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Edexcel GCSE Maths revision notes

Probability

Section P
9 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 1MA1 section P

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

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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 7070 people. 4343 choose tea and the rest choose coffee. Of the tea drinkers, 1818 add sugar. Complete the missing frequencies.

  1. 1.Coffee drinkers: 7043=2770-43=27.
  2. 2.Tea drinkers without sugar: 4318=2543-18=25.
  3. 3.Check the first split: 43+27=7043+27=70.

Answer: 2727 choose coffee and 2525 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

ORIGINAL

1

A spinner is used 3030 times. It lands on red 1818 times and on blue 1212 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)

Tier 2 · Standard

ORIGINAL

1

A frequency tree describes 8080 visits to a kiosk. On 5252 visits a hot drink is chosen and on the remaining visits a cold drink is chosen. A snack is also bought on 3131 hot-drink visits and on 77 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)

Tier 3 · Hard

ORIGINAL

1

Two weeks of trials are combined. In week 1, 7272 of 120120 trials succeed; 4545 successful trials and 1818 unsuccessful trials are fast. In week 2, 9999 of 180180 trials succeed; 6666 successful trials and 2727 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)

Your progress and exam materials
P2

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

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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 nn repeated trials, expected frequency is n×pn\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 360360 times. Decide whether the game is fair and find each player's expected number of wins.

  1. 1.P(Player 1 wins)=24=12P(\text{Player 1 wins})=\dfrac{2}{4}=\dfrac{1}{2}.
  2. 2.P(Player 2 wins)=24=12P(\text{Player 2 wins})=\dfrac{2}{4}=\dfrac{1}{2}, so the game is fair.
  3. 3.Expected wins for each player =360×12=180=360\times\dfrac{1}{2}=180.

Answer: The game is fair; each player is expected to win 180180 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

ORIGINAL

1

A fair six-sided die is rolled 240240 times. Work out the expected number of rolls on which the score is greater than 44.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

A fair spinner has four equal sectors labelled 11, 11, 22 and 33. It is spun 600600 times. Work out the expected total of all the scores.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

A box contains 33 green, 22 yellow and 11 red token. A token is chosen at random and replaced. A player receives £4\pounds4 for green, £1\pounds1 for yellow and nothing for red. The entry fee is £2.20\pounds2.20. Work out the organiser's expected profit from 900900 plays and decide whether the game is fair to the player.

(5)

(Total for Question 1 is 5 marks)

P3

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

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Probabilities lie on a scale from 00 to 11: 00 is impossible, 11 is certain, and 0.50.5 means equally likely to happen or not happen.
  • Theoretical probability comes from a model, while relative frequency comes from results: relative frequency=frequencynumber of trials\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 156156 times in 300300 tosses. Find the relative frequency of heads and compare it with the theoretical probability.

  1. 1.Relative frequency =156300=0.52=\dfrac{156}{300}=0.52.
  2. 2.For a fair coin, the theoretical probability is 0.50.5.
  3. 3.0.520.52 is close to 0.50.5; the difference can be due to random variation.

Answer: The relative frequency is 0.520.52, close to the theoretical value 0.50.5.

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

ORIGINAL

1

Place these events in order from least likely to most likely: event A has probability 0.720.72, event B has probability 14\dfrac{1}{4}, and event C has probability 0.50.5.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1

An outcome occurs 8484 times in 140140 trials. Its theoretical probability is 58\dfrac{5}{8}. Work out the relative frequency and the expected frequency in 560560 further trials. Comment on the experimental result.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

A model says that a device flashes with probability 0.50.5 on each trial. In a short run it flashes 4141 times in 8080 trials. After 500500 trials in total it has flashed 247247 times. Calculate both relative frequencies and decide which result gives stronger evidence about the model.

(4)

(Total for Question 1 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

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • An exhaustive set covers every possible outcome.
  • If those outcomes are mutually exclusive, exactly one occurs, so their probabilities add to 11.
  • For mutually exclusive events, P(A or B)=P(A)+P(B)P(A\text{ or }B)=P(A)+P(B).
  • An event and its complement are exhaustive and mutually exclusive, so P(not A)=1P(A)P(\text{not }A)=1-P(A).
  • If events overlap, subtract the intersection: P(A or B)=P(A)+P(B)P(A and B)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(red)=0.28P(\text{red})=0.28 and P(blue)=0.41P(\text{blue})=0.41. Find P(green)P(\text{green}).

  1. 1.The three colours are exhaustive, so their probabilities total 11.
  2. 2.P(green)=10.280.41P(\text{green})=1-0.28-0.41.
  3. 3.P(green)=0.31P(\text{green})=0.31.

Answer: 0.310.31.

Common mistakes

  • Don't fall into the trap of forgetting that all exhaustive outcomes must total exactly 11.
  • 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

ORIGINAL

1

The probability that a parcel arrives early is 0.370.37. Work out the probability that it does not arrive early.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1

A trial has four mutually exclusive outcomes A, B, C and D. Their probabilities are 0.280.28, 0.410.41, xx and 0.190.19 respectively. Work out xx.

(2)

(Total for Question 1 is 2 marks)

Tier 3 · Hard

ORIGINAL

1

For two events A and B, P(A)=0.58P(A)=0.58, P(B)=0.47P(B)=0.47 and P(AB)=0.21P(A\cap B)=0.21. Work out the probability that neither A nor B occurs.

(4)

(Total for Question 1 is 4 marks)

P5

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

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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 162162 heads in 300300 tosses and 10381038 heads in 20002000 tosses. Compare the two relative frequencies with 0.50.5.

  1. 1.For 300300 tosses: 162300=0.54\dfrac{162}{300}=0.54.
  2. 2.For 20002000 tosses: 10382000=0.519\dfrac{1038}{2000}=0.519.
  3. 3.0.5190.519 is closer to the theoretical probability 0.50.5 than 0.540.54.

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

ORIGINAL

1

A fair coin gives a relative frequency of heads of 0.640.64 after 2525 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)

Tier 2 · Standard

ORIGINAL

1

A spinner is designed to land on green with probability 0.30.3. It lands on green 88 times in 2020 spins and 6363 times in 200200 spins. Compare the two empirical probabilities with the theoretical probability.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

A fair die is tested. In the first 6060 rolls, a six appears 1616 times. After 600600 rolls in total, a six has appeared 108108 times. A student says the die must be biased because neither relative frequency equals 16\dfrac{1}{6}. Evaluate the student's claim.

(4)

(Total for Question 1 is 4 marks)

P6

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

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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 11 to 2020 that are multiples of 33 or 55 by separating the intersection and the two 'only' regions.

  1. 1.Both: {15}\{15\}.
  2. 2.Multiples of 33 only: {3,6,9,12,18}\{3,6,9,12,18\}.
  3. 3.Multiples of 55 only: {5,10,20}\{5,10,20\}.

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

ORIGINAL

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)

Tier 2 · Standard

ORIGINAL

1

The universal set is the integers from 11 to 2020. Set A contains the multiples of 33 and set B contains the factors of 1818. Enumerate the four regions of a Venn diagram for A and B.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1

A three-digit number is formed from three different digits chosen from 11, 22, 33 and 44. Enumerate all the numbers that are greater than 230230 and even. How many of these numbers do not contain the digit 11?

(4)

(Total for Question 1 is 4 marks)

P7

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

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A possibility space lists every permitted outcome once, often as ordered pairs in a grid.
  • If two experiments are independent, with mm and nn equally likely outcomes, the complete space has mnmn 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)(a,b) and (b,a)(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 99.

  1. 1.There are 6×6=366\times6=36 equally likely ordered pairs.
  2. 2.The favourable pairs are (3,6),(4,5),(5,4),(6,3)(3,6),(4,5),(5,4),(6,3).
  3. 3.Probability =436=19=\dfrac{4}{36}=\dfrac{1}{9}.

Answer: 19\dfrac{1}{9}.

Common mistakes

  • Don't fall into the trap of counting (3,6)(3,6) and (6,3)(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 mnmn outcomes.

Tier 1 · Easy

ORIGINAL

1

A fair coin is tossed and a fair spinner labelled 11, 22, 33 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)

Tier 2 · Standard

ORIGINAL

1

Spinner A has equal sectors labelled 11, 22, 33. Spinner B has equal sectors labelled 11, 22, 33, 44. Construct a possibility space and find the probability that the two scores have a sum greater than 55.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

Two different cards are chosen in order from cards labelled 11, 22, 33, 44 and 66. 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 33.

(4)

(Total for Question 1 is 4 marks)

P8

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

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

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 33 red and 22 blue counters. Two counters are taken without replacement. Find the probability that they are different colours.

  1. 1.P(RB)=35×24=310P(RB)=\dfrac{3}{5}\times\dfrac{2}{4}=\dfrac{3}{10}.
  2. 2.P(BR)=25×34=310P(BR)=\dfrac{2}{5}\times\dfrac{3}{4}=\dfrac{3}{10}.
  3. 3.Add the two routes: 310+310=35\dfrac{3}{10}+\dfrac{3}{10}=\dfrac{3}{5}.

Answer: 35\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

ORIGINAL

1

A fair coin is tossed and an independent spinner lands on red with probability 35\dfrac{3}{5}. Work out the probability of a head and red.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

A bag contains 44 blue counters and 33 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)

Tier 3 · Hard

ORIGINAL

1

A bag contains 77 green and 44 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)

P9

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

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Higher tier only.
  • A conditional probability restricts the sample space to outcomes satisfying the stated condition.
  • In P(AB)P(A\mid B), read 'the probability of AA given BB': the denominator is the total frequency in BB, and the numerator is the frequency in both AA and BB.
  • 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.
Another way to see this:
Worked example

A Venn diagram shows 1212 people in A only, 1818 in both A and B, 2727 in B only and 99 outside both sets. Find P(AB)P(A\mid B).

  1. 1.The condition restricts the sample space to set B.
  2. 2.There are 18+27=4518+27=45 people in B.
  3. 3.1818 of these are also in A, so P(AB)=1845=25P(A\mid B)=\dfrac{18}{45}=\dfrac{2}{5}.

Answer: 25\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(AB)P(A\mid B) and P(BA)P(B\mid A).

Exam tip

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

Tier 1 · Easy

ORIGINAL

1

A two-way table records 6060 students. Of the 2424 who travel by bus, 99 arrive late. Find the probability that a randomly chosen bus traveller arrives late.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

Of 500500 expected customers, 60%60\% are members. Of the members, 70%70\% order online. Of the non-members, 35%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)

Tier 3 · Hard

ORIGINAL

1

In an expected-frequency Venn diagram for 240240 people, 138138 are in set A, 102102 are in set B and 5454 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)

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