P Probability — revision question pack

9 specification points · notes, questions, answers and worked methods

Checked against Edexcel 1MA1 section P. Review basis: the qualification registry sourced from the Pearson Edexcel Level 1/Level 2 GCSE (9-1) in Mathematics (1MA1) specification; registry verification recorded 9 July 2026.

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P1 · Record, describe and analyse the frequency of outcomes of probability experiments using tables and frequency trees

Explanation

A frequency is a count of how many times an outcome occurs.
In a frequency table, all the frequencies should add to the total number of trials.
A frequency tree splits a total into groups: the child branches at every split must add back to their parent.
Complete a missing branch by subtraction, then use the finished tree to compare counts or calculate relative frequencies.
When groups have different totals, compare proportions rather than raw frequencies.

Worked example

A frequency tree starts with $70$ people. $43$ choose tea and the rest choose coffee. Of the tea drinkers, $18$ add sugar. Complete the missing frequencies.

1.Coffee drinkers: $70-43=27$ .
2.Tea drinkers without sugar: $43-18=25$ .
3.Check the first split: $43+27=70$ .

Answer: $27$ choose coffee and $25$ choose tea without sugar.

Common mistakes

Don't fall into the trap of writing probabilities on a tree that asks for frequencies.
Don't fall into the trap of subtracting a branch from the overall total instead of its parent.

Exam tip

Check every split by adding its two branches back to the parent frequency.

Tier 1 · Easy

1
A spinner is used $30$ times. It lands on red $18$ times and on blue $12$ times. Complete a frequency table for the two outcomes and work out the relative frequency of red.
(2)
(Total for Question 1 is 2 marks)
2
A weather station records one result on each of $28$ days. There are $11$ dry days and $9$ wet days. All the other days are windy. Work out the number of windy days.
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
A frequency tree describes $80$ visits to a kiosk. On $52$ visits a hot drink is chosen and on the remaining visits a cold drink is chosen. A snack is also bought on $31$ hot-drink visits and on $7$ cold-drink visits. Complete all terminal frequencies and work out the percentage of visits on which a snack is bought.
(3)
(Total for Question 1 is 3 marks)
2
A frequency tree represents $96$ sports-club members. Of the junior members, $28$ train on weekdays and $17$ train at weekends. Of the adult members, $33$ train on weekdays. Work out the three missing frequencies.
(3)
(Total for Question 2 is 3 marks)
3
A wildlife centre records $180$ animal sightings. There are $48$ morning sightings. The afternoon frequency is $24$ more than the morning frequency. All the other sightings are in the evening. Two thirds of the evening sightings are birds and the rest are mammals. Work out the missing frequencies and the relative frequency of an evening mammal sighting among all $180$ sightings.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1
Two weeks of trials are combined. In week 1, $72$ of $120$ trials succeed; $45$ successful trials and $18$ unsuccessful trials are fast. In week 2, $99$ of $180$ trials succeed; $66$ successful trials and $27$ unsuccessful trials are fast. Construct a combined frequency table for success or failure against fast or not fast. Compare the proportions that are fast in the two outcome groups.
(5)
(Total for Question 1 is 5 marks)
2
A factory checks containers made by two machines. For Machine A, $21$ containers are faulty and $259$ are not faulty. For Machine B, $30$ containers are faulty and $470$ are not faulty. Work out the relative frequency of faulty containers for each machine and compare the results.
(3)
(Total for Question 2 is 3 marks)
3
A travel survey records $250$ journeys by car, bus or walking. There are $140$ morning journeys and the rest are evening journeys. In total, $92$ journeys are by car and $101$ are by bus. In the morning, $56$ journeys are by car and the bus frequency is twice the walking frequency. Complete a two-way frequency table for time of day against method of travel.
(5)
(Total for Question 3 is 5 marks)
4
A frequency tree records whether parcels are domestic or international and whether they are delayed. There are $42$ international parcels. Of the domestic parcels, $18$ are delayed. Of the international parcels, $9$ are delayed. The relative frequency of a delayed parcel among all the parcels is $0.15$ . Work out the total number of parcels and all the missing frequencies in the tree.
(4)
(Total for Question 4 is 4 marks)
5
A workshop tests $120$ components and $78$ pass. Components for a new order will be made in full trays of $80$ . The workshop wants the expected number that pass to be at least $500$ . Using the test result, decide the minimum number of full trays the workshop should make. Show that one fewer tray is not enough.
(4)
(Total for Question 5 is 4 marks)

P2 · Apply ideas of randomness, fairness and equally likely events to calculate expected outcomes of multiple future experiments

Explanation

A random experiment has an uncertain individual result, even though its long-run behaviour can be predicted.
If outcomes are equally likely, probability is the number of favourable outcomes divided by the total number of outcomes.
For $n$ repeated trials, expected frequency is $n\times p$ .
This is a prediction, not a guarantee.
To judge whether a game is fair, compare each player's chances or calculate the expected winnings after any cost; equal-looking sectors or prizes do not by themselves prove fairness.

Worked example

A fair spinner has four equal sectors labelled A, A, B and C. Player 1 wins on A; Player 2 wins on B or C. The spinner is used $360$ times. Decide whether the game is fair and find each player's expected number of wins.

1. $P(\text{Player 1 wins})=\dfrac{2}{4}=\dfrac{1}{2}$ .
2. $P(\text{Player 2 wins})=\dfrac{2}{4}=\dfrac{1}{2}$ , so the game is fair.
3.Expected wins for each player $=360\times\dfrac{1}{2}=180$ .

Answer: The game is fair; each player is expected to win $180$ times.

Common mistakes

Don't fall into the trap of treating an expected frequency as the exact result that must occur.
Don't fall into the trap of calling a game fair without considering both probabilities and rewards.

Exam tip

Write the probability first, then multiply it by the number of future trials.

Tier 1 · Easy

1
A fair six-sided die is rolled $240$ times. Work out the expected number of rolls on which the score is greater than $4$ .
(2)
(Total for Question 1 is 2 marks)
2
A fair coin is tossed $90$ times. Work out an estimate of how many times the coin will land on tails. Give a reason why the actual number of tails may be different.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1
A fair spinner has four equal sectors labelled $1$ , $1$ , $2$ and $3$ . It is spun $600$ times. Work out the expected total of all the scores.
(3)
(Total for Question 1 is 3 marks)
2
A fair spinner has six equal sectors labelled A, A, A, B, C and D. It is spun $420$ times. Work out an estimate of how many times it lands on A and an estimate of how many times it lands on B. Hence estimate the difference between these numbers.
(3)
(Total for Question 2 is 3 marks)
3
A biased six-sided die has probability $0.08$ of landing on $6$ . The expected number of sixes in a planned experiment is $36$ . Work out the number of rolls planned and the expected number of rolls that do not land on $6$ .
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
A box contains $3$ green, $2$ yellow and $1$ red token. A token is chosen at random and replaced. A player receives $\pounds4$ for green, $\pounds1$ for yellow and nothing for red. The entry fee is $\pounds2.20$ . Work out the organiser's expected profit from $900$ plays and decide whether the game is fair to the player.
(5)
(Total for Question 1 is 5 marks)
2
A fair spinner has five equal sectors, two red and three blue. In a contest, Ruby scores one point for red on each of her three spins. Ben scores one point for blue on each of his two spins. Work out each player's expected score and decide whether the contest is fair in terms of expected score.
(4)
(Total for Question 2 is 4 marks)
3
A processing job is chosen at random from ten equally likely job cards. Five cards show a time of $2$ minutes, three cards show $4$ minutes and two cards show $x$ minutes. The expected processing time is $3.4$ minutes. Work out $x$ . A new method reduces each $x$ -minute job by $1.5$ minutes. Work out the expected total processing time for $800$ jobs after this change.
(4)
(Total for Question 3 is 4 marks)
4
A company routes each new request using a wheel with sectors of $72^\circ$ , $108^\circ$ and $180^\circ$ for teams A, B and C respectively. The company expects $2500$ requests. One worker can handle $125$ requests. Teams A, B and C are assigned $4$ , $6$ and $9$ workers respectively. Work out the expected number of requests for each team and decide whether every team has enough workers for its expected workload.
(4)
(Total for Question 4 is 4 marks)
5
A box holds $15$ equally likely job cards labelled A, B or C. The number of A cards is twice the number of B cards. Over $840$ selections with replacement, the expected number of A selections is $224$ more than the expected number of B selections. Work out how many cards have each label and the expected number of C selections.
(4)
(Total for Question 5 is 4 marks)

P3 · Relate relative expected frequencies to theoretical probability, using appropriate language and the 0-1 probability scale

Explanation

Probabilities lie on a scale from $0$ to $1$ : $0$ is impossible, $1$ is certain, and $0.5$ means equally likely to happen or not happen.
Theoretical probability comes from a model, while relative frequency comes from results: $\text{relative frequency}=\dfrac{\text{frequency}}{\text{number of trials}}$ .
In many unbiased trials, relative frequency is expected to settle near the theoretical probability, but random variation means the two values need not be exactly equal.

Worked example

A coin lands heads $156$ times in $300$ tosses. Find the relative frequency of heads and compare it with the theoretical probability.

1.Relative frequency $=\dfrac{156}{300}=0.52$ .
2.For a fair coin, the theoretical probability is $0.5$ .
3. $0.52$ is close to $0.5$ ; the difference can be due to random variation.

Answer: The relative frequency is $0.52$ , close to the theoretical value $0.5$ .

Common mistakes

Don't fall into the trap of dividing by the wrong total number of trials.
Don't fall into the trap of claiming a small experimental difference proves that the model is wrong.

Exam tip

Use probability language such as impossible, unlikely, even chance, likely and certain accurately.

Tier 1 · Easy

1
Place these events in order from least likely to most likely: event A has probability $0.72$ , event B has probability $\dfrac{1}{4}$ , and event C has probability $0.5$ .
(1)
(Total for Question 1 is 1 mark)
2
A card is chosen at random from cards labelled CAT, DOG and OWL. Write down the probability that the chosen word has three letters and the probability that it has four letters.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1
An outcome occurs $84$ times in $140$ trials. Its theoretical probability is $\dfrac{5}{8}$ . Work out the relative frequency and the expected frequency in $560$ further trials. Comment on the experimental result.
(3)
(Total for Question 1 is 3 marks)
2
An event occurs $18$ times in $80$ trials. Its theoretical probability is $0.2$ . Work out the relative frequency and choose the best description of the theoretical probability: impossible, unlikely, even chance, likely or certain.
(2)
(Total for Question 2 is 2 marks)
3
The theoretical probability of event A is $0.18$ . In an experiment, event A occurs $23$ times in $100$ trials. Event B is the complement of event A in the theoretical model. Work out the experimental relative frequency of A and the theoretical probability of B. Put these two values and $0.18$ in increasing order, then describe A and B as likely or unlikely using the theoretical model.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
A model says that a device flashes with probability $0.5$ on each trial. In a short run it flashes $41$ times in $80$ trials. After $500$ trials in total it has flashed $247$ times. Calculate both relative frequencies and decide which result gives stronger evidence about the model.
(4)
(Total for Question 1 is 4 marks)
2
A spinner is designed to land on blue with probability $0.35$ . In a test it lands on blue $52$ times in $160$ spins. Use the relative frequency to estimate the number of blues in the next $600$ spins. Compare this with the number predicted by the theoretical probability.
(4)
(Total for Question 2 is 4 marks)
3
A probability model predicts that an event will occur $144$ times in $360$ trials. In an experiment, the event occurs $93$ times in $240$ trials. Work out the theoretical probability and the experimental relative frequency. For $800$ further trials, work out the expected frequency from each value and compare the predictions.
(4)
(Total for Question 3 is 4 marks)
4
A probability model gives an event a theoretical probability of $0.4$ . The event occurs $57$ times in the first $150$ trials and $105$ times in the next $250$ trials. Work out the relative frequency for each block of trials and for all $400$ trials. Which result is closest to the theoretical probability, and which result gives the strongest overall evidence about the model?
(4)
(Total for Question 4 is 4 marks)
5
An event occurs $44$ times in the first $110$ trials. It then occurs $32$ more times. The relative frequency after all the trials is $0.38$ . Work out how many further trials were carried out. The theoretical probability of the event is $0.375$ . Compare the first and final relative frequencies with this value.
(4)
(Total for Question 5 is 4 marks)

P4 · Apply the property that the probabilities of an exhaustive set of outcomes sum to one; apply the property that the probabilities of an exhaustive set of mutually exclusive events sum to one

Explanation

An exhaustive set covers every possible outcome.
If those outcomes are mutually exclusive, exactly one occurs, so their probabilities add to $1$ .
For mutually exclusive events, $P(A\text{ or }B)=P(A)+P(B)$ .
An event and its complement are exhaustive and mutually exclusive, so $P(\text{not }A)=1-P(A)$ .
If events overlap, subtract the intersection: $P(A\text{ or }B)=P(A)+P(B)-P(A\text{ and }B)$ .

Worked example

A spinner can land on red, blue or green. $P(\text{red})=0.28$ and $P(\text{blue})=0.41$ . Find $P(\text{green})$ .

1.The three colours are exhaustive, so their probabilities total $1$ .
2. $P(\text{green})=1-0.28-0.41$ .
3. $P(\text{green})=0.31$ .

Answer: $0.31$ .

Common mistakes

Don't fall into the trap of forgetting that all exhaustive outcomes must total exactly $1$ .
Don't fall into the trap of adding probabilities for events that are not mutually exclusive.

Exam tip

Write a probability-sum equation before substituting the known values.

Tier 1 · Easy

1
The probability that a parcel arrives early is $0.37$ . Work out the probability that it does not arrive early.
(1)
(Total for Question 1 is 1 mark)
2
Events A and B are mutually exclusive. $P(A)=0.32$ and $P(B)=0.45$ . Work out $P(A\text{ or }B)$ .
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
A trial has four mutually exclusive outcomes A, B, C and D. Their probabilities are $0.28$ , $0.41$ , $x$ and $0.19$ respectively. Work out $x$ .
(2)
(Total for Question 1 is 2 marks)
2
A delivery is early, on time or late. These outcomes are mutually exclusive and exhaustive. The probability that it is early or on time is $0.74$ . The probability that it is on time is $0.31$ . Work out the probability that it is early and the probability that it is late.
(3)
(Total for Question 2 is 3 marks)
3
A spinner can land on $1$ , $2$ , $3$ or $4$ . The probabilities are $0.15$ , $0.25$ , $0.35$ and $0.25$ respectively. Event E is landing on an even number. Event F is landing on a number greater than $2$ . Work out $P(E\text{ or }F)$ . Explain why adding $P(E)$ and $P(F)$ would not give the correct answer.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
For two events A and B, $P(A)=0.58$ , $P(B)=0.47$ and $P(A\cap B)=0.21$ . Work out the probability that neither A nor B occurs.
(4)
(Total for Question 1 is 4 marks)
2
The mutually exclusive outcomes A, B and C are exhaustive. $P(A)=0.28$ and $P(B)=2P(C)$ . Work out $P(B)$ and $P(A\text{ or }C)$ .
(4)
(Total for Question 2 is 4 marks)
3
A survey of $250$ households finds that $64\%$ use a streaming service and $46\%$ use cable television. There are $35$ households that use neither. Work out the number of households that use both services, the number that use exactly one service and the probability that a randomly chosen household uses exactly one service.
(4)
(Total for Question 3 is 4 marks)
4
A machine gives exactly one result code, W, X, Y or Z, on every run. $P(\text{W or X})=0.45$ , $P(X)=2P(W)$ and $P(\text{X or Y})=0.60$ . Work out the probability of each result code.
(4)
(Total for Question 4 is 4 marks)
5
Events A and B can occur together. $P(B)=0.50$ , the probability that neither event occurs is $0.30$ , and $P(A)=3P(A\cap B)$ . Work out the probability that A or B, but not both, occurs.
(4)
(Total for Question 5 is 4 marks)

P5 · Understand that empirical unbiased samples tend towards theoretical probability distributions, with increasing sample size

Explanation

An empirical distribution is based on observed relative frequencies; a theoretical distribution is predicted by a probability model.
With unbiased, independent trials, a larger sample usually gives more stable relative frequencies that are closer to the theoretical probabilities.
This is a long-run tendency, not a promise that every larger sample will be closer.
Increasing sample size reduces random variation, but it does not remove bias caused by an unfair device or poor sampling method.

Worked example

A fair coin gives $162$ heads in $300$ tosses and $1038$ heads in $2000$ tosses. Compare the two relative frequencies with $0.5$ .

1.For $300$ tosses: $\dfrac{162}{300}=0.54$ .
2.For $2000$ tosses: $\dfrac{1038}{2000}=0.519$ .
3. $0.519$ is closer to the theoretical probability $0.5$ than $0.54$ .

Answer: The larger sample gives the closer relative frequency in this experiment.

Common mistakes

Don't fall into the trap of saying that a larger sample guarantees an exact theoretical distribution.
Don't fall into the trap of assuming more trials can correct a biased experiment.

Exam tip

When comparing experiments, calculate relative frequencies rather than comparing raw counts.

Tier 1 · Easy

1
A fair coin gives a relative frequency of heads of $0.64$ after $25$ tosses. Explain what is likely to happen to the relative frequency as many more unbiased tosses are made.
(1)
(Total for Question 1 is 1 mark)
2
Kai tosses an unbiased coin many times and recalculates the relative frequency of heads after every $10$ tosses. Give a reason why the first few relative frequencies are likely to fluctuate more than the later ones.
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
A spinner is designed to land on green with probability $0.3$ . It lands on green $8$ times in $20$ spins and $63$ times in $200$ spins. Compare the two empirical probabilities with the theoretical probability.
(3)
(Total for Question 1 is 3 marks)
2
A spinner has theoretical probabilities $0.2$ , $0.3$ and $0.5$ for red, yellow and blue. In $50$ spins the frequencies are $8$ , $19$ and $23$ . In $500$ spins the frequencies are $102$ , $146$ and $252$ . Work out the two empirical distributions and decide which is closer to the theoretical distribution.
(4)
(Total for Question 2 is 4 marks)
3
A fair spinner has probability $0.25$ of landing on blue. It lands on blue $5$ times in its first $20$ spins and $56$ times after $200$ spins in total. A student says a larger sample must always give a relative frequency closer to the theoretical probability. Work out both relative frequencies and decide whether these results support the student's claim. Give a reason for your answer.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
A fair die is tested. In the first $60$ rolls, a six appears $16$ times. After $600$ rolls in total, a six has appeared $108$ times. A student says the die must be biased because neither relative frequency equals $\dfrac{1}{6}$ . Evaluate the student's claim.
(4)
(Total for Question 1 is 4 marks)
2
A council wants to estimate the proportion of all teenagers who play chess online. It first asks $30$ teenage chess-club members and $18$ say yes. It later asks $900$ teenage chess-club members at several tournaments and $540$ say yes. Work out both empirical probabilities. Is the second result necessarily a reliable estimate for all teenagers because its sample is larger? Give a reason for your answer.
(4)
(Total for Question 2 is 4 marks)
3
A wheel has four sectors with theoretical probabilities $0.15$ , $0.25$ , $0.30$ and $0.30$ for red, blue, green and white. In $40$ spins the frequencies are $9$ , $8$ , $13$ and $10$ . In $400$ spins the frequencies are $58$ , $104$ , $126$ and $112$ . Work out both empirical distributions. For the $40$ -spin distribution, write down the colour whose empirical probability differs most from its theoretical probability. State, with a reason, which of the two distributions you expect to be closer to the theoretical distribution.
(5)
(Total for Question 3 is 5 marks)
4
A spinner lands on red $34$ times in its first $160$ spins. It is then spun a further $80$ times. After all $240$ spins, the relative frequency of red is exactly $0.25$ . A maintenance warning is issued if the relative frequency of red during the further $80$ spins is greater than $0.30$ . Decide whether a warning should be issued.
(4)
(Total for Question 4 is 4 marks)
5
Two wheels are designed to land on red with probability $0.25$ . Wheel A lands on red $17$ times in its first $60$ spins and $156$ times after $600$ spins in total. Wheel B lands on red $16$ times in its first $60$ spins and $204$ times after $600$ spins in total. Work out the four relative frequencies. Which wheel gives stronger evidence that it may be biased? Explain your answer.
(5)
(Total for Question 5 is 5 marks)

P6 · Enumerate sets and combinations of sets systematically, using tables, grids, Venn diagrams and tree diagrams

Explanation

Systematic enumeration means listing every permitted outcome once in a clear order.
Use a table or grid when two quantities vary, a tree for successive choices, and a Venn diagram for overlapping sets.
For a two-set Venn diagram, fill the intersection first, then the regions belonging only to each set, then the outside region.
A list of possible outcomes does not automatically make them equally likely; probability still depends on how the experiment works.

Worked example

List the numbers from $1$ to $20$ that are multiples of $3$ or $5$ by separating the intersection and the two 'only' regions.

1.Both: $\{15\}$ .
2.Multiples of $3$ only: $\{3,6,9,12,18\}$ .
3.Multiples of $5$ only: $\{5,10,20\}$ .

Answer: $\{3,5,6,9,10,12,15,18,20\}$ .

Common mistakes

Don't fall into the trap of missing an outcome or listing the same outcome twice.
Don't fall into the trap of placing intersection values in both 'only' regions as well.

Exam tip

Choose a fixed order and count your final outcomes as a check.

Tier 1 · Easy

1
A uniform is made from one of two shirts, blue or white, and one of three ties, red, silver or green. List all possible shirt-and-tie combinations.
(2)
(Total for Question 1 is 2 marks)
2
Write down all the ordered pairs of positive integers that have a total of $7$ and have the first number less than the second number.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1
The universal set is the integers from $1$ to $20$ . Set A contains the multiples of $3$ and set B contains the factors of $18$ . Enumerate the four regions of a Venn diagram for A and B.
(4)
(Total for Question 1 is 4 marks)
2
A journey uses either a bus or a train, followed by walking, cycling or taking a taxi. A bus journey cannot be followed by a taxi. List all the possible journeys and write down how many there are.
(3)
(Total for Question 2 is 3 marks)
3
The first coordinate of an ordered pair is chosen from $\{-2,1,4\}$ and the second coordinate is chosen from $\{-1,3\}$ . List all the possible ordered pairs. Then list the pairs for which the sum of the coordinates is greater than $2$ .
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
A three-digit number is formed from three different digits chosen from $1$ , $2$ , $3$ and $4$ . Enumerate all the numbers that are greater than $230$ and even. How many of these numbers do not contain the digit $1$ ?
(4)
(Total for Question 1 is 4 marks)
2
The universal set is the integers from $1$ to $30$ . Set A contains multiples of $2$ , set B contains square numbers, and set C contains factors of $24$ . List the integers in each region that is in exactly two of the three sets and work out how many integers are in exactly two sets.
(4)
(Total for Question 2 is 4 marks)
3
A ticket code consists of one letter, A, B or C, followed by two different digits chosen from $1$ , $2$ , $3$ and $4$ . For an A-code, the final digit must be even. For a B-code, the two digits must have a total of $5$ . For a C-code, the first digit must be less than the second digit. List every permitted code and work out the total number of codes.
(4)
(Total for Question 3 is 4 marks)
4
A robot makes exactly five moves. Each move is either one square east, E, or one square north, N. It must make three east moves and two north moves, but it must not make two north moves consecutively. List every permitted route and work out how many there are.
(3)
(Total for Question 4 is 3 marks)
5
A club chooses a captain and a deputy from students A, B, C, D and E. The same student cannot hold both roles. Student C cannot be captain, and students A and B cannot both hold a role. List every possible ordered captain-deputy pair and work out the total number of choices.
(5)
(Total for Question 5 is 5 marks)

P7 · Construct theoretical possibility spaces for single and combined experiments with equally likely outcomes and use these to calculate theoretical probabilities

Explanation

A possibility space lists every permitted outcome once, often as ordered pairs in a grid.
If two experiments are independent, with $m$ and $n$ equally likely outcomes, the complete space has $mn$ equally likely ordered pairs.
Probability is then favourable pairs divided by all pairs.
Order matters when there is a first and second result: $(a,b)$ and $(b,a)$ are different.
Do not use favourable outcomes over total outcomes unless the combined outcomes are equally likely.

Worked example

Two fair six-sided dice are rolled. Use the possibility space to find the probability that the total is $9$ .

1.There are $6\times6=36$ equally likely ordered pairs.
2.The favourable pairs are $(3,6),(4,5),(5,4),(6,3)$ .
3.Probability $=\dfrac{4}{36}=\dfrac{1}{9}$ .

Answer: $\dfrac{1}{9}$ .

Common mistakes

Don't fall into the trap of counting $(3,6)$ and $(6,3)$ as one outcome.
Don't fall into the trap of using favourable outcomes over an incomplete possibility space.

Exam tip

Label both axes of a grid and check that it contains the expected $mn$ outcomes.

Tier 1 · Easy

1
A fair coin is tossed and a fair spinner labelled $1$ , $2$ , $3$ is spun. Write the six outcomes as ordered pairs and find the probability of getting a head and an even number.
(2)
(Total for Question 1 is 2 marks)
2
A fair five-sided spinner is labelled $2$ , $3$ , $5$ , $8$ and $10$ . Write down its possibility space and work out the probability that it lands on a prime number.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1
Spinner A has equal sectors labelled $1$ , $2$ , $3$ . Spinner B has equal sectors labelled $1$ , $2$ , $3$ , $4$ . Construct a possibility space and find the probability that the two scores have a sum greater than $5$ .
(3)
(Total for Question 1 is 3 marks)
2
Fair spinner A is labelled $1$ , $4$ and $7$ . Fair spinner B is labelled $2$ and $5$ . Write down the possibility space of the total score and work out the probability that the total is even.
(3)
(Total for Question 2 is 3 marks)
3
A fair coin is tossed and a fair spinner labelled $2$ , $4$ and $6$ is spun. For a head, the score is one more than the spinner number. For a tail, the score is one less than the spinner number. Construct the possibility space for the score and work out the probability that the score is $3$ or $5$ .
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
Two different cards are chosen in order from cards labelled $1$ , $2$ , $3$ , $4$ and $6$ . The first card is the tens digit and the second is the units digit. Construct the theoretical possibility space and find the probability that the two-digit number is divisible by $3$ .
(4)
(Total for Question 1 is 4 marks)
2
Fair spinner A is labelled $1$ , $2$ and $5$ . Fair spinner B is labelled $2$ , $3$ , $4$ and $6$ . The two scores are multiplied together. Write down the possibility space and work out the probability that the result is a multiple of $4$ .
(4)
(Total for Question 2 is 4 marks)
3
A numerator is chosen at random from $\{2,5,8\}$ and a denominator is chosen independently at random from $\{3,6,10,12\}$ . Each choice in a set is equally likely. Construct the possibility space of fractions. Work out the probability that the fraction is greater than $\dfrac{1}{2}$ and the probability that its simplest form has denominator $3$ . Which event is more likely?
(5)
(Total for Question 3 is 5 marks)
4
A fair six-sided die is rolled once. If the die score is $n$ , a game awards $n(8-n)$ points. Construct the theoretical possibility space for the points awarded. Work out the probability that the points awarded are a multiple of $3$ . Explain why the four different possible point scores are not equally likely.
(4)
(Total for Question 4 is 4 marks)
5
A fair coin is tossed. After a head, a fair two-sided spinner labelled $2$ and $4$ is spun. After a tail, a fair three-sided spinner labelled $1$ , $3$ and $5$ is spun. Draw a possibility tree and work out the probability that the spinner score is greater than $2$ . Explain why counting three favourable scores out of the five listed outcomes would not give the correct probability.
(5)
(Total for Question 5 is 5 marks)

P8 · Calculate the probability of independent and dependent combined events, including using tree diagrams and other representations, and know the underlying assumptions

Explanation

On a probability tree, multiply probabilities along one route and add the products of alternative, mutually exclusive routes.
Independent events leave later probabilities unchanged.
Dependent events change later probabilities because an earlier result changes what remains.
Without replacement, reduce the total and update the relevant outcome count after each draw; with replacement, the original probabilities repeat.
Always check whether the context justifies equal or repeated branch probabilities.

Worked example

A bag contains $3$ red and $2$ blue counters. Two counters are taken without replacement. Find the probability that they are different colours.

1. $P(RB)=\dfrac{3}{5}\times\dfrac{2}{4}=\dfrac{3}{10}$ .
2. $P(BR)=\dfrac{2}{5}\times\dfrac{3}{4}=\dfrac{3}{10}$ .
3.Add the two routes: $\dfrac{3}{10}+\dfrac{3}{10}=\dfrac{3}{5}$ .

Answer: $\dfrac{3}{5}$ .

Common mistakes

Don't fall into the trap of keeping the denominator unchanged when there is no replacement.
Don't fall into the trap of adding probabilities down one route instead of multiplying them.

Exam tip

Annotate every second-stage branch before calculating; this exposes replacement errors.

Tier 1 · Easy

1
A fair coin is tossed and an independent spinner lands on red with probability $\dfrac{3}{5}$ . Work out the probability of a head and red.
(2)
(Total for Question 1 is 2 marks)
2
A token is chosen from a bag, replaced and then a second token is chosen. The probability of choosing a purple token each time is $0.3$ . Work out the probability that the first token is purple and the second is not purple.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1
A bag contains $4$ blue counters and $3$ amber counters. Two counters are chosen at random without replacement. Work out the probability that both counters are blue.
(3)
(Total for Question 1 is 3 marks)
2
A box contains $6$ winning tickets and $4$ losing tickets. Two tickets are chosen at random without replacement. Work out the probability that exactly one ticket is a winning ticket.
(3)
(Total for Question 2 is 3 marks)
3
Two independent sensors test the same item. Sensor A detects a fault with probability $0.7$ and sensor B detects a fault with probability $0.4$ . Work out the probability that exactly one sensor detects a fault.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1
A bag contains $7$ green and $4$ yellow counters. Four counters are chosen at random without replacement. Work out the probability that exactly three are green. State why the branch probabilities change after each choice.
(5)
(Total for Question 1 is 5 marks)
2
A player takes two penalties. The probability that the player scores the first penalty is $\dfrac{2}{3}$ . If the first penalty is scored, the probability of scoring the second is $\dfrac{3}{4}$ . If the first penalty is missed, the probability of scoring the second is $\dfrac{1}{2}$ . Work out the probability that the results of the two penalties are the same.
(3)
(Total for Question 2 is 3 marks)
3
A player takes three shots. The probability of scoring the first shot is $0.6$ . After a score, the probability of scoring the next shot is $0.7$ . After a miss, the probability of scoring the next shot is $0.4$ . Work out the probability that the player scores exactly two of the three shots.
(4)
(Total for Question 3 is 4 marks)
4
Three independent alarms test the same fault. The probabilities that alarms A, B and C activate are $0.8$ , $0.6$ and $0.5$ respectively. Work out the probability that at least two alarms activate. State the assumption represented by using unchanged probabilities on every branch.
(5)
(Total for Question 4 is 5 marks)
5
Box A contains $3$ red and $2$ blue counters. Box B contains $2$ red and $4$ blue counters. One counter is chosen at random from box A and transferred to box B. A counter is then chosen at random from box B. Work out the probability that one counter in this two-stage process is red and the other is blue.
(4)
(Total for Question 5 is 4 marks)

P9 · Calculate and interpret conditional probabilities through representation using expected frequencies with two-way tables, tree diagrams and Venn diagrams [Higher only]

Explanation

Higher tier only.
A conditional probability restricts the sample space to outcomes satisfying the stated condition.
In $P(A\mid B)$ , read 'the probability of $A$ given $B$ ': the denominator is the total frequency in $B$ , and the numerator is the frequency in both $A$ and $B$ .
Two-way tables, Venn diagrams and expected-frequency trees can all supply these restricted counts.
The overall total is not the denominator unless it is also the conditioning group.

Worked example

A Venn diagram shows $12$ people in A only, $18$ in both A and B, $27$ in B only and $9$ outside both sets. Find $P(A\mid B)$ .

1.The condition restricts the sample space to set B.
2.There are $18+27=45$ people in B.
3. $18$ of these are also in A, so $P(A\mid B)=\dfrac{18}{45}=\dfrac{2}{5}$ .

Answer: $\dfrac{2}{5}$ .

Common mistakes

Don't fall into the trap of dividing by the overall total instead of the conditioning-group total.
Don't fall into the trap of reversing $P(A\mid B)$ and $P(B\mid A)$ .

Exam tip

Circle the words after 'given that' before choosing the denominator.

Tier 1 · Easy

1
A two-way table records $60$ students. Of the $24$ who travel by bus, $9$ arrive late. Find the probability that a randomly chosen bus traveller arrives late.
(2)
(Total for Question 1 is 2 marks)
2
$P(A\cap B)=0.14$ and $P(B)=0.35$ . Use expected frequencies out of $100$ . Given that B occurs, work out the probability that A occurs.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1
Of $500$ expected customers, $60\%$ are members. Of the members, $70\%$ order online. Of the non-members, $35\%$ order online. Use expected frequencies to find the probability that an online customer is a member.
(4)
(Total for Question 1 is 4 marks)
2
A library records $180$ loans. Of these, $105$ are fiction, $72$ are audiobooks and $48$ are fiction audiobooks. Given that a loan is not fiction, work out the probability that it is an audiobook.
(3)
(Total for Question 2 is 3 marks)
3
An app has $160$ subscribers. Of these, $96$ pay monthly and the rest pay annually. In total, $72$ subscribers renew, including $54$ monthly subscribers. Complete a two-way table for payment plan against whether a subscriber renews. Given that a subscriber renews, work out the probability that the subscriber pays monthly. Given that a subscriber pays annually, work out the probability that the subscriber does not renew.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1
In an expected-frequency Venn diagram for $240$ people, $138$ are in set A, $102$ are in set B and $54$ are in both sets. A person is chosen from those who are in exactly one of the sets. Find the conditional probability that the person is in A.
(4)
(Total for Question 1 is 4 marks)
2
A bag contains $7$ red counters and $3$ blue counters. Two counters are chosen at random without replacement. Given that at least one counter is red, work out the probability that both counters are red. You must show all your working.
(4)
(Total for Question 2 is 4 marks)
3
A hospital receives $600$ blood samples. Laboratory A tests $40\%$ of them and laboratory B tests the rest. Of the samples tested by laboratory A, $15\%$ need a repeat test. In total, $60$ samples need a repeat test. Given that a sample does not need a repeat test, work out the probability that it was tested by laboratory B. Also work out the probability that a sample was tested by laboratory A given that it needs a repeat test, and compare this with the overall probability that a sample was tested by laboratory A.
(5)
(Total for Question 3 is 5 marks)
4
A ticket is made by choosing a letter from A, B, C and D, followed by a digit selected from the list $1$ , $2$ , $3$ , $4$ . All $16$ ordered pairs are equally likely. Give the letters the values A $=1$ , B $=2$ , C $=3$ and D $=4$ . A ticket is valid when its letter value plus its digit is at least $6$ . Given that a ticket is valid, work out the probability that its digit is even.
(4)
(Total for Question 4 is 4 marks)
5
A sorting system sends $30\%$ of items to line X and the rest to line Y. Of the items sent to line X, $12\%$ need a manual check. Of all items that need a manual check, $45\%$ were sent to line X. Use expected frequencies for $1000$ items to work out the probability that an item was sent to line Y, given that it does not need a manual check.
(5)
(Total for Question 5 is 5 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

P1 · Record, describe and analyse the frequency of outcomes of probability experiments using tables and frequency trees

Tier 1 · Easy

Mark scheme for P1 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	Red: $18$ ; blue: $12$ . Relative frequency of red $=\dfrac{18}{30}=0.6$ .	2	Record the observed counts in the table and check that $18+12=30$ . Divide the red frequency by the total number of spins: $18\div30=0.6$ .
2	$8$ windy days.	1	The three frequencies must total $28$ . The windy frequency is $28-11-9=8$ . Check that $11+9+8=28$ .

Tier 2 · Standard

Mark scheme for P1 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	Cold drink: $28$ ; hot drink and no snack: $21$ ; cold drink and no snack: $21$ . Snack percentage $=47.5\%$ .	3	The cold-drink branch has frequency $80-52=28$ . The no-snack frequencies are $52-31=21$ and $28-7=21$ . Snacks are bought on $31+7=38$ visits, so the percentage is $\dfrac{38}{80}\times100=47.5\%$ .
2	Junior: $45$ ; adult: $51$ ; adult and weekend: $18$ .	3	The junior frequency is $28+17=45$ . The adult frequency is $96-45=51$ . The adult weekend branch is $51-33=18$ .
3	Afternoon: $72$ ; evening: $60$ ; evening birds: $40$ ; evening mammals: $20$ . Relative frequency of an evening mammal sighting $=\dfrac{20}{180}=\dfrac{1}{9}$ .	4	The afternoon frequency is $48+24=72$ , so the evening frequency is $180-48-72=60$ . Two thirds of $60$ is $40$ , leaving $20$ evening mammal sightings. Divide this frequency by all $180$ sightings to get $20/180=1/9$ .

Tier 3 · Hard

Mark scheme for P1 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	Success: $111$ fast and $60$ not fast; failure: $45$ fast and $84$ not fast. Proportion of successful trials that are fast $=\dfrac{111}{171}\approx64.9\%$ ; proportion of failed trials that are fast $=\dfrac{45}{129}\approx34.9\%$ . A successful trial is more likely to be fast.	5	Combine like branches: successful trials total $72+99=171$ , of which $45+66=111$ are fast, so $60$ are not fast. Failures total $(120-72)+(180-99)=129$ , of which $18+27=45$ are fast, so $84$ are not fast. Compare within each outcome group: $111/171\approx0.649$ and $45/129\approx0.349$ .
2	Machine A: $0.075$ ; Machine B: $0.06$ . Machine A has the greater relative frequency of faulty containers, by $0.015$ (or $1.5$ percentage points).	3	For Machine A, $21\div280=0.075$ . For Machine B, $30\div500=0.06$ . Since $0.075>0.06$ , Machine A has the greater relative frequency; the difference is $0.075-0.06=0.015$ .
3	Morning: car $56$ , bus $56$ , walking $28$ , total $140$ . Evening: car $36$ , bus $45$ , walking $29$ , total $110$ . Totals: car $92$ , bus $101$ , walking $57$ , all journeys $250$ .	5	Let the morning walking frequency be $x$ , so the morning bus frequency is $2x$ . Then $56+2x+x=140$ , giving $x=28$ and a morning bus frequency of $56$ . Subtract the morning entries from the column totals: evening car is $92-56=36$ and evening bus is $101-56=45$ . The evening total is $250-140=110$ , so evening walking is $110-36-45=29$ .
4	Total parcels: $180$ ; domestic parcels: $138$ . Domestic and not delayed: $120$ ; international and not delayed: $33$ .	4	There are $18+9=27$ delayed parcels. If $N$ is the total number, then $27/N=0.15$ , so $N=180$ . There are $180-42=138$ domestic parcels. The remaining terminal frequencies are $138-18=120$ and $42-9=33$ .
5	$10$ full trays (or $800$ components). $9$ trays give an expected $468$ passes, whereas $10$ trays give an expected $520$ passes.	4	The test gives a pass relative frequency of $78/120=13/20$ . One tray therefore gives an expected $80\times13/20=52$ passes. Nine trays give $9\times52=468$ , which is below $500$ , while ten trays give $10\times52=520$ . Therefore the minimum is $10$ full trays.

P2 · Apply ideas of randomness, fairness and equally likely events to calculate expected outcomes of multiple future experiments

Tier 1 · Easy

Mark scheme for P2 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$80$ rolls	2	The favourable scores are $5$ and $6$ , so the probability is $2/6=1/3$ . The expected frequency is $240\times\dfrac{1}{3}=80$ .
2	$45$ tails. An expected frequency is a prediction, so random variation may give a different result.	2	For a fair coin, $P(\text{tail})=\dfrac{1}{2}$ . The expected frequency is $90\times\dfrac{1}{2}=45$ , but individual results are random.

Tier 2 · Standard

Mark scheme for P2 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$1050$	3	The expected score per spin is $\dfrac{1+1+2+3}{4}=\dfrac{7}{4}$ . Over $600$ spins, the expected total is $600\times\dfrac{7}{4}=1050$ .
2	A: $210$ times; B: $70$ times; difference: $140$ .	3	The expected frequency of A is $420\times\dfrac{3}{6}=210$ . The expected frequency of B is $420\times\dfrac{1}{6}=70$ . The expected difference is $210-70=140$ .
3	$450$ rolls are planned; the expected number of non-sixes is $414$ .	3	If $n$ rolls are planned, $0.08n=36$ , so $n=36\div0.08=450$ . The probability of a non-six is $0.92$ , giving an expected frequency of $450\times0.92=414$ ; equivalently, subtract the $36$ expected sixes from $450$ .

Tier 3 · Hard

Mark scheme for P2 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	Expected payout per play $=\pounds\dfrac{7}{3}$ . Expected organiser profit from $900$ plays $=-\pounds120$ . The game is not fair; the player has an expected gain of about $13.3$ pence per play.	5	The expected payout is $4\times\dfrac{3}{6}+1\times\dfrac{2}{6}=2+\dfrac{1}{3}=\dfrac{7}{3}$ pounds. The organiser receives $\pounds2.20=\pounds\dfrac{11}{5}$ per play, so expected profit per play is $\dfrac{11}{5}-\dfrac{7}{3}=-\dfrac{2}{15}$ pounds. Over $900$ plays this is $900\times(-2/15)=-120$ pounds. A fair entry fee would equal the expected payout, so this game favours the player.
2	Ruby's expected score is $\dfrac{6}{5}$ and Ben's expected score is $\dfrac{6}{5}$ . The contest is fair in terms of expected score because their expected scores are equal.	4	Ruby scores on a spin with probability $2/5$ , so her expected score is $3\times\dfrac{2}{5}=\dfrac{6}{5}$ . Ben scores with probability $3/5$ , so his expected score is $2\times\dfrac{3}{5}=\dfrac{6}{5}$ . Equal expected scores make the contest fair in terms of expected score.
3	$x=6$ minutes; expected total processing time $=2480$ minutes.	4	Use the expected time: $\dfrac{5(2)+3(4)+2x}{10}=3.4$ . Hence $22+2x=34$ and $x=6$ . The reduced long-job time is $4.5$ minutes, so the new expected time is $\dfrac{10+12+2(4.5)}{10}=3.1$ minutes. For $800$ jobs, the expected total is $800\times3.1=2480$ minutes.
4	Expected requests: A $500$ , B $750$ , C $1250$ . No. Teams A and B have exactly enough capacity, but team C can handle only $1125$ requests, an expected shortfall of $125$ .	4	Use each sector angle as a fraction of $360^\circ$ . The expected numbers are $2500(72/360)=500$ , $2500(108/360)=750$ and $2500(180/360)=1250$ . The assigned capacities are $4(125)=500$ , $6(125)=750$ and $9(125)=1125$ , so only team C is short.
5	A cards: $8$ ; B cards: $4$ ; C cards: $3$ . Expected C selections: $168$ .	4	Let the number of B cards be $b$ , so there are $2b$ A cards. The difference in their expected frequencies is $840(2b/15-b/15)=56b$ . Hence $56b=224$ , giving $b=4$ . There are $8$ A cards and $15-8-4=3$ C cards, so the expected C frequency is $840(3/15)=168$ .

P3 · Relate relative expected frequencies to theoretical probability, using appropriate language and the 0-1 probability scale

Tier 1 · Easy

Mark scheme for P3 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	B, C, A	1	Write every probability as a decimal: $\dfrac{1}{4}=0.25$ , then compare $0.25<0.5<0.72$ .
2	Three letters: $1$ ; four letters: $0$ .	2	Every word has three letters, so that event is certain and has probability $1$ . No word has four letters, so that event is impossible and has probability $0$ .

Tier 2 · Standard

Mark scheme for P3 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	Relative frequency $=0.6$ . Expected frequency $=350$ . $0.6$ is reasonably close to the theoretical probability $0.625$ .	3	The relative frequency is $84/140=0.6$ . The theoretical probability is $5/8=0.625$ , so the expected frequency is $560\times5/8=350$ . The observed value differs by only $0.025$ , which is plausible experimental variation.
2	Relative frequency $=0.225$ ; the event is unlikely.	2	The relative frequency is $18\div80=0.225$ . A probability of $0.2$ lies between $0$ and $0.5$ , so 'unlikely' is the best description.
3	Experimental relative frequency of A $=0.23$ ; theoretical $P(B)=0.82$ . $0.18<0.23<0.82$ ; A is unlikely and B is likely in the theoretical model.	3	The experimental relative frequency is $23/100=0.23$ . Since B is the complement of A, its theoretical probability is $1-0.18=0.82$ . Order the decimals directly. A theoretical probability below $0.5$ is described as unlikely, while one above $0.5$ is described as likely.

Tier 3 · Hard

Mark scheme for P3 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	Short-run relative frequency $=0.5125$ . Overall relative frequency $=0.494$ . The $500$ -trial result gives stronger evidence and is very close to $0.5$ , so it supports the model.	4	For the short run, calculate $41/80=0.5125$ . For all trials, calculate $247/500=0.494$ . Both are close to $0.5$ , but the larger sample is less affected by random variation, so the overall relative frequency is the stronger comparison.
2	Relative frequency $=0.325$ ; estimate from the test $=195$ blues. The theoretical prediction is $210$ blues, which is $15$ more than the test-based estimate.	4	The observed relative frequency is $52/160=0.325$ . This gives $600\times0.325=195$ blues. The theoretical model gives $600\times0.35=210$ , and $210-195=15$ .
3	Theoretical probability $=0.4$ ; experimental relative frequency $=0.3875$ . Model prediction: $320$ ; experiment-based prediction: $310$ ; the model predicts $10$ more occurrences.	4	The model probability is $144/360=0.4$ . The experimental relative frequency is $93/240=0.3875$ . Multiplying each by $800$ gives $320$ from the model and $310$ from the experiment, a difference of $10$ .
4	First block: $0.38$ ; second block: $0.42$ ; all $400$ trials: $0.405$ . The combined result is closest to $0.4$ and gives the strongest overall evidence because it uses all $400$ trials.	4	Calculate $57/150=0.38$ and $105/250=0.42$ . Across both blocks the event occurs $162$ times, so the combined relative frequency is $162/400=0.405$ . Its distance from $0.4$ is $0.005$ , compared with $0.02$ for either block, and it uses the largest sample.
5	$90$ further trials were carried out. First relative frequency: $0.4$ ; final relative frequency: $0.38$ . The final value is closer to $0.375$ .	4	The event occurs $44+32=76$ times altogether. Since $76/N=0.38$ , the total number of trials is $N=200$ , so there were $200-110=90$ further trials. The first relative frequency is $44/110=0.4$ , which is $0.025$ from $0.375$ ; the final value is only $0.005$ away.

P4 · Apply the property that the probabilities of an exhaustive set of outcomes sum to one; apply the property that the probabilities of an exhaustive set of mutually exclusive events sum to one

Tier 1 · Easy

Mark scheme for P4 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$0.63$	1	Early and not early are exhaustive and mutually exclusive, so use the complement: $1-0.37=0.63$ .
2	$0.77$	1	Mutually exclusive events cannot happen together, so add their probabilities: $0.32+0.45=0.77$ .

Tier 2 · Standard

Mark scheme for P4 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$x=0.12$	2	The outcomes are exhaustive, so their probabilities sum to $1$ . Therefore $x=1-(0.28+0.41+0.19)=1-0.88=0.12$ .
2	$P(\text{early})=0.43$ and $P(\text{late})=0.26$ .	3	As early and on time are mutually exclusive, $P(\text{early})=0.74-0.31=0.43$ . The probability of being late is the complement of being early or on time, so $P(\text{late})=1-0.74=0.26$ .
3	$P(E\text{ or }F)=0.25+0.35+0.25=0.85$ . Landing on $4$ is in both E and F, so adding $P(E)$ and $P(F)$ would count its probability twice.	3	The outcomes in E or F are $2$ , $3$ and $4$ , so add their probabilities once: $0.25+0.35+0.25=0.85$ . Event E contains $2$ and $4$ , while F contains $3$ and $4$ ; the shared outcome $4$ causes double-counting if the two event probabilities are simply added.

Tier 3 · Hard

Mark scheme for P4 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$0.16$	4	Split the possibility space into mutually exclusive regions. The probability of $A\cup B$ is $0.58+0.47-0.21=0.84$ , because the intersection was counted twice. Neither event is the complement of the union, so its probability is $1-0.84=0.16$ .
2	$P(B)=0.48$ and $P(A\text{ or }C)=0.52$ .	4	Let $P(C)=x$ , so $P(B)=2x$ . Since the outcomes are exhaustive, $0.28+2x+x=1$ . Hence $3x=0.72$ and $x=0.24$ , giving $P(B)=0.48$ . As A and C are mutually exclusive, $P(A\text{ or }C)=0.28+0.24=0.52$ .
3	$60$ households use both services; $155$ use exactly one service. Probability of exactly one service $=\dfrac{155}{250}=\dfrac{31}{50}$ .	4	There are $160$ streaming households and $115$ cable households. Since $35$ use neither, $250-35=215$ use at least one. The intersection is $160+115-215=60$ . Exactly one service is used by $215-60=155$ households, so the probability is $155/250=31/50$ .
4	$P(W)=0.15$ , $P(X)=0.30$ , $P(Y)=0.30$ and $P(Z)=0.25$ .	4	Let $P(W)=w$ . Since W and X cannot occur together, $w+2w=0.45$ , so $w=0.15$ and $P(X)=0.30$ . Then $P(Y)=0.60-0.30=0.30$ . The four codes cover every result, so $P(Z)=1-0.15-0.30-0.30=0.25$ .
5	$0.60$	4	Let $P(A\cap B)=x$ , so $P(A)=3x$ . Since neither occurs with probability $0.30$ , $P(A\cup B)=0.70$ . Therefore $3x+0.50-x=0.70$ , giving $x=0.10$ and $P(A)=0.30$ . The A-only probability is $0.30-0.10=0.20$ and the B-only probability is $0.50-0.10=0.40$ , so exactly one occurs with probability $0.60$ .

P5 · Understand that empirical unbiased samples tend towards theoretical probability distributions, with increasing sample size

Tier 1 · Easy

Mark scheme for P5 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	It is likely to move closer to the theoretical probability $0.5$ , although it need not do so after every extra toss.	1	Use the long-run tendency: for unbiased coin tosses the empirical proportion tends towards the theoretical probability $1/2$ as the sample becomes large.
2	The early relative frequencies use small samples, so each new result has a larger effect. With more tosses, the relative frequency tends to become more stable and approach $0.5$ .	1	For a small total, one extra head changes the fraction by a relatively large amount. As the denominator grows, one result changes the fraction less, so random variation has less effect.

Tier 2 · Standard

Mark scheme for P5 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	For $20$ spins, the relative frequency is $0.4$ . For $200$ spins, the relative frequency is $0.315$ . The larger sample gives the value closer to $0.3$ .	3	Calculate $8/20=0.4$ and $63/200=0.315$ . Their distances from $0.3$ are $0.1$ and $0.015$ , so the empirical result from $200$ spins is closer to the theoretical probability.
2	$50$ spins: red $0.16$ , yellow $0.38$ , blue $0.46$ . $500$ spins: red $0.204$ , yellow $0.292$ , blue $0.504$ . The $500$ -spin distribution is closer to the theoretical distribution.	4	Divide each frequency by its experiment total. For $50$ spins this gives $8/50=0.16$ , $19/50=0.38$ and $23/50=0.46$ . For $500$ spins it gives $102/500=0.204$ , $146/500=0.292$ and $252/500=0.504$ . Every value in the larger sample is closer to its theoretical probability.
3	No, the results do not support the claim. The relative frequencies are $0.25$ and $0.28$ . The larger-sample value is farther from $0.25$ . A larger unbiased sample tends to be more stable, but it is not guaranteed to be closer every time.	3	Calculate $5/20=0.25$ and $56/200=0.28$ . The first value matches the theoretical probability, whereas the later value differs by $0.03$ . Long-run convergence is a tendency rather than a guarantee for every growing sample.

Tier 3 · Hard

Mark scheme for P5 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	The relative frequencies are $\dfrac{16}{60}\approx0.267$ and $\dfrac{108}{600}=0.18$ . The theoretical probability is $\dfrac{1}{6}\approx0.167$ . The larger-sample result is closer to the theoretical value, and an exact match is not expected, so the evidence does not establish bias.	4	Compare each empirical probability with $1/6$ . The differences are about $0.100$ for the first $60$ rolls and $0.013$ after $600$ rolls. Random variation explains a non-zero difference, and the movement towards $1/6$ as the sample grows is consistent with a fair die.
2	No. Both empirical probabilities are $0.6$ . The second result is more stable for chess-club members, but the sample is biased towards teenagers interested in chess. A larger biased sample need not estimate the proportion for all teenagers reliably.	4	The first empirical probability is $18/30=0.6$ and the second is $540/900=0.6$ . The larger result has less random variation for the sampled group, but every person sampled is a chess-club member, so increasing the sample size does not remove this selection bias.
3	$40$ spins: $(0.225,0.20,0.325,0.25)$ ; $400$ spins: $(0.145,0.26,0.315,0.28)$ . Red differs most in the $40$ -spin distribution (difference $0.075$ ). The $400$ -spin distribution, because a larger number of trials gives relative frequencies closer to the theoretical probabilities.	5	Divide the first four frequencies by $40$ and the second four by $400$ . For $40$ spins the absolute differences are $0.075,0.05,0.025,0.05$ , so red differs most. For $400$ spins the differences are $0.005,0.01,0.015,0.02$ , all smaller, and the larger number of trials is expected to lie closer to the theoretical distribution.
4	$26$ of the further spins land on red. Yes. Their relative frequency is $26/80=0.325$ , which is greater than $0.30$ .	4	A relative frequency of $0.25$ after $240$ spins means there are $240\times0.25=60$ red results altogether. Therefore $60-34=26$ of the further $80$ spins land on red. Since $26/80=0.325>0.30$ , the maintenance warning should be issued.
5	Wheel A: $0.283$ (to 3 d.p.) initially and $0.26$ overall. Wheel B: $0.267$ (to 3 d.p.) initially and $0.34$ overall. Wheel B gives stronger evidence that it may be biased, because its larger-sample relative frequency remains $0.09$ above $0.25$ , whereas wheel A's is only $0.01$ above it.	5	Divide each red frequency by its number of spins. The early results are both fairly close to $0.25$ , but the $600$ -spin results are more stable evidence. Wheel A moves closer to $0.25$ , while wheel B moves much farther away and remains there over the larger sample.

P6 · Enumerate sets and combinations of sets systematically, using tables, grids, Venn diagrams and tree diagrams

Tier 1 · Easy

Mark scheme for P6 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	Blue-red, blue-silver, blue-green, white-red, white-silver, white-green.	2	Hold the shirt colour fixed and list every tie, then repeat for the other shirt. This gives $2\times3=6$ combinations, each appearing once.
2	$(1,6),(2,5),(3,4)$	2	Start with first number $1$ and increase it systematically: $(1,6),(2,5),(3,4)$ . The next pair would have the first number greater than the second, so the list is complete.

Tier 2 · Standard

Mark scheme for P6 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$A\cap B=\{3,6,9,18\}$ . A only $=\{12,15\}$ ; B only $=\{1,2\}$ . Neither $=\{4,5,7,8,10,11,13,14,16,17,19,20\}$ .	4	List $A=\{3,6,9,12,15,18\}$ and $B=\{1,2,3,6,9,18\}$ . Put their common values in the intersection, remove these to find each 'only' region, then place every unused integer from $1$ to $20$ outside both circles.
2	Bus-walk, bus-cycle, train-walk, train-cycle, train-taxi. $5$ journeys.	3	List every second stage after bus, omitting taxi, then list every second stage after train. This gives $2+3=5$ permitted journeys.
3	$(-2,-1),(-2,3),(1,-1),(1,3),(4,-1),(4,3)$ . Sum greater than $2$ : $(1,3),(4,-1),(4,3)$ .	3	Fix the first coordinate at $-2$ , then $1$ , then $4$ , pairing each with both possible second coordinates. Check the coordinate sums; only $1+3$ , $4+(-1)$ and $4+3$ are greater than $2$ .

Tier 3 · Hard

Mark scheme for P6 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$234,312,314,324,342,412,432$ . $4$ of these numbers do not contain the digit $1$ .	4	Work systematically by hundreds digit. Starting with $2$ gives $234$ ; starting with $3$ gives $312,314,324,342$ ; starting with $4$ gives $412,432$ . The values without digit $1$ are $234,324,342,432$ , so there are $4$ .
2	A and B only: $\{16\}$ ; A and C only: $\{2,6,8,12,24\}$ ; B and C only: $\{1\}$ . $7$ integers are in exactly two sets.	4	The only number in all three sets is $4$ , so exclude it from the exactly-two regions. The remaining common values are $16$ in A and B, $2,6,8,12,24$ in A and C, and $1$ in B and C. There are $1+5+1=7$ integers.
3	A-codes: A₁₂, A₃₂, A₄₂, A₁₄, A₂₄, A₃₄. B-codes: B₁₄, B₄₁, B₂₃, B₃₂. C-codes: C₁₂, C₁₃, C₁₄, C₂₃, C₂₄, C₃₄. $16$ permitted codes.	4	For A, fix the last digit as $2$ or $4$ and choose any different first digit, giving $6$ codes. The ordered digit pairs totalling $5$ are $14,41,23,32$ , giving $4$ B-codes. The increasing pairs are $12,13,14,23,24,34$ , giving $6$ C-codes. The total is $6+4+6=16$ .
4	NENEE, NEENE, NEEEN, ENENE, ENEEN, EENEN; $6$ routes.	3	Place the two N moves systematically in non-adjacent positions among the five moves, then fill the other positions with E. The possible N-position pairs are $(1,3),(1,4),(1,5),(2,4),(2,5),(3,5)$ , giving the six listed routes and no repeats.
5	AC, AD, AE; BC, BD, BE; DA, DB, DC, DE; EA, EB, EC, ED. $14$ choices.	5	List by captain. With A as captain, B is excluded and the deputy can be C, D or E; similarly there are three choices with B as captain. With D or E as captain, any of the other four students can be deputy because A and B are not then serving together. This gives $3+3+4+4=14$ .

P7 · Construct theoretical possibility spaces for single and combined experiments with equally likely outcomes and use these to calculate theoretical probabilities

Tier 1 · Easy

Mark scheme for P7 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$(H,1),(H,2),(H,3),(T,1),(T,2),(T,3)$ . $P(\text{head and even})=\dfrac{1}{6}$ .	2	Pair each coin result with each spinner result. Only $(H,2)$ satisfies both conditions, so one of the six equally likely outcomes is favourable.
2	Possibility space: $\{2,3,5,8,10\}$ ; probability $=\dfrac{3}{5}$ .	2	The five equally likely outcomes are $2,3,5,8,10$ . Three of them, $2,3,5$ , are prime, so the probability is $3/5$ .

Tier 2 · Standard

Mark scheme for P7 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	Possibility space: $(1,1),(1,2),(1,3),(1,4),(2,1),(2,2),(2,3),(2,4),(3,1),(3,2),(3,3),(3,4)$ . The favourable ordered pairs are $(2,4),(3,3),(3,4)$ . $P(\text{sum greater than }5)=\dfrac{3}{12}=\dfrac{1}{4}$ .	3	A $3$ by $4$ grid contains $12$ equally likely ordered pairs. Mark the entries whose sums exceed $5$ : $6,6,7$ at $(2,4),(3,3),(3,4)$ . Therefore the probability is $3/12=1/4$ .
2	Totals: for A= $1$ : $3,6$ ; for A= $4$ : $6,9$ ; for A= $7$ : $9,12$ . $P(\text{even total})=\dfrac{3}{6}=\dfrac{1}{2}$ .	3	The six equally likely totals are $3,6,6,9,9,12$ . Three of these totals are even, so the probability is $3/6=1/2$ .
3	Scores: for H, $3,5,7$ ; for T, $1,3,5$ . $P(\text{score is }3\text{ or }5)=\dfrac{4}{6}=\dfrac{2}{3}$ .	3	Pair each coin result with each spinner result. The six equally likely scores are $3,5,7,1,3,5$ . Four outcomes give $3$ or $5$ , so the probability is $4/6=2/3$ .

Tier 3 · Hard

Mark scheme for P7 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	Possibility space: $12,13,14,16,21,23,24,26,31,32,34,36,41,42,43,46,61,62,63,64$ . There are $5\times4=20$ equally likely ordered outcomes. The favourable numbers are $12,21,24,42,36,63$ . Probability $=\dfrac{6}{20}=\dfrac{3}{10}$ .	4	Make a $5$ by $5$ grid and exclude its diagonal because the cards must be different. A number is divisible by $3$ when its digit sum is divisible by $3$ . Testing the $20$ allowed ordered pairs gives the six listed numbers, so the probability is $6/20=3/10$ .
2	Products: for A= $1$ : $2,3,4,6$ ; for A= $2$ : $4,6,8,12$ ; for A= $5$ : $10,15,20,30$ . $P(\text{multiple of }4)=\dfrac{5}{12}$ .	4	The $3$ by $4$ possibility space has $12$ equally likely products. The multiples of $4$ are $4$ in the first row, $4,8,12$ in the second row and $20$ in the third row. There are $5$ favourable outcomes, so the probability is $5/12$ .
3	Possibility space: $\dfrac23,\dfrac26,\dfrac2{10},\dfrac2{12},\dfrac53,\dfrac56,\dfrac5{10},\dfrac5{12},\dfrac83,\dfrac86,\dfrac8{10},\dfrac8{12}$ . $P(\text{fraction}>\dfrac12)=\dfrac{7}{12}$ . $P(\text{simplest denominator is }3)=\dfrac{6}{12}=\dfrac{1}{2}$ . A fraction greater than $\dfrac12$ is more likely.	5	A $3$ by $4$ grid gives $12$ equally likely fractions. The seven fractions greater than $1/2$ are $2/3,5/3,5/6,8/3,8/6,8/10,8/12$ . After simplifying, six outcomes have denominator $3$ : $2/3,2/6,5/3,8/3,8/6,8/12$ . Compare $7/12$ with $6/12$ .
4	For $n=1,2,3,4,5,6$ , the point scores are $7,12,15,16,15,12$ . $P(\text{multiple of }3)=\dfrac{4}{6}=\dfrac{2}{3}$ . The scores $12$ and $15$ each come from two die results, while $7$ and $16$ each come from one.	4	Substitute each equally likely die result into $n(8-n)$ . Four of the six results give a multiple of $3$ : $12,15,15,12$ . Therefore the probability is $4/6=2/3$ . Repeated point scores have twice the probability of scores produced by only one face.
5	Tree outcomes: H₂ and H₄ each have probability $\dfrac{1}{4}$ ; T₁, T₃ and T₅ each have probability $\dfrac{1}{6}$ . $P(\text{spinner score}>2)=\dfrac{1}{4}+\dfrac{1}{6}+\dfrac{1}{6}=\dfrac{7}{12}$ . The five outcomes are not equally likely because the head branch splits into two outcomes while the tail branch splits into three.	5	The coin gives probability $1/2$ to each first branch. Therefore H₂ and H₄ each have probability $1/2\times1/2=1/4$ , while T₁, T₃ and T₅ each have probability $1/2\times1/3=1/6$ . The favourable routes are H₄, T₃ and T₅, so add $1/4+1/6+1/6=7/12$ rather than using $3/5$ .

P8 · Calculate the probability of independent and dependent combined events, including using tree diagrams and other representations, and know the underlying assumptions

Tier 1 · Easy

Mark scheme for P8 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$\dfrac{3}{10}$	2	The events are independent, so multiply their probabilities: $\dfrac{1}{2}\times\dfrac{3}{5}=\dfrac{3}{10}$ .
2	$0.21$	2	Replacement makes the choices independent. The probability of not purple is $1-0.3=0.7$ , so the required probability is $0.3\times0.7=0.21$ .

Tier 2 · Standard

Mark scheme for P8 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$\dfrac{2}{7}$	3	The first blue probability is $4/7$ . After a blue is removed, $3$ of the $6$ remaining counters are blue. Multiply along the route: $\dfrac{4}{7}\times\dfrac{3}{6}=\dfrac{12}{42}=\dfrac{2}{7}$ .
2	$\dfrac{8}{15}$	3	The two possible orders are winning then losing and losing then winning. Their probabilities are $\dfrac{6}{10}\times\dfrac{4}{9}=\dfrac{4}{15}$ and $\dfrac{4}{10}\times\dfrac{6}{9}=\dfrac{4}{15}$ . Adding the two routes gives $\dfrac{8}{15}$ .
3	$0.54$	3	There are two mutually exclusive routes. The probability that only A detects the fault is $0.7\times0.6=0.42$ . The probability that only B detects it is $0.3\times0.4=0.12$ . Add the routes to get $0.42+0.12=0.54$ .

Tier 3 · Hard

Mark scheme for P8 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$P(\text{exactly three green})=\dfrac{14}{33}$ . The probabilities change because counters are not replaced, so the composition and total in the bag change.	5	There are four possible colour orders: GGGY, GGYG, GYGG and YGGG. Each has probability $\dfrac{7}{11}\times\dfrac{6}{10}\times\dfrac{5}{9}\times\dfrac{4}{8}=\dfrac{7}{66}$ , with the factors in the corresponding order. Add the four mutually exclusive routes: $4\times\dfrac{7}{66}=\dfrac{14}{33}$ .
2	$\dfrac{2}{3}$	3	The probability of scoring both penalties is $\dfrac{2}{3}\times\dfrac{3}{4}=\dfrac{1}{2}$ . The probability of missing both is $\dfrac{1}{3}\times\dfrac{1}{2}=\dfrac{1}{6}$ . These routes are mutually exclusive, so the required probability is $\dfrac{1}{2}+\dfrac{1}{6}=\dfrac{2}{3}$ .
3	$0.31$	4	The three possible routes are score-score-miss, score-miss-score and miss-score-score. Their probabilities are $0.6\times0.7\times0.3=0.126$ , $0.6\times0.3\times0.4=0.072$ and $0.4\times0.4\times0.7=0.112$ . Adding the mutually exclusive routes gives $0.126+0.072+0.112=0.31$ .
4	$P(\text{at least two activate})=0.70$ . The calculation assumes that whether one alarm activates does not affect either of the other alarms.	5	The four qualifying routes are ABC, AB not C, A not B C and not A BC. Their probabilities are $0.8(0.6)(0.5)=0.24$ , $0.8(0.6)(0.5)=0.24$ , $0.8(0.4)(0.5)=0.16$ and $0.2(0.6)(0.5)=0.06$ . Add them to get $0.70$ . Independence keeps each alarm's branch probability unchanged.
5	$\dfrac{16}{35}$	4	If a red counter is transferred, the probability of then choosing blue from box B is $4/7$ , giving $3/5\times4/7=12/35$ . If a blue counter is transferred, the probability of then choosing red is $2/7$ , giving $2/5\times2/7=4/35$ . Add the two mutually exclusive routes: $12/35+4/35=16/35$ .

P9 · Calculate and interpret conditional probabilities through representation using expected frequencies with two-way tables, tree diagrams and Venn diagrams [Higher only]

Tier 1 · Easy

Mark scheme for P9 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$P(\text{late}\mid\text{bus})=\dfrac{9}{24}=\dfrac{3}{8}$	2	The condition restricts the group to the $24$ bus travellers. Of these, $9$ are late, so divide $9$ by $24$ and simplify.
2	$P(A\mid B)=\dfrac{14}{35}=\dfrac{2}{5}$	2	Out of $100$ expected outcomes, $35$ are in B and $14$ are in both A and B. Restricting to B gives $P(A\mid B)=14/35=2/5$ .

Tier 2 · Standard

Mark scheme for P9 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$P(\text{member}\mid\text{online})=\dfrac{3}{4}$	4	There are $0.60\times500=300$ members and $200$ non-members. Expected online frequencies are $0.70\times300=210$ members and $0.35\times200=70$ non-members. Among $210+70=280$ online customers, $210$ are members, so the conditional probability is $210/280=3/4$ .
2	$\dfrac{24}{75}=\dfrac{8}{25}$	3	There are $180-105=75$ non-fiction loans. Of the $72$ audiobooks, $72-48=24$ are non-fiction. Restricting to the $75$ non-fiction loans gives $24/75=8/25$ .
3	Monthly: $54$ renew and $42$ do not; annual: $18$ renew and $46$ do not. $P(\text{monthly}\mid\text{renews})=\dfrac{54}{72}=\dfrac{3}{4}$ . $P(\text{does not renew}\mid\text{annual})=\dfrac{46}{64}=\dfrac{23}{32}$ .	4	There are $160-96=64$ annual subscribers. Annual renewals total $72-54=18$ . The non-renewal frequencies are $96-54=42$ monthly and $64-18=46$ annual. Restricting to renewals gives $54/72=3/4$ ; restricting to annual subscribers gives $46/64=23/32$ .

Tier 3 · Hard

Mark scheme for P9 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$\dfrac{7}{11}$	4	The A-only frequency is $138-54=84$ and the B-only frequency is $102-54=48$ . Exactly one set contains $84+48=132$ people. Restricting to this group, the required probability is $84/132=7/11$ .
2	$\dfrac{1}{2}$	4	The probability of two red counters is $\dfrac{7}{10}\times\dfrac{6}{9}=\dfrac{7}{15}$ . The probability of at least one red is $1-\left(\dfrac{3}{10}\times\dfrac{2}{9}\right)=1-\dfrac{1}{15}=\dfrac{14}{15}$ . Restricting to outcomes with at least one red gives $\dfrac{7}{15}\div\dfrac{14}{15}=\dfrac{1}{2}$ .
3	$P(\text{lab B}\mid\text{no repeat})=\dfrac{336}{540}=\dfrac{28}{45}$ . $P(\text{lab A}\mid\text{repeat})=\dfrac{36}{60}=\dfrac{3}{5}$ , compared with overall $P(\text{lab A})=\dfrac{2}{5}$ . A sample that needs a repeat test is more likely to have been tested by laboratory A than a sample chosen without that condition.	5	Laboratory A tests $0.40\times600=240$ samples, so laboratory B tests $360$ . Laboratory A has $0.15\times240=36$ repeat tests, leaving $60-36=24$ from laboratory B. There are $600-60=540$ samples with no repeat test, of which $360-24=336$ were tested by laboratory B, giving $336/540=28/45$ . Among samples needing a repeat test, $36/60=3/5$ were tested by laboratory A; this exceeds the overall proportion $240/600=2/5$ .
4	Valid tickets: B₄, C₃, C₄, D₂, D₃, D₄; probability $=\dfrac{4}{6}=\dfrac{2}{3}$ .	4	Restrict the possibility space to the six valid tickets. Four of these, B₄, C₄, D₂ and D₄, have an even digit. The conditional probability is therefore $4/6=2/3$ .
5	$P(\text{line Y}\mid\text{no manual check})=\dfrac{656}{920}=\dfrac{82}{115}$ .	5	Out of $1000$ items, $300$ go to X and $700$ to Y. Line X has $0.12\times300=36$ checked items. Since these $36$ are $45\%$ of all checked items, the checked total is $36\div0.45=80$ , so Y has $80-36=44$ checked items. Therefore $700-44=656$ unchecked items came from Y, out of $1000-80=920$ unchecked items altogether. The conditional probability is $656/920=82/115$ .