What is S3.2: Conditional probability?

Conditional probability restricts the sample space: P(A| B)=P(A∩ B)/P(B) for P(B)>0. On a tree diagram, multiply probabilities along a path and add the probabilities of mutually exclusive paths that satisfy the event.

How many marks are S3.2 questions worth?

S3.2 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has S3.2 appeared in?

We have not yet indexed a verified past-paper appearance for S3.2 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

S3.2

Understand and use conditional probability, including the use of tree diagrams, Venn diagrams and two-way tables; understand and use the conditional probability formula P(A|B) = P(A∩B)/P(B).

Draft — not yet indexed

Conditional probability

Worked answers and methods for S3.2 on Edexcel A-level Maths 9MA0.

Explanation

Conditional probability restricts the sample space: $P(A\mid B)=\frac{P(A\cap B)}{P(B)}$ for $P(B)>0$ . On a tree diagram, multiply probabilities along a path and add the probabilities of mutually exclusive paths that satisfy the event.
Without replacement, later branch probabilities change because both the total and the relevant category count may have changed.
In a Venn diagram or two-way table, use the condition as the denominator; reversing $P(A\mid B)$ to $P(B\mid A)$ is a common error.
There are two valid tree paths.
Amber then blue has probability $\frac58\cdot\frac37=\frac{15}{56}$ .

Worked example

A bag contains $5$ amber counters and $3$ blue counters. Two counters are taken without replacement. Find the probability that exactly one counter of each colour is taken.

1.There are two valid tree paths.
2.Amber then blue has probability $\frac58\cdot\frac37=\frac{15}{56}$ .
3.Blue then amber has probability $\frac38\cdot\frac57=\frac{15}{56}$ .
4.Adding gives $\frac{30}{56}=\frac{15}{28}$ .

Answer: $\frac{15}{28}$

Common mistakes

Don't multiply probabilities from mutually exclusive branches and add probabilities along a single tree path.
Don't use replacement probabilities on a without-replacement tree, leaving second-stage denominators unchanged.

Exam tip

Label every branch with the conditional probability after the first outcome, then add the required disjoint paths.

Worked practice

Tier 1 · Easy

60

students,

24

travel by bus and

15

of those bus travellers arrive before

8{:}30

. Find the probability that a randomly chosen bus traveller arrives before

8{:}30

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$\frac{15}{24}=\frac58$	2
Notes
The condition restricts the denominator to the $24$ bus travellers. Of these, $15$ arrived before $8{:}30$ , so the probability is $15/24=5/8$ .

(2 marks)

Tier 2 · Standard

In a group of

60

students,

35

study Mathematics,

28

study Physics and

18

study both. Find the probability that a randomly chosen Physics student studies Mathematics, and the probability that a randomly chosen student studies neither subject.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$P(M\mid P)=\dfrac{9}{14}$ $P(\text{neither})=\dfrac14$	4
Notes
Among the $28$ Physics students, $18$ also study Mathematics, so $P(M\mid P)=18/28=9/14$ . The number studying at least one subject is $35+28-18=45$ , leaving $60-45=15$ studying neither. Thus $P(\text{neither})=15/60=1/4$ .

(4 marks)

Tier 3 · Hard

For events

A

and

B

P(A)=0.60

P(B)=0.50

and

P(A\mid B)=0.70

. Find

P(A'\mid B')

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$0.50$	5
Notes
From the conditional probability formula, $P(A\cap B)=P(A\mid B)P(B)=0.70(0.50)=0.35$ . Hence $P(A\cup B)=0.60+0.50-0.35=0.75$ , so $P(A'\cap B')=1-0.75=0.25$ . Also $P(B')=0.50$ . Therefore $P(A'\mid B')=0.25/0.50=0.50$ .

(5 marks)

Tier 1 · Easy

For events

A

and

B

P(A\cap B)=0.18

and

P(B)=0.45

. Find

P(A\mid B)

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$0.4$	2
Notes
Use the conditional probability formula: $P(A\mid B)=P(A\cap B)/P(B)=0.18/0.45=0.4$ .

(2 marks)

Tier 2 · Standard

In a shipment,

2\%

of devices are defective,

4\%

are flagged by an automated check, and

1.5\%

are both defective and flagged. Find the probability that a flagged device is defective and the probability that a defective device is flagged.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$P(D\mid F)=0.375$ $P(F\mid D)=0.75$	4
Notes
$P(D\mid F)=P(D\cap F)/P(F)=0.015/0.04=0.375$ . Reversing the condition changes the denominator: $P(F\mid D)=0.015/0.02=0.75$ .

(4 marks)

Tier 3 · Hard

Events

A

and

B

satisfy

P(A)=0.4

P(B\mid A)=0.7

and

P(A\mid B)=0.5

. Given that

P(B\mid A')=q

, find

q

and

P(A\cup B)

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$q=\dfrac{7}{15}$ $P(A\cup B)=0.68$	5
Notes
$P(A\cap B)=P(A)P(B\mid A)=0.4(0.7)=0.28$ . Since $P(A\mid B)=0.5$ , $P(B)=0.28/0.5=0.56$ . Total probability gives $0.56=0.4(0.7)+0.6q$ , so $q=0.28/0.6=7/15$ . Finally $P(A\cup B)=0.4+0.56-0.28=0.68$ .

(5 marks)

Tier 2 · Standard

Machine A produces

60\%

of a factory's components and machine B produces the rest. Their defective rates are

2\%

and

5\%

respectively. After learning that a randomly selected component is defective, calculate the conditional probability that its source was machine B.

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$P(B\mid D)=0.625$	4
Notes
The joint probabilities are $P(A\cap D)=0.60(0.02)=0.012$ and $P(B\cap D)=0.40(0.05)=0.020$ . Hence $P(D)=0.012+0.020=0.032$ . Restricting the sample space to defective components gives $P(B\mid D)=0.020/0.032=0.625$ .

(4 marks)

Tier 3 · Hard

Box A is selected with probability

0.7

and box B with probability

0.3

. Box A contains

4

red and

2

blue counters; box B contains

3

red and

1

blue counter. Two counters are drawn without replacement from the selected box and both are red. Find the probability that box B was selected.

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$P(B\mid RR)=\dfrac{15}{43}$	5
Notes
For box A, $P(RR\mid A)=\frac46\frac35=\frac25$ , so $P(A\cap RR)=\frac7{10}\frac25=\frac7{25}$ . For box B, $P(RR\mid B)=\frac34\frac23=\frac12$ , so $P(B\cap RR)=\frac3{10}\frac12=\frac3{20}$ . Therefore $P(RR)=\frac7{25}+\frac3{20}=\frac{43}{100}$ , and $P(B\mid RR)=(3/20)/(43/100)=15/43$ .

(5 marks)

Tier 3 · Hard

In a group of

100

people,

40

use service

A

and

30

use service

B

. Let

x

be the number who use both services. Complete the four cell counts of a two-way table in terms of

x

. Given that using

A

and using

B

are independent, find

x

and then calculate

P(A'\mid B')

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	The cell counts for $(A\cap B),(A\cap B'),(A'\cap B),(A'\cap B')$ are $x,40-x,30-x,30+x$ . $x=12$ $P(A'\mid B')=\dfrac{42}{70}=0.6$	5
Notes
The margins give the remaining cells by subtraction, and the neither cell is $100-[x+(40-x)+(30-x)]=30+x$ . Independence requires $x/100=(40/100)(30/100)$ , so $x=12$ . There are $70$ people in $B'$ , of whom $30+12=42$ are also in $A'$ , giving $42/70=0.6$ .

(5 marks)

Q10

Tier 3 · Hard

10.

A customer is a premium member with probability

0.30

and otherwise is a standard member. A premium member renews with probability

0.80

, while a standard member renews with probability

q

. The overall renewal probability is

0.59

. Find

q

. Given that a customer does not renew, find the probability that the customer is a standard member. Two customers are then selected independently from this population; find the probability that exactly one renews.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$q=0.50$ $P(\text{standard}\mid\text{not renew})=\dfrac{35}{41}$ $P(\text{exactly one renewal})=0.4838$	6
Notes
Total probability gives $0.59=0.30(0.80)+0.70q$ , so $q=0.50$ . The joint probability of being standard and not renewing is $0.70(0.50)=0.35$ , while $P(\text{not renew})=0.41$ , so the conditional probability is $35/41$ . For two independently selected customers, exactly one renewal has probability $2(0.59)(0.41)=0.4838$ .

(6 marks)

Verified exam appearances

We have not yet indexed a verified real-paper appearance for S3.2. Browse the Edexcel A-level Maths 9MA0 past papers directly.

Other points in S3 Probability

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