P Probability — 1MA1 revision notes

P1

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

Notes

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 $70$ 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

ORIGINAL

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)

Tier 2 · Standard

ORIGINAL

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)

Tier 3 · Hard

ORIGINAL

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)

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 $n$ 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

ORIGINAL

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)

Tier 2 · Standard

ORIGINAL

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)

Tier 3 · Hard

ORIGINAL

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)

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 $0$ 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

ORIGINAL

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)

Tier 2 · Standard

ORIGINAL

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)

Tier 3 · Hard

ORIGINAL

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)

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 $1$ .
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

ORIGINAL

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)

Tier 2 · Standard

ORIGINAL

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)

Tier 3 · Hard

ORIGINAL

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)

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 $162$ 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

ORIGINAL

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)

Tier 2 · Standard

ORIGINAL

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)

Tier 3 · Hard

ORIGINAL

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)

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 $1$ 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

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

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)

Tier 3 · Hard

ORIGINAL

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)

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 $m$ 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

ORIGINAL

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)

Tier 2 · Standard

ORIGINAL

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)

Tier 3 · Hard

ORIGINAL

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)

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.

Definition

Specification 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 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

1

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%

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.

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.

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

\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

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)

Tier 3 · Hard

ORIGINAL

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)

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 $P(A\mid B)$ , 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.

Another way to see this:StandardVisual

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

ORIGINAL

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)

Tier 2 · Standard

ORIGINAL

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)

Tier 3 · Hard

ORIGINAL

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)

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

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

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

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

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

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

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

Tier 2 · Standard

Tier 3 · Hard

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

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

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

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

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

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

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

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

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

Tier 1 · Easy

Tier 2 · Standard

Tier 3 · Hard

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