P Probability — notes on one page

Every Edexcel GCSE Maths 1MA1 specification point in P, read straight through: 9 points with definitions, explanations, mark-scheme wording and worked examples. Practice questions are in the printable question pack.

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.

How this checking works

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

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.

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

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.

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

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.

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

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.

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

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.

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

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.

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

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.

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

Definition. Specification P8: calculate the probability of independent and dependent combined events, including using tree diagrams and other representations, and know the underlying assumptions.

On a tree diagram, multiply along a branch for the probability of that sequence and add the products of the branches that satisfy the condition.
Each pair of branches adds to $1$ , so a “do not win” branch carries $1-0.3=0.7$ .
Independent events keep the same probabilities on the second set of branches; dependent events, such as taking counters without replacement, reduce the total and the count of the colour already taken.
“Different colours” needs both orders added, and a single product scores the method mark only if nothing is then done to it.
Higher tier: an unknown number of counters gives probabilities in $x$ , and equating a product or sum to a given value yields a quadratic.

Worked example

A bag contains

3

red counters and

2

blue counters. Two counters are taken from the bag at random without replacement. Work out the probability that the two counters are different colours.

(3)

(Total for Question 1 is 3 marks)

Working

1.Red then blue: $\dfrac{3}{5}\times\dfrac{2}{4}=\dfrac{6}{20}$ ; the second denominator is $4$ because one counter has gone.
2.Blue then red: $\dfrac{2}{5}\times\dfrac{3}{4}=\dfrac{6}{20}$ .
3.Add the two routes: $\dfrac{6}{20}+\dfrac{6}{20}=\dfrac{12}{20}=\dfrac{3}{5}$ .

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

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Additional guidance
1	for one correct product, eg $\dfrac{3}{5}\times\dfrac{2}{4}$ or $\dfrac{2}{5}\times\dfrac{3}{4}$	P1
	for adding both orders, eg $\dfrac{3}{5}\times\dfrac{2}{4}+\dfrac{2}{5}\times\dfrac{3}{4}$	P1	oe or ft probabilities from a diagram
	for $\dfrac{3}{5}$	A1	oe; accept 0.6 or 60%

Mark-scheme wording to use

“B1 for 0.7 on the first branch … B1 for 0.65, 0.65 on the second branches” — Pearson 1MA1, November 2022 Paper 3H, Q6(a)
“P1 for x/10 or (10 − x)/10 or (x − 1)/9 or (10 − x)/9 or x/9 or (9 − x)/9 seen on diagram or in a calculation P1 for x/10 × (10 − x)/9 or (10 − x)/10 × x/9 … P1 for x/10 × (10 − x)/9 + (10 − x)/10 × x/9 … P1 (dep on P3) for beginning to process the algebra A1 (10x − x^2)/45 oe” — Pearson 1MA1, Specimen 1 Paper 1H, Q21
“M1 for 0.3 × 0.35 A1 oe” — Pearson 1MA1, November 2022 Paper 3H, Q6(b)

What the examiner wants. A product left alone. The November 2024 Paper 3H report says the method mark was for $0.7\times0.7$ as a standalone product, so students who then halved it, subtracted it from $1$ or added another product scored nothing; the June 2024 Paper 3H report notes a few students seemed unsure of the probabilities for the second coin, seemingly thinking it was a conditional probability.

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.

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

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.