Edexcel A-level Maths · Topic S3
Practice questions covering probability from this specification. Every question below is the real Pearson question as it is printed on the paper, with its mark scheme.
6 questions on Probability, each with its mark scheme and worked method. Reveal the answer when you have had a go.
1.
(3)
(Total for Question 1 is 3 marks)
2.
(2)
(Total for Question 2 is 2 marks)
3.
(4)
(Total for Question 3 is 4 marks)
4.
(2)
(Total for Question 4 is 2 marks)
5.
(3)
(Total for Question 5 is 3 marks)
6.
(2)
(Total for Question 6 is 2 marks)
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Printable question pack — S3 ProbabilityAll 30 questions across 3 specification points — 24 more than the 6 above — with the full mark scheme. Guest accessible · browser print.11 questions shown as printed, 6 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 9MA0 · June 2024 · Paper 31 · Q5 · 10 marks
10 marks
Examiner report
Part (a) was answered very well by most students and in part (b) a good proportion also recognised the need for the events to be independent. Some students failed to score the mark in part (b) because they did not use correct mathematical language: we saw comments such as “they are not affected by each other” or “the events are not mutually exclusive” and a few simply said “it is continuous data” or “they must take part in both events”. Many students answered part (c) successfully, but some made errors when trying to find P(T < 300) either by using a continuity correction and finding P(T < 299.5) or failing to convert minutes to seconds and trying to find P(T < 5). There were many good attempts at part (d). Most students could start the problem and write down a pair of linear equations in and with suitable values of z. Many students showed some working when solving their equations whilst others simply wrote down solutions using their calculators. This is fine if the equations are correct but if there are errors in the equations then the method mark cannot be awarded if no method for solving the equations is seen. The front of the paper indicates that if students use their calculators instead of the tables, then they should use an equivalent degree of accuracy which in this case, for finding the z values, was 4 decimal places. Some failed to do this and lost one of the accuracy marks but there were a good number of students scoring full marks for this part. A common error here was missing the minus sign on the z value for the lower tail and a few students assigned the z values the wrong way around.
For a school athletics club, high-jump height H is normally distributed with mean 1.4 m and standard deviation 0.15 m. The 1500 m time T is normally distributed with mean 330 s and standard deviation 26 s.
Edexcel 9MA0 · June 2024 · Paper 31 · Q6 · 11 marks
11 marks
Examiner report
It was encouraging to see most students attempting this final question and many made good progress with the first two parts. The major obstacles to success here were poor use of, sometimes invisible, brackets and weaknesses in manipulating fractions involving algebra and decimals. Part (a) was answered very well with many giving all 3 pairs which was fine provided they were all correct. Part (b) was answered well too with most successfully finding p = 0.13 although a few students omitted the 0.08 from P(C). Part (c) proved more discriminating. Whilst many students knew ( ) ( ) ( ) P P | P A B A B B = they were not able to find the relevant values from the Venn diagram. Some thought ( ) ( ) ( ) P P P A B A B = and quite a few were unable to establish the relationship q + r = 0.22 and some of those who did get this far believed that r couldn’t be 0 and gave an answer of 0.21 0.55 . A common error was to write ( ) ( ) ( ) P 0.05 P | P 0.25 0.08 A q A B B q r + = = + + + . In part (d), having been given an equation with which to work, students were much more successful with many gaining all 3 marks. There were errors with the conditional probability again and many slips in adding probabilities, but it was common to see 0.4 0.5 0.73 r = + leading to the correct values for r and q. When part (e) was attempted, it was usually correct however a few students multiplied 0.25 and 0.08 instead of adding them. Part (f) proved to be one of the most discriminating parts of the paper. There were many blank responses, and some simply gave an answer of p = 0.13. Some though could find the correct region but gave the answer as a probability e.g. ( ) P B A C rather than an event which was unfortunate as they had usually not made this mistake in part (a).
A Venn diagram has overlapping sets A-B-C in a chain, with D wholly inside C and disjoint from A and B. Region probabilities are: A only = q; A intersection B = 0.05; B only = p; B intersection C = 0.27; C only outside D = 0.25; D = 0.08; outside all sets = r.
Edexcel 9MA0 · June 2023 · Paper 31 · Q1 · 6 marks
6 marks
Examiner report
This proved to be a positive and accessible start to the paper. Part (a) was nearly always correct and most knew how to write down a suitable expression for independence for part (b) which usually resulted in a correct equation for p. Those that didn’t achieve this often thought that P(C) = 0.3 or 0.3p or even just p. Solving for p was mostly completed correctly and even those who had an incorrect value for p could often score the follow through for q provided their p + q = 0.27 In part (c), some used the simple conditional probability formula [ P( ) P( )P( | ) A B A A B = ] in the formula booklet to get started whereas others just wrote down a correct ratio of probabilities from their Venn diagram. A fairly common error was to have P(B) = 0.6 as the denominator and a few misinterpreted the formula and used ( ) ( ) P . P A B B Those using the more complicated formula for P(A|B) in the formula booklet rarely managed to complete part (c) and students should be encouraged to know how to use the formulae in the formula booklet efficiently.
A Venn diagram shows three events A, B and C. The mutually exclusive regions have probabilities: A only 0.13; A intersect B 0.25; B only 0.05; B intersect C 0.30; C only p; and outside A union B union C is q. Events A and C do not overlap.
Edexcel 9MA0 · June 2023 · Paper 31 · Q5 · 8 marks
8 marks
Examiner report
This question proved to be the most challenging on the paper for students. In part (a) few could see how to use the two bullet points to reach the required result. A good number realised that P(S |{X = x}) = k x but couldn’t always use the conditional probability formula to reach ( ) P 50 50 k S X b = = . Those that did get this far were often able to give a convincing proof of the given result. A common misconception was believing that , etc 20 50 k k a b = = and then using the sum of the probabilities equalling 1 but this scored no marks. It was good to see a number of students using the result from part (a) to form similar relationships between the probabilities, such as d = 2b and going on to solve their equations to find the probability distribution. The final part was particularly challenging with many misinterpreting the 30% probability given in the stem to part (c). Many confused this 0.3 [= P(S | X = 100) ] with the 0.4 [= P(X = 100)] from the probability distribution and simply compared these two values. Those that did realise that Nav’s 30% meant that k = 30 often went on to show that when x = 20 this would give a probability [P(S | X = 20)] > 1 and so the model was not suitable.
Tisam plays a game using a spinner, a ball and a cup. The random variable X has values 20, 50, 80 and 100 with probabilities a, b, c and d respectively. After spinning, Tisam stands X cm from the cup and tries to throw the ball into it. Let S be the event that the throw is successful. Her model assumes P(S | X = x) = k/x, where k is constant, and that P(S intersect {X = x}) is the same for every possible x.
Edexcel 9MA0 · June 2022 · Paper 31 · Q5 · 10 marks
10 marks
Examiner report
Part (a) was well answered with most students able to gain the mark for a correct answer. A simplified fraction or the correct decimal value was often seen. Whilst the majority of students continued to score well in part (b) the most common incorrect answer came from students who thought the question was conditional, giving 170 550 , i.e. the probability of not being a professional given they live in area B. Some students attempted a product of probabilities assuming independence, which was not given and not correct. In part (c) there was mixed success in the interpretation of the information given in the question with only about half of students getting the Venn diagram completely correct. Some correct diagrams followed minimal working whilst others filled at least a page. A common omission was to leave the region H R F ′∩′∩ ′ blank. The method mark was most commonly given when implied by one correct value in the Venn diagram, rather than for the student’s working, which was commonly unclear. Some working indicated incorrect use of the given percentages. Where only one value was correct in the Venn diagram it was usually the 481 or 259, for ‘professionals from area A’. When none or only one of the four correct values was seen many students benefited from being able to achieve 1 mark for the sums of certain pairs of values in their diagram. There were a number of students who simply abandoned the Venn diagram which often resulted in no attempts being made at the remaining parts of the question. Part (d) had a mixed response with many correct answers. This part could be answered independently of whether or not (c) had been attempted and it was good to see most students doing so successfully Whilst many realised the 133 given in the Venn diagram was required, the follow- through mark could only be scored by students who had some value in the region 𝐻𝐻′ ∩ 𝑅𝑅′ ∩𝐹𝐹′ so that this value could be added to the 133 in their calculation. A common non-scoring answer of 133 1825 was often seen. Most students realised that part (d) was a conditional probability question with many taking the simpler approach of using numbers rather than probabilities in their ratio. Those with a correct Venn diagram usually scored both marks here. As is often the case, a noticeable number of students had the incorrect event, namely F, as the denominator in their probability expression rather than H.
A company has 1825 employees. Counts by classification and home area are: professional, area A 740 and area B 380; skilled, area A 275 and area B 90; elementary, area A 260 and area B 80. A random employee is chosen. Let F mean professional, H mean works from home and R mean lives in area A. In each area, 65% of professional, 40% of skilled and 5% of elementary employees work from home.
Edexcel 9MA0 · October 2021 · Paper 31 · Q1 · 7 marks
7 marks
Examiner report
In part (a) many scored the mark by referring to the possibility of bias being introduced by the interviewer or the lack of randomness when using quota sampling. The most common answers that did not score the mark referred to cost or time failing to appreciate that quota sampling is generally quicker and therefore cheaper to implement than simple random sampling. Some thought that quota sampling required a sampling frame which suggested they were confused about the different types of sampling. Part (b) was generally answered very well with most stating the distribution, and scoring the method mark, and then evaluating part (i) correctly on their calculator. Part (ii) met with less success as many could not deal with the inequality or were unsure how to use their calculator correctly. Some seemed to confuse significant figures with decimal places: the front of the exam paper instructs then to use 3 significant figures but some only gave an answer of 0.022 here and lost this mark. Part (c) was usually answered correctly and many went on to use the correct binomial distribution in part (d) though some found interpreting “fewer than 3” a problem and a common incorrect answer was 0.924 (from P( 3 X ). Overall though this proved a good start to the paper and over 40% scored 6 or 7 marks here.
In a university, 8% of students are members of the university dance club. A random sample of 36 students is taken, and X is the number who are members of the dance club. Only 40% of dance club members can dance the tango.
Edexcel 9MA0 · October 2021 · Paper 31 · Q4 · 11 marks
11 marks
Examiner report
This proved to be a good source of marks for most candidates with only 2% failing to score and over 18% scoring full marks. Most students navigated parts (a) and (b) successfully but a few missed the information that “No students read all three magazines” and wasted a lot of time trying to find two equations in p and q. The conditional probability in part (c) proved to be a challenge for some; there were those who effectively used P(E | S) and others were miscopying values from their Venn diagram. Those who failed to make any progress in (i) were often able to pick up a mark for a correct follow through when finding the value of t. Part (d) was the least successful part of the question often due to poor use of the notation. There were a curiously large number of students who confused events with the probability of events; for example we saw a number of cases of students writing S E′ ∩ = 0.48 instead of P( S E′ ∩ ) = 0.48. Another source of errors was confusing the symbols and ∩ ∪ in their expressions. There was a demand to show clear working here and this meant that we needed to see probabilities correctly labelled and an appropriate test clearly shown and a number of attempts lacked these features but there were plenty of good solutions to this part too.
A college produces magazines about green issues, equality and sports. For a randomly selected student, let G, E and S be the events that the student reads the corresponding magazine. A three-set Venn diagram has regions: G only 0.08; E only 0.09; S only 0.36; G intersect E but not S is 0.05; G intersect S but not E is q; E intersect S but not G is r; G intersect E intersect S is p; and outside all three sets is t.
Edexcel 9MA0 · October 2021 · Paper 31 · Q5 · 11 marks
11 marks
Examiner report
Most students used their calculators in part (a) with reasonable success but a frequent error was to use 0.1 instead of 0.01 and without any working being shown there was no opportunity to award the method mark. Part (b) was a straightforward probability calculation designed to be completed on a calculator as the specification requires and so it was disappointing that nearly 25% of the candidates scored no marks at all on this question. Identifying the conditional probability in part (c) defeated many students and some of those who seemed to be attempting to use a conditional probability were unable to use the correct notation to write their expression. Those who did start with a correct conditional probability and moved onto a ratio of probabilities often had a correct denominator but a common error was to have P(F > 160) as the numerator rather than P(160 < F < 175). Part (d) was a familiar hypothesis test and many students scored something here. The hypotheses were usually stated in terms of µ and most of the time 166.5 rather than 164.6 was used. Selecting the correct model caused a number of problems: some used 164.6 as the mean and others used 7.4 (or occasionally 6.1) as the standard deviation, simply stating the correct model will score the method mark here but a number of students didn’t show us the model they were using and we had to award this mark by implication from correct working. Most students successfully used the probability approach (as we would recommend) with a few attempting a critical region or using a z value and comparing with 1.6449 from the tables. The final mark required a correct interpretation of their calculation in context and some students simply stopped after stating that the result was significant and failed to score this final mark.
The heights F cm of females from a country are normally distributed with mean 166.5 cm and standard deviation 6.1 cm. For a different country, female heights are normally distributed with standard deviation 7.4 cm. Mia believes the mean height in the different country is less than 166.5 cm. She takes a random sample of 50 females there and obtains a sample mean of 164.6 cm.
Edexcel 9MA0 · October 2020 · Paper 31 · Q1 · 8 marks
8 marks
Examiner report
This proved to be a popular start to the paper with over 35% of students scoring full marks. Part (a) was answered well with most understanding the term “mutually exclusive” and giving their answer as events not probabilities. Part (b) was very straightforward and nearly all could interpret the Venn diagram correctly to answer this. Many knew a suitable rule for independence in part (c) and could use this to find P(A) = 0.6 or write down a suitable equation for q which was then solved correctly. Part (d) was more challenging as many could not use the given conditional probability correctly to obtain an equation for r but most knew how to use this answer to find s and were able to gain at least one mark in this part.
A Venn diagram represents events A, B, C and D. Sets A and C are disjoint. Set D lies wholly inside A and is disjoint from B. Set B overlaps A and C. The labelled regions are: A intersect B = 0.24; B only = 0.07; B intersect C = p; A only but outside B and D = 0.16; D = q; C only = r; and outside A union B union C = s.
Edexcel 9MA0 · October 2020 · Paper 31 · Q4 · 10 marks
10 marks
Examiner report
Part (a) was generally answered very well. Being a “show that” question meant that we needed to see a full explanation and most clearly showed the sum of the probabilities equal to 1 and then solved the resulting equation to find k. Failing to show this key step usually meant that no marks could be awarded. A significant number of students left the rest of the question blank but those who did make an attempt were usually able to identify some, though rarely all, the cases in (b). Finding the probabilities was more of a problem with some using P(D = 50) + P(D = 30) rather than the product and those having correct probabilities for all 3 cases were quite rare. It was good to see many students who had faltered on (b) still making a start on part (c). Many identified the 4 angles as terms of an arithmetic series and some were able to reach the condition 2a + 3d = 180 (or equivalent) but many could not see how to complete the problem. Some used the information that a > 50 to reach d < 80 3 but were unable to make the connection that this meant that only the two cases d = 10 and d = 20 were required and so the probability could simply be written down.
The discrete random variable D takes values 10, 20, 30, 40 and 50, with respective probabilities k/10, k/20, k/30, k/40 and k/50, where k is a constant.
Edexcel 9MA0 · October 2020 · Paper 31 · Q5 · 15 marks
15 marks
Examiner report
It was disappointing that over 20% of candidates scored no marks here as there were some straightforward marks available to those who had studied this part of the specification. Part (a) was a very straightforward opening part and the vast majority of students could use their calculator efficiently to score this mark. Part (b) was a fairly standard hypothesis test and a good number made progress here. Some still did not use µ for the hypotheses and a null hypothesis of 10 T = was all too common. Most attempts used a normal distribution with a correct variance but a significant number of students used a mean of 11.5 rather than 10. This is not a correct method for a hypothesis test (the probabilities should be calculated on the assumption that H0 is true) but, on this occasion, for those who still managed to reach a correct conclusion a special case enabled them to secure one mark. A variety of different approaches were used in part (b), some finding the critical value for X and others calculating the test statistic of 1.677… but some of these candidates compared this value to the significance level rather than the critical value of 1.6669 from the tables. The first part of (c) was another fairly straightforward calculation which many scoring this mark, sometimes including those who failed to score the mark in part (a). Students need to be aware of which “tail” their calculator gives them. Beyond (c) (i) the question was more challenging but it is encouraging that over 50% of students scored some of these first 6 marks. In (c)(ii) although a good number of students recognised that a conditional probability was required, few could find a correct expression for the numerator and a common error was to see (c)(i) P( 0) T > used rather than a correct numerator of P(0 < T < 2). In part (iii) many realised that the model would assign some probability to negative values of T but this will always be true when a normal distribution is used as a model. The point here was that a significant proportion of the values for T were negative and so this means that the normal model is not very good. Most responses seemed to believe that a model must match reality exactly but a few did comment that P(T > 0) should be closer to 1 or that the difference between (i) and (ii) was too big and therefore the model was not a good one to use. Fully correct answers to part (d) were rare and this part gave the A* students an opportunity to shine. Successful students realised that they needed to look at P(T > 2) and often found this by taking 1 – (c)(i) but they rarely realised that the median splits this probability into 2 so they were seeking the value of t such that P(T > t) = 0.40…Some who reached this stage tried, incorrectly to use their calculator, failing to realise that their calculator gave t for probabilities of the form P(T < t), a few used the tables provided to find z = 0.2533 and some did prepare correctly to use their calculator and found P(T < t ) = 0.5978… (the 0.402… + (c)(i)) from whence the correct value of t = 5.9 could be found. It was encouraging to see some good use of diagrams in this question being used to help students identify which probabilities they needed.
A health centre claims that the time X minutes a doctor spends with a patient is modelled by N(10, 4^2). A receptionist samples 20 patients and obtains a mean time of 11.5 minutes. The centre also claims that the time T minutes a dentist spends on a routine appointment is modelled by T ~ N(5, 3.5^2).
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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.