How many marks are P9 questions worth?

In verified past-paper appearances, P9 questions on Edexcel GCSE Maths 1MA1 have been worth 4–6 marks.

Which past exam papers has P9 appeared in?

P9 has appeared in 2 verified past-paper questions indexed on this page, with direct links to the official question paper and mark scheme for each.

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

Higher only

Draft — not yet indexed

Conditional probability

Worked answers, methods and verified real exam appearances for P9 on Edexcel GCSE Maths 1MA1.

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.

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.

Worked practice

Tier 1 · Easy

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)

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Mark scheme
1	$P(\text{late}\mid\text{bus})=\dfrac{9}{24}=\dfrac{3}{8}$	2	The condition restricts the group to the $24$ bus travellers. Of these, $9$ are late, so divide $9$ by $24$ and simplify.

Tier 2 · Standard

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 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Answer	Mark	Mark scheme
2	$P(\text{member}\mid\text{online})=\dfrac{3}{4}$	4	There are $0.60\times500=300$ members and $200$ non-members. Expected online frequencies are $0.70\times300=210$ members and $0.35\times200=70$ non-members. Among $210+70=280$ online customers, $210$ are members, so the conditional probability is $210/280=3/4$ .

Tier 3 · Hard

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 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Answer	Mark	Mark scheme
3	$\dfrac{7}{11}$	4	The A-only frequency is $138-54=84$ and the B-only frequency is $102-54=48$ . Exactly one set contains $84+48=132$ people. Restricting to this group, the required probability is $84/132=7/11$ .

Tier 1 · Easy

P(A\cap B)=0.14

and

P(B)=0.35

. Use expected frequencies out of

100

. Given that B occurs, work out the probability that A occurs.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Answer	Mark	Mark scheme
4	$P(A\mid B)=\dfrac{14}{35}=\dfrac{2}{5}$	2	Out of $100$ expected outcomes, $35$ are in B and $14$ are in both A and B. Restricting to B gives $P(A\mid B)=14/35=2/5$ .

Tier 2 · Standard

A library records

180

loans. Of these,

105

are fiction,

72

are audiobooks and

48

are fiction audiobooks. Given that a loan is not fiction, work out the probability that it is an audiobook.

(3)

(Total for Question 5 is 3 marks)

Mark scheme

Mark scheme for question 5
Question	Answer	Mark	Mark scheme
5	$\dfrac{24}{75}=\dfrac{8}{25}$	3	There are $180-105=75$ non-fiction loans. Of the $72$ audiobooks, $72-48=24$ are non-fiction. Restricting to the $75$ non-fiction loans gives $24/75=8/25$ .

Tier 3 · Hard

A bag contains

7

red counters and

3

blue counters. Two counters are chosen at random without replacement. Given that at least one counter is red, work out the probability that both counters are red. You must show all your working.

(4)

(Total for Question 6 is 4 marks)

Mark scheme

Mark scheme for question 6
Question	Answer	Mark	Mark scheme
6	$\dfrac{1}{2}$	4	The probability of two red counters is $\dfrac{7}{10}\times\dfrac{6}{9}=\dfrac{7}{15}$ . The probability of at least one red is $1-\left(\dfrac{3}{10}\times\dfrac{2}{9}\right)=1-\dfrac{1}{15}=\dfrac{14}{15}$ . Restricting to outcomes with at least one red gives $\dfrac{7}{15}\div\dfrac{14}{15}=\dfrac{1}{2}$ .

Tier 2 · Standard

An app has

160

subscribers. Of these,

96

pay monthly and the rest pay annually. In total,

72

subscribers renew, including

54

monthly subscribers. Complete a two-way table for payment plan against whether a subscriber renews. Given that a subscriber renews, work out the probability that the subscriber pays monthly. Given that a subscriber pays annually, work out the probability that the subscriber does not renew.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Answer	Mark	Mark scheme
7	Monthly: $54$ renew and $42$ do not; annual: $18$ renew and $46$ do not. $P(\text{monthly}\mid\text{renews})=\dfrac{54}{72}=\dfrac{3}{4}$ . $P(\text{does not renew}\mid\text{annual})=\dfrac{46}{64}=\dfrac{23}{32}$ .	4	There are $160-96=64$ annual subscribers. Annual renewals total $72-54=18$ . The non-renewal frequencies are $96-54=42$ monthly and $64-18=46$ annual. Restricting to renewals gives $54/72=3/4$ ; restricting to annual subscribers gives $46/64=23/32$ .

Tier 3 · Hard

A hospital receives

600

blood samples. Laboratory A tests

40\%

of them and laboratory B tests the rest. Of the samples tested by laboratory A,

15\%

need a repeat test. In total,

60

samples need a repeat test. Given that a sample does not need a repeat test, work out the probability that it was tested by laboratory B. Also work out the probability that a sample was tested by laboratory A given that it needs a repeat test, and compare this with the overall probability that a sample was tested by laboratory A.

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Answer	Mark	Mark scheme
8	$P(\text{lab B}\mid\text{no repeat})=\dfrac{336}{540}=\dfrac{28}{45}$ . $P(\text{lab A}\mid\text{repeat})=\dfrac{36}{60}=\dfrac{3}{5}$ , compared with overall $P(\text{lab A})=\dfrac{2}{5}$ . A sample that needs a repeat test is more likely to have been tested by laboratory A than a sample chosen without that condition.	5	Laboratory A tests $0.40\times600=240$ samples, so laboratory B tests $360$ . Laboratory A has $0.15\times240=36$ repeat tests, leaving $60-36=24$ from laboratory B. There are $600-60=540$ samples with no repeat test, of which $360-24=336$ were tested by laboratory B, giving $336/540=28/45$ . Among samples needing a repeat test, $36/60=3/5$ were tested by laboratory A; this exceeds the overall proportion $240/600=2/5$ .

Tier 3 · Hard

A ticket is made by choosing a letter from A, B, C and D, followed by a digit selected from the list

1

2

3

4

. All

16

ordered pairs are equally likely. Give the letters the values A

=1

, B

=2

, C

=3

and D

=4

. A ticket is valid when its letter value plus its digit is at least

6

. Given that a ticket is valid, work out the probability that its digit is even.

(4)

(Total for Question 9 is 4 marks)

Mark scheme

Mark scheme for question 9
Question	Answer	Mark	Mark scheme
9	Valid tickets: B₄, C₃, C₄, D₂, D₃, D₄; probability $=\dfrac{4}{6}=\dfrac{2}{3}$ .	4	Restrict the possibility space to the six valid tickets. Four of these, B₄, C₄, D₂ and D₄, have an even digit. The conditional probability is therefore $4/6=2/3$ .

Q10

Tier 3 · Hard

A sorting system sends

30\%

of items to line X and the rest to line Y. Of the items sent to line X,

12\%

need a manual check. Of all items that need a manual check,

45\%

were sent to line X. Use expected frequencies for

1000

items to work out the probability that an item was sent to line Y, given that it does not need a manual check.

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Answer	Mark	Mark scheme
10	$P(\text{line Y}\mid\text{no manual check})=\dfrac{656}{920}=\dfrac{82}{115}$ .	5	Out of $1000$ items, $300$ go to X and $700$ to Y. Line X has $0.12\times300=36$ checked items. Since these $36$ are $45\%$ of all checked items, the checked total is $36\div0.45=80$ , so Y has $80-36=44$ checked items. Therefore $700-44=656$ unchecked items came from Y, out of $1000-80=920$ unchecked items altogether. The conditional probability is $656/920=82/115$ .

Verified exam appearances

Series	Paper	Question	Marks	Calculator	Tier	Links
2022-06	2H	Q16	6	Allowed	Higher	QP MS
2023-11	2H	Q21	4	Allowed	Higher	QP MS

Other points in P Probability

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