What is P8: Combined events?

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.

How many marks are P8 questions worth?

In verified past-paper appearances, P8 questions on Edexcel GCSE Maths 1MA1 have been worth 2–5 marks.

Which past exam papers has P8 appeared in?

P8 has appeared in 29 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

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

Combined events

Worked answers, methods and verified real exam appearances for P8 on Edexcel GCSE Maths 1MA1.

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

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.

Worked practice

Tier 1 · Easy

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)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$\dfrac{3}{10}$	2	The events are independent, so multiply their probabilities: $\dfrac{1}{2}\times\dfrac{3}{5}=\dfrac{3}{10}$ .

Tier 2 · Standard

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 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$\dfrac{2}{7}$	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: $\dfrac{4}{7}\times\dfrac{3}{6}=\dfrac{12}{42}=\dfrac{2}{7}$ .

Tier 3 · Hard

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 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$P(\text{exactly three green})=\dfrac{14}{33}$ . 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: $4\times\dfrac{7}{66}=\dfrac{14}{33}$ .

Tier 1 · Easy

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 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$0.21$	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

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 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$\dfrac{8}{15}$	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 $\dfrac{8}{15}$ .

Tier 3 · Hard

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 6 is 3 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$\dfrac{2}{3}$	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 $\dfrac{1}{2}+\dfrac{1}{6}=\dfrac{2}{3}$ .

Tier 2 · Standard

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 7 is 3 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	$0.54$	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

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 8 is 4 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$0.31$	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 $0.126+0.072+0.112=0.31$ .

Tier 3 · Hard

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 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	$P(\text{at least two activate})=0.70$ . 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.

Q10

Tier 3 · Hard

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 10 is 4 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$\dfrac{16}{35}$	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: $12/35+4/35=16/35$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2023-06	2H	Q22	2	Allowed	Higher	QP MS
2019-11	3H	Q21	2	Allowed	Higher	QP MS
2022-11	1H	Q10	4	Non-calculator	Higher	QP MS
2024-06	3F	Q27	4	Allowed	Foundation	QP MS
2024-11	1H	Q19	4	Non-calculator	Higher	QP MS
2024-06	3H	Q6	4	Allowed	Higher	QP MS
2022-11	1H	Q20	3	Non-calculator	Higher	QP MS
2019-11	2H	Q16	3	Allowed	Higher	QP MS
2023-11	1H	Q18	4	Non-calculator	Higher	QP MS
2022-11	3F	Q26	4	Allowed	Foundation	QP MS
2023-11	2H	Q10	3	Allowed	Higher	QP MS
2024-06	2H	Q18	5	Allowed	Higher	QP MS
2023-06	1F	Q31	2	Non-calculator	Foundation	QP MS
2024-06	1H	Q16	2	Non-calculator	Higher	QP MS
2019-06	2H	Q10	5	Allowed	Higher	QP MS
2022-11	3H	Q6	4	Allowed	Higher	QP MS
2024-11	3F	Q20	4	Allowed	Foundation	QP MS
2021-11	1H	Q16	4	Non-calculator	Higher	QP MS
2023-06	1H	Q12	3	Non-calculator	Higher	QP MS
2022-06	3H	Q21	4	Allowed	Higher	QP MS
2023-06	3H	Q24	5	Allowed	Higher	QP MS
2024-11	2H	Q22	5	Allowed	Higher	QP MS
2021-11	3H	Q20	2	Allowed	Higher	QP MS
2019-11	3H	Q11	3	Allowed	Higher	QP MS
2022-06	2F	Q20	4	Allowed	Foundation	QP MS
2024-11	3H	Q4	4	Allowed	Higher	QP MS
2023-06	2H	Q10	4	Allowed	Higher	QP MS
2021-11	2H	Q20	4	Allowed	Higher	QP MS
2022-06	1H	Q16	4	Non-calculator	Higher	QP MS

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