, 31 speak French; 2 speak all three of French, German and Spanish; 4 speak French and Spanish but not German; 7 speak German and Spanish; 8 speak none; all 10 German speakers speak at least one other language. Two people are chosen without replacement.
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Edexcel 1MA1 · November 2017 · Paper 1F · Q12 · 4 marks
4 marks
Examiner report
This was a well answered question. The majority of students were able to answer part (a) scoring full marks and then able to go on to answer part (b) and answer that correctly also. If a slip was seen in part (a), 2 marks were often awarded and the follow through applied to part (b). Most candidates were comfortable with two way tables but a small minority did reverse the entries so careful reading of the headings is required. If part (b) was incorrect often 8 was seen as the incorrect answer, some students wrote unlikely when a numerical answer was required.
Emma has 45 rabbits. 30 of the rabbits are male. 8 of the female rabbits have short hair. 12 of the rabbits with long hair are male. (a) Use the information to complete the two-way table. Male Female Total Long hair Short hair Total One of Emma's rabbits is chosen at random. (b) Write down the probability that this rabbit is a female with short hair.
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Edexcel 1MA1 · November 2017 · Paper 1F · Q17 · 2 marks
2 marks
Examiner report
This question was answered well by many candidates. The most successful approach was to write a list of possibilities and then extract the number of successful combination and place this number over the total number of combinations. Students who attempted a systematic listing of outcomes scored well. Some problems arose when an unsystematic listing was attempted or only a partial list was given. was a common incorrect answer. Some mixed up the concept of odd and even and so 5 was another common answer although sometimes this answer was due to a basic adding error.
There are 3 cards in Box A and 3 cards in Box B. There is a number on each card. Box A Box B 9 3 4 2 5 3 Ryan takes at random a card from Box A and a card from Box B. He adds together the numbers on the two cards to get a total score. Work out the probability that the total score is an odd number.
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Edexcel 1MA1 · November 2017 · Paper 1H · Q13 · 4 marks
4 marks
Examiner report
The concepts of proportion within part (a) of this question were too complicated for most students; many started with the fraction of 15 rather than 4 , and as a result failed to score marks. Those who started correctly normally scored full marks. Part (b) was very poorly answered with almost no one scoring two marks. For those who scored 1 mark, it was normally for stating a correct bound (5.5 or 6.5).
A factory makes 450 pies. A representative sample has 15 pies. (a) If 4 sample pies are chicken, find total chicken pies. (b) If the number of steak pies in a sample is 6 correct to the nearest whole number, find the lower bound for the probability a random pie is steak.
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Edexcel 1MA1 · November 2017 · Paper 2F · Q12 · 5 marks
5 marks
Examiner report
This was very well understood and full marks were gained by the vast majority of students in both parts. In part (a) a minority of students put the frequencies in the wrong order in (usually) two of the right-hand boxes; some gave the frequencies incorrectly as probabilities of 200. In part (b) the most common incorrect answer was 13 (which gained 1 mark) where students had not read the question properly. Very rarely did students use incorrect notation for the probability.
There are 200 people: 95 male; 23 have no garden, including 10 males. (a) Complete the frequency tree. (b) Given a person has no garden, find P(female).
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Edexcel 1MA1 · November 2017 · Paper 2F · Q19 · 3 marks
3 marks
Examiner report
When students worked with numbers of cubes initially they were more successful. Some students began by incorrectly writing the ratio of Y:B the wrong way around; another common mistake was to interpret the green cubes as 4 × yellow (instead of 4 × blue). Some algebraic attempts were seen but these were rare and lead to many mistakes in calculating G. The most successful methods tended to use possible numbers of each colour, eg 4, 2, 16. It was not uncommon for students to use their own chosen values to represent the ratios which could also lead to the correct answer. Some marks were lost when working was not clear, or was sometimes contradictory when multiple methods were presented.
There are only blue cubes, yellow cubes and green cubes in a bag. There are twice as many blue cubes as yellow cubes and four times as many green cubes as blue cubes. Hannah takes at random a cube from the bag. Work out the probability that Hannah takes a yellow cube.
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Edexcel 1MA1 · November 2017 · Paper 2H · Q4 · 3 marks
3 marks
Examiner report
This question discriminated well between students of different abilities. The weakest students merely rephrased the question, for example by writing B = 2Y. Students going no further than this could not be awarded any marks. Most students did gain some credit for their attempts for writing down a ratio linking the number of cubes for at least two of the three colours. More able students wrote down a correct ratio linking all three colours. A common error was the use of the incorrect ratio Y : B : G = 1 : 2 : 4 leading to the incorrect answer, ଵ . An alternative strategy used by many students was to assign particular numbers of cubes to satisfy the conditions given and work from these. ଶ ଶଶ was a commonly seen acceptable answer as an alternative for the correct probability in its simplest form, ଵ ଵଵ . The general marking guidance given in the mark scheme states that probabilities given correct to at least 2 decimal places are also acceptable, so 0.09 was also accepted as a final answer.
Blue:yellow=2:1 and green:blue=4:1. Find the probability of yellow.
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Edexcel 1MA1 · November 2017 · Paper 2H · Q21 · 4 marks
4 marks
Examiner report
Only a small minority of students calculated the correct probability in part (a). In fact, not many students were even able get as far as multiplying three probabilities together and those that did often calculated ( ଵ ଷ)³ or equivalent. Some students attempted to use tree diagrams but these were usually incomplete or incorrect. In part (b) a few students were able to set up a hypothetical number of counters in the bag, usually 5 red, 5 blue and 5 yellow counters and then calculate a probability for comparison. Most of these students clearly stated their decision based on a correct comparison of probabilities. However, when comparing fractions, some students did not write them in a suitable form by using the same numerators or the same denominators or by converting the fractions to decimals.
There are 4 counters of each of three colours. (a) Find P(three red without replacement). More equal numbers of each colour are added. (b) Is three red less, equally or more likely? Show working.
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Edexcel 1MA1 · November 2017 · Paper 3F · Q26 · 3 marks
3 marks
Examiner report
Both parts to this question were poorly answered. In part (a), clearly the demand for the best estimate for the drawing pin to land “point up” was a distractor as many students gave their reason for selecting Mel to be because she/he had the greatest number for the times the pin landed point up. This was not acceptable. A number of students felt the pin should have a 50% chance of landing point up, presumably because there were two outcomes, and selected Tom accordingly. In part (b), the very few students that separately found the probability of the drawing pin landing point up and the probability of the drawing pin landing point down rarely multiplied them together to answer the question.
Lucy records 31 down,14 up; Mel 53 down,27 up; Tom 16 down,9 up. (a) Whose gives the best P(up) estimate? (b) Using all results, estimate P(up then down).
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Edexcel 1MA1 · November 2017 · Paper 3H · Q8 · 3 marks
3 marks
Examiner report
Students who identified Mel in part (a) did not always give a correct reason. Some referred to her having the greatest number of points up, not to her having done the most trials. Tom was identified by some students because his results give the greatest probability of getting point up or because his results have the smallest difference between the number of points down and the number of points up. In part (b) students were expected to use all the results to find the fractions and 50 and to then multiply these two fractions. Many students did find these two fractions. Those who simplified them to 2 and 1 often went on to multiply but most students did not multiply the two fractions. Many students added rather than multiplied. A few students used the results from just one of the three people but they could be awarded the method mark if they multiplied their two fractions to find the probability of point up followed by point down. Some students lost the accuracy mark because they gave the answer as 0.2. A decimal equivalent to a probability should be written to at least 2 decimal places (unless tenths).
Lucy records 31 down,14 up; Mel 53 down,27 up; Tom 16 down,9 up. Find the best P(up) estimate and then P(up followed by down) using all results.
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Edexcel 1MA1 · November 2017 · Paper 3H · Q10 · 3 marks
3 marks
Examiner report
In part (a) many students worked out the probability of getting a red counter as 0.05. A common incorrect answer was 0.5, often with 0.95 or 1 – 0.95 shown in the working. Part (b) asks for the least possible number of counters in the bag. Students are advised to read the question carefully as a surprisingly large number gave a colour, not a number, as the answer. Sometimes they gave the lowest probability. Some students worked out the least possible number of counters as 20 but gave no reason for their answer; they scored one of the two marks. The most common correct reasons given referred to the numbers of counters having to be whole numbers. Some students gave a number greater than 20 as the least possible number of counters but scored one mark for a correct reason.
There are only blue counters, yellow counters, green counters and red counters in a bag. A counter is taken at random from the bag. The table shows the probabilities of getting a blue counter or a yellow counter or a green counter. Colour blue yellow green red Probability 0.2 0.35 0.4 (a) Work out the probability of getting a red counter. (b) What is the least possible number of counters in the bag? You must give a reason for your answer.
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Edexcel 1MA1 · November 2017 · Paper 3H · Q12 · 3 marks
3 marks
Examiner report
Working out 0.75 × 0.4 to get the probability of both spinners landing on white and working out 0.25 × 0.6 to get the probability of both spinners landing on red gained the two method marks. Some students found only one of these probabilities, usually the former, and scored one mark only. A number of those students who did work out both probabilities failed to spot that 0.3 is double 0.15 and therefore the answer will be double 24. Some students used 0.15 and 24 to work out that the total number of spins is 160 and were then often able to get the correct answer. A common error was to add the probabilities instead of multiplying them.
Alan has two spinners, spinner A and spinner B. Each spinner can land on only red or white. The probability that spinner A will land on red is 0.25 The probability that spinner B will land on red is 0.6 The probability tree diagram shows this information. Spinner A Spinner B red 0.6 red 0.25 0.4 white red 0.6 0.75 white 0.4 white Alan spins spinner A once and he spins spinner B once. He does this a number of times. The number of times both spinners land on red is 24 Work out an estimate for the number of times both spinners land on white.
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Edexcel 1MA1 · June 2017 · Paper 1F · Q6 · 2 marks
2 marks
Examiner report
A surprising number of students answered part (a) incorrectly, normally marking at around ଵ ସ instead of ଵ ଶ. This must have come from not reading the question carefully enough and just seeing a single “B” rather than noticing there was two of them. Almost every student was able to answer part (b) correctly.
Sammy spins a fair 4-sided spinner. A B C B (i) On the probability scale, mark with a cross (x) the probability that the spinner will land on B. 0 1 1 2 (1) (ii) On the probability scale, mark with a cross (x) the probability that the spinner will land on F. 0 1 1 2 (1)
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Edexcel 1MA1 · June 2017 · Paper 1F · Q12 · 2 marks
2 marks
Examiner report
This is a familiar question and students generally performed well and often got both marks. Those who didn’t very commonly got one mark for either a correct numerator or denominator in a proper fraction. There are still students using incorrect notation for probabilities such as ratio, or words and centres should be aware that this will always result in dropped marks.
There are only 7 blue pens, 4 green pens and 6 red pens in a box. One pen is taken at random from the box. Write down the probability that this pen is blue.
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Edexcel 1MA1 · June 2017 · Paper 1H · Q17 · 4 marks
4 marks
Examiner report
A correct answer of 28 or 7 was obtained by a large proportion of candidates gaining full marks; incorrect simplification of an otherwise correct answer was not penalised. Candidates were usually able to pick one mark up for writing a fraction over 8, showing understanding that the counter was not replaced. Where marks were lost this tended to be for common errors such as using a method that involved replacement, or incorrect processing of the fraction such as addition of the fractions done incorrectly, or writing only one of the fractions rather than adding them. Some candidates used fractions with a denominator of 9 on their first branches and then a denominator of 7 on their second set of branches which then failed to score them any marks.
There are 9 counters in a bag. 7 of the counters are green. 2 of the counters are blue. Ria takes at random two counters from the bag. Work out the probability that Ria takes one counter of each colour. You must show your working.
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Edexcel 1MA1 · June 2017 · Paper 2F · Q7 · 1 mark
1 mark
Examiner report
This question was often correctly answered but there were some students who gave answers such as 9.985 and 99.985, which are not appropriate as probabilities. Examiners were left wondering whether all students had access to a calculator to answer this question.
The probability that a new fridge has a fault is 0.015 What is the probability that a new fridge does not have a fault?
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Edexcel 1MA1 · June 2017 · Paper 2F · Q17 · 3 marks
3 marks
Examiner report
This question was a good discriminator. The majority of students successfully calculated the probability of the dice landing on 1. A small number of weaker students added the given probabilities but did not subtract their result from 1. It was encouraging to see that a large proportion of students also showed they knew how to find an expected frequency by multiplying a probability by 200. Some students interpreted the instruction to work out an estimate as a need to round either the probabilities or expected frequencies. This often resulted in a final answer of 100. They apparently did not appreciate that working out the outcome of 200 trials by using a theoretical model will inevitability give an estimate. Some students gave an answer in the form 200 and so lost the accuracy mark.
The table shows the probabilities that a biased dice will land on 2, on 3, on 4, on 5 and on 6 Number on dice 1 2 3 4 5 6 Probability 0.17 0.18 0.09 0.15 0.1 Neymar rolls the biased dice 200 times. Work out an estimate for the total number of times the dice will land on 1 or on 3
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Edexcel 1MA1 · June 2017 · Paper 2H · Q1 · 3 marks
3 marks
Examiner report
The great majority of students were able to correctly find the probability of the dice landing on “1”. Many went on to correctly solve the problem but many stopped after just finding the probability of the dice landing on 1 or 3, eg 0.31 + 0.18 = 0.49 was a common answer. Some students made errors when adding probabilities, but would have received the first process mark if they had shown that they were subtracting their sum from 1. Students should be encouraged to show this important line in their working, even though they may think it is trivial. It was not uncommon here for students, misinterpreting the word ‘estimate’, to round their answer to 0.5 before multiplying by 200. Some did this earlier in their solution. Some students found the expected number of 1s and 3s separately and failed to add them together to find the total number. Having correctly solved the problem, too many students gave their answer as a probability, 200 , or a simplified version of this. Quite a few students multiplied P(1) and P(3) before multiplying by 200 as well.
The table shows the probabilities that a biased dice will land on 2, on 3, on 4, on 5 and on 6 Number on dice 1 2 3 4 5 6 Probability 0.17 0.18 0.09 0.15 0.1 Neymar rolls the biased dice 200 times. Work out an estimate for the total number of times the dice will land on 1 or on 3
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Edexcel 1MA1 · June 2017 · Paper 2H · Q12 · 2 marks
2 marks
Examiner report
In part (a), the vast majority of students recognised the error in the diagram but many failed to articulate their explanation satisfactorily, often incorrectly stating that they should add up to 30 or 29 instead of 1 In part (b), most students explained that the statement was incorrect because multiplication of the probabilities was required. Some students incorrectly argued that adding the probabilities gave the probability of either boy scoring a goal. Also many noticed that they expected the probability of both to be lower than either of the original probabilities.
There are 30 students in Mr Lear's class. 16 of the students are boys. Two students from the class are chosen at random. Mr Lear draws this probability tree diagram for this information. 1st student 2nd student 15 30 16 30 14 30 16 30 14 30 13 30 (a) Write down one thing that is wrong with the probabilities in the probability tree diagram. Owen and Wasim play for the school football team. The probability that Owen will score a goal in the next match is 0.4 The probability that Wasim will score a goal in the next match is 0.25 Mr Slater says, "The probability that both boys will score a goal in the next match is 0.4 + 0.25" (b) Is Mr Slater right? Give a reason for your answer.
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Edexcel 1MA1 · June 2017 · Paper 3F · Q15 · 6 marks
6 marks
Examiner report
Students who started by putting 15 in the intersection generally went on to answer part (a) quite well and often placed all seven numbers correctly inside the circles. Some students, however, wrote two 15s in the intersection or wrote 15 in more than one region. The outside region, ) ( B A , proved to be much more problematic. It was very common to see either no numbers at all in this region or duplicates of the numbers that had already been placed inside the circles. Those who did attempt to put the rest of the odd numbers in the outside region sometimes failed to include all eight numbers. It should be emphasised to students that each number in the universal set should appear just once in a Venn diagram. Many students scored the one mark for labelling the circles, usually with A and B but occasionally with “multiples of 3” and “multiples of 5”. In part (b) many students scored one mark for a correct denominator of 15 or, more usually, for a denominator that followed through correctly from their Venn diagram. A correct numerator was seen far less frequently and it was evident that many students confused B A with B A . Some incorrect notation for probability, eg ratio, was seen.
E is the odd positive integers below 30. A={3,9,15,21,27}; B={5,15,25}. (a) Place all elements in a two-set Venn diagram. (b) Find P(A union B).
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Edexcel 1MA1 · June 2017 · Paper 3F · Q17 · 2 marks
2 marks
Examiner report
Part (a) was not well answered. Many students chose to write out a list of the 25 dress sizes and these students were usually able to find the median dress size. Common incorrect answers were 11 (the median of 8, 10, 12 and 14) and 7 (the median of the four frequencies). Part (b) was answered very poorly indeed with relatively few students able to explain that Zoe is not correct because some of the women could have both a shoe size of 7 and a dress size of 14. The majority of those who said that Zoe is not correct gave explanations that focused on the fraction calculation rather than on the context, eg stating that she should have multiplied the fractions. Many students thought that Zoe is correct.
The table shows some information about the dress sizes of 25 women. Dress size Number of women 8 2 10 9 12 8 14 6. (a) Find the median dress size. Three of the 25 women have a shoe size of 7. Zoe says that, if one of the 25 women is chosen at random, the probability that she has either a shoe size of 7 or a dress size of 14 is 9/25 because 3/25 + 6/25 = 9/25. (b) Is Zoe correct? You must give a reason for your answer.
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Edexcel 1MA1 · June 2017 · Paper 3H · Q1 · 6 marks
6 marks
Examiner report
This question was accessible to almost all students with the modal mark being 4 out of 6. Most students gained at least two marks on part (a). They were able to list the numbers correctly in the various sections of the Venn diagram but the commons errors seen were a failure to use labels or to place the remaining numbers in the universal set correctly. Students who performed best wrote out all potential values and ticked them off to ensure all were included in the Venn diagram. Part (b) was generally well answered with most students able to follow through their Venn diagram correctly.
E is the odd positive integers below 30. A={3,9,15,21,27}; B={5,15,25}. (a) Place all elements in a two-set Venn diagram. (b) Find P(A union B).
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Edexcel 1MA1 · June 2017 · Paper 3H · Q3 · 2 marks
2 marks
Examiner report
Part (a) was well answered by the majority of students. Some wrote out all the data and then found the median from their list. Incorrect answers seen were often when the student tried to work out the mean instead of stating the median. Part (b) was intended to assess one of the new assessment objectives and was not well answered by students. Many felt that any combination of probabilities required either adding or multiplying dependent upon the ‘or’ or the ‘and’ in the question. Few looked at the concept being described and considered that the qualities being discussed were not mutually exclusive. This phrase was not essential for the mark but a description of a person being able to have both attributes was required.
The table shows some information about the dress sizes of 25 women. Dress size Number of women 8 2 10 9 12 8 14 6. (a) Find the median dress size. Three of the 25 women have a shoe size of 7. Zoe says that, if one of the 25 women is chosen at random, the probability that she has either a shoe size of 7 or a dress size of 14 is 9/25 because 3/25 + 6/25 = 9/25. (b) Is Zoe correct? You must give a reason for your answer.
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Edexcel 1MA1 · sample · Paper 1F · Q6 · 2 marks
2 marks
Greg rolls a fair ordinary dice once.
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Edexcel 1MA1 · sample · Paper 1F · Q9 · 1 mark
1 mark
There are 3 red beads and 1 blue bead in a jar.
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Edexcel 1MA1 · sample · Paper 1F · Q17 · 5 marks
5 marks
100 students had some homework. 42 of these students are boys. 8 of the 100 students did not do their homework. 53 of the girls did do their homework.
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Edexcel 1MA1 · sample · Paper 1H · Q24 · 4 marks
4 marks
John has an empty box.
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Edexcel 1MA1 · sample · Paper 2F · Q5 · 1 mark
1 mark
There are boys and girls in a classroom.
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Edexcel 1MA1 · sample · Paper 2F · Q29 · 2 marks
2 marks
A probability tree describes games A and B.
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Edexcel 1MA1 · sample · Paper 2H · Q12 · 5 marks
5 marks
A and B are two sets of traffic lights on a road. The probability that a car is stopped by lights A is 0.4 If a car is stopped by lights A, then the probability that the car is not stopped by lights B is 0.7 If a car is not stopped by lights A, then the probability that the car is not stopped by lights B is 0.2
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Edexcel 1MA1 · sample · Paper 3F · Q15 · 3 marks
3 marks
There are 25 boys and 32 girls in a club.
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Edexcel 1MA1 · sample · Paper 3H · Q12 · 6 marks
6 marks
Sami asked 50 people which drinks they liked from tea, coffee and milk. All 50 people like at least one of the drinks 19 people like all three drinks. 16 people like tea and coffee but do not like milk. 21 people like coffee and milk. 24 people like tea and milk. 40 people like coffee. 1 person likes only milk. Sami selects at random one of the 50 people.
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Edexcel 1MA1 · specimen2 · Paper 1F · Q7 · 4 marks
4 marks
A box contains 6 plain chocolates, 8 milk chocolates and 10 white chocolates.
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Edexcel 1MA1 · specimen2 · Paper 1F · Q21 · 5 marks
5 marks
A game uses a fair six-sided die numbered 1–6 and a fair five-sided spinner numbered 1–5. A player wins £5 by rolling 5 and spinning 5; wins £2 by rolling 1 or spinning 1 (including both); otherwise wins nothing. Thirty people each pay £1 to play.
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Edexcel 1MA1 · specimen2 · Paper 1H · Q3 · 5 marks
5 marks
A game uses a fair six-sided die numbered 1–6 and a fair five-sided spinner numbered 1–5. A player wins £5 by rolling 5 and spinning 5; wins £2 by rolling 1 or spinning 1 (including both); otherwise wins nothing. Thirty people each pay £1 to play.
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Edexcel 1MA1 · specimen2 · Paper 1H · Q16 · 1 mark
1 mark
P(Sanay late)=0.05 and P(Jaden late)=0.15. Alfie multiplies these to get P(both late)=0.0075.
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Edexcel 1MA1 · specimen2 · Paper 2F · Q19 · 3 marks
3 marks
Fair spinners have outcomes 1,2,3,4 and 1,2,3,4,5; score is sum.
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Edexcel 1MA1 · specimen2 · Paper 2F · Q26 · 4 marks
4 marks
A Venn diagram contains A only = {15}, A ∩ B = {12, 18}, B only = {10, 14, 16}, and outside both sets = {11, 13, 17, 19}.
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Edexcel 1MA1 · specimen2 · Paper 2H · Q5 · 4 marks
4 marks
A Venn diagram contains A only = {15}, A ∩ B = {12, 18}, B only = {10, 14, 16}, and outside both sets = {11, 13, 17, 19}.
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Edexcel 1MA1 · specimen2 · Paper 3F · Q10 · 4 marks
4 marks
A stem-and-leaf diagram gives these speeds: 38; 41, 43, 44, 46, 47, 48, 48, 49, 49; 52, 52, 54, 56, 57, 57, 58, 58, 59; 61, 61, 62, 62, 62, 62, 63; 70.
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Edexcel 1MA1 · specimen2 · Paper 3F · Q19 · 3 marks
3 marks
A bag contains 120 counters. The probabilities of choosing a blue, green or yellow counter are 0.35, 0.40 and 0.20 respectively; the remaining counters are red.
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Edexcel 1MA1 · specimen2 · Paper 3H · Q22 · 7 marks
7 marks
There are y black socks and 5 white socks in a drawer. Joshua takes two socks at random without replacement. The probability that he takes one white sock and one black sock is 6/11.
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Everything here stays free. If probability is where the marks keep going, I teach it one to one from the mark schemes. Free intro call, then a free first lesson.