What is P4: Probabilities summing to one?

An exhaustive set covers every possible outcome.

How many marks are P4 questions worth?

In verified past-paper appearances, P4 questions on Edexcel GCSE Maths 1MA1 have been worth 2–5 marks.

Which past exam papers has P4 appeared in?

P4 has appeared in 24 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

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

Probabilities summing to one

Worked answers, methods and verified real exam appearances for P4 on Edexcel GCSE Maths 1MA1.

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.

Worked practice

Tier 1 · Easy

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)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$0.63$	1	Early and not early are exhaustive and mutually exclusive, so use the complement: $1-0.37=0.63$ .

Tier 2 · Standard

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 2 is 2 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$x=0.12$	2	The outcomes are exhaustive, so their probabilities sum to $1$ . Therefore $x=1-(0.28+0.41+0.19)=1-0.88=0.12$ .

Tier 3 · Hard

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 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$0.16$	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 $1-0.84=0.16$ .

Tier 1 · Easy

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 4 is 1 mark)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$0.77$	1	Mutually exclusive events cannot happen together, so add their probabilities: $0.32+0.45=0.77$ .

Tier 2 · Standard

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 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$P(\text{early})=0.43$ 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 $P(\text{late})=1-0.74=0.26$ .

Tier 3 · Hard

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 6 is 4 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$P(B)=0.48$ 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, $P(A\text{ or }C)=0.28+0.24=0.52$ .

Tier 2 · Standard

A spinner can land on

1

2

3

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 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$P(E\text{ or }F)=0.25+0.35+0.25=0.85$ . 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

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 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$60$ 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 $155/250=31/50$ .

Tier 3 · Hard

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 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$P(W)=0.15$ , $P(X)=0.30$ , $P(Y)=0.30$ 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 $P(Z)=1-0.15-0.30-0.30=0.25$ .

Q10

Tier 3 · Hard

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 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$0.60$	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 $0.60$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2023-06	2H	Q22	2	Allowed	Higher	QP MS
2023-06	1F	Q24	5	Non-calculator	Foundation	QP MS
2019-11	3F	Q16	4	Allowed	Foundation	QP MS
2022-11	3H	Q19	3	Allowed	Higher	QP MS
2021-11	3H	Q6	4	Allowed	Higher	QP MS
2019-06	2F	Q16	5	Allowed	Foundation	QP MS
2023-11	1H	Q18	4	Non-calculator	Higher	QP MS
2024-06	1F	Q13	3	Non-calculator	Foundation	QP MS
2022-11	3F	Q26	4	Allowed	Foundation	QP MS
2019-06	1H	Q1	4	Non-calculator	Higher	QP MS
2023-06	3F	Q7	3	Allowed	Foundation	QP MS
2022-11	1F	Q17	5	Non-calculator	Foundation	QP MS
2024-11	3F	Q20	4	Allowed	Foundation	QP MS
2023-11	1F	Q9	3	Non-calculator	Foundation	QP MS
2024-06	2H	Q4	4	Allowed	Higher	QP MS
2023-06	1H	Q12	3	Non-calculator	Higher	QP MS
2024-11	1H	Q2	4	Non-calculator	Higher	QP MS
2021-11	3H	Q20	2	Allowed	Higher	QP MS
2022-06	2F	Q20	4	Allowed	Foundation	QP MS
2019-06	1F	Q22	4	Non-calculator	Foundation	QP MS
2022-06	1F	Q13	2	Non-calculator	Foundation	QP MS
2024-11	3H	Q4	4	Allowed	Higher	QP MS
2021-11	3F	Q26	4	Allowed	Foundation	QP MS
2024-11	1F	Q19	4	Non-calculator	Foundation	QP MS

Other points in P Probability

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