What is S3.1: Mutually exclusive and independent events?

Two events are mutually exclusive if they cannot happen together, so P(A∩ B)=0 and P(A B)=P(A)+P(B); on a Venn diagram their regions do not overlap. Two events are independent if one occurring does not change the probability of the other: P(A∩ B)=P(A)P(B), equivalently P(A| B)=P(A).

How many marks are S3.1 questions worth?

S3.1 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has S3.1 appeared in?

We have not yet indexed a verified past-paper appearance for S3.1 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

S3.1

Understand and use mutually exclusive and independent events when calculating probabilities; link to discrete and continuous distributions.

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Mutually exclusive and independent events

Worked answers and methods for S3.1 on Edexcel A-level Maths 9MA0.

Definition

Specification S3.1: understand and use mutually exclusive and independent events when calculating probabilities; link to discrete and continuous distributions. Events $A$ and $B$ are mutually exclusive when $P(A\cap B)=0$ and independent when $P(A\cap B)=P(A)P(B)$ .

Explanation

Two events are mutually exclusive if they cannot happen together, so $P(A\cap B)=0$ and $P(A\cup B)=P(A)+P(B)$ ; on a Venn diagram their regions do not overlap. Two events are independent if one occurring does not change the probability of the other: $P(A\cap B)=P(A)P(B)$ , equivalently $P(A\mid B)=P(A)$ .
To test independence, compute $P(A)P(B)$ and compare it with $P(A\cap B)$ , quoting both numbers; to use independence, write the product equation with the unknown region and solve.
The general addition rule $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ holds for any two events.
The same ideas apply to random variables: events on independent Normal variables multiply, and the probability of an interval for a continuous variable is an area under its curve.
State the property being used in words the examiner recognises.

Worked example

The events

A

and

B

are independent, with

P(A)=0.45

and

P(A\cap B)=0.18

. Find

P(A\cup B)

(3)

(Total for Question 1 is 3 marks)

Working

1.Independence gives $P(A\cap B)=P(A)\,P(B)$ , so $0.18=0.45\,P(B)$ and $P(B)=0.4$ .
2.Addition rule: $P(A\cup B)=P(A)+P(B)-P(A\cap B)=0.45+0.4-0.18$ .
3. $P(A\cup B)=0.67$ .

Answer: $0.67$

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1.	Uses independence: $0.18=0.45\times P(B)$ or $P(B)=\dfrac{0.18}{0.45}$	M1
	$P(B)=0.4$	A1
	$0.45+0.4-0.18=0.67$	A1
Notes
A1: oe

(3 marks)

Common mistakes

Don't add $P(A)$ and $P(B)$ for overlapping events without subtracting $P(A\cap B)$ .
Don't treat mutually exclusive events as independent, making their intersection both zero and a product.

Mark-scheme wording to use

“P(C) = 0.6 and P(B) = p + 0.32 and P(B∩C) = 0.27 … M1 for all relevant labelled probabilities listed or a correct equation/expression for p … A1 for p = 0.13” — Pearson 9MA0, June 2024 Paper 31, Q6(b)
“B1 for a fully correct explanation using correct probabilities and a valid comparison, such as 0.13 != 0.35(0.45), or P(B | A) != P(B)” — Pearson 9MA0, Sample Paper 31, Q4(b)
“Need H and T to be independent or events {H > 1.6} and {T < 300} are independent … B1 for a suitable reason mentioning or implying H and T are independent” — Pearson 9MA0, June 2024 Paper 31, Q5(b)

What the examiner wants

The word “independent”, used correctly. The June 2024 Paper 31 report says a good proportion recognised the need for the events to be independent, but some scored nothing because they did not use correct mathematical language, writing “they are not affected by each other” or “the events are not mutually exclusive” instead.

Exam tip

State whether the events are independent or mutually exclusive before choosing the intersection and union formulae.

Worked practice

Tier 1 · Easy

Events

A

and

B

are mutually exclusive, with

P(A)=0.38

and

P(B)=0.27

. Find

P(A\cup B)

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$0.65$	2
Notes
Mutually exclusive events have no overlap, so $P(A\cup B)=P(A)+P(B)=0.38+0.27=0.65$ .

(2 marks)

Tier 2 · Standard

Events

A

and

B

are mutually exclusive, with

P(A)=0.3

and

P(B)=0.4

. Find

P(A\cup B)

and determine whether

A

and

B

are independent.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$P(A\cup B)=0.7$ $A$ and $B$ are not independent.	3
Notes
Mutual exclusivity gives $P(A\cap B)=0$ , so $P(A\cup B)=0.3+0.4=0.7$ . Independence would require $P(A\cap B)=P(A)P(B)=0.12$ . Since $0\neq0.12$ , the events are not independent.

(3 marks)

Tier 3 · Hard

A discrete random variable

X

has

P(X=0)=0.2

P(X=1)=0.5

and

P(X=2)=0.3

. Independently,

Y

is uniformly distributed on

0\leq y\leq5

. Let

A

be the event

X\geq1

and

B

the event

Y<2

. Find

P(A\cap B)

and

P(A\cup B)

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	$P(A\cap B)=0.32$ . $P(A\cup B)=0.88$ .	5
Notes
$P(A)=0.5+0.3=0.8$ . Since $Y$ is uniform, $P(B)=2/5=0.4$ . The variables are independent, so $P(A\cap B)=0.8(0.4)=0.32$ . Therefore $P(A\cup B)=0.8+0.4-0.32=0.88$ .

(5 marks)

Tier 1 · Easy

Events

C

and

D

satisfy

P(C)=0.6

P(D)=0.5

and

P(C\cap D)=0.3

. Determine whether they are independent and whether they are mutually exclusive.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$C$ and $D$ are independent because $P(C)P(D)=0.3=P(C\cap D)$ . $C$ and $D$ are not mutually exclusive because $P(C\cap D)\neq0$ .	2
Notes
Independence requires the intersection probability to equal the product $0.6(0.5)=0.3$ , which it does. Mutual exclusivity requires a zero intersection, which it does not have.

(2 marks)

Tier 2 · Standard

Events

C

and

D

satisfy

P(C)=0.6

P(D\mid C)=0.25

and

P(D\mid C')=0.50

. Find

P(D)

and determine whether

C

and

D

are independent.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$P(D)=0.35$ $C$ and $D$ are not independent.	4
Notes
Split according to $C$ : $P(D)=0.6(0.25)+0.4(0.50)=0.35$ . If $C$ and $D$ were independent, $P(D\mid C)$ would equal $P(D)$ , but $0.25\neq0.35$ . Therefore they are not independent.

(4 marks)

Tier 3 · Hard

Events

A

and

B

are independent. Given that

P(B)=2P(A)

and

P(A\cup B)=\frac{21}{25}

, find the exact values of

P(A)

and

P(A\cap B)

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	$P(A)=\dfrac{15-\sqrt{57}}{20}$ $P(A\cap B)=\dfrac{141-15\sqrt{57}}{100}$	5
Notes
Let $P(A)=x$ , so $P(B)=2x$ and independence gives $P(A\cap B)=2x^2$ . The union formula gives $3x-2x^2=21/25$ , hence $50x^2-75x+21=0$ . Therefore $x=(15\pm\sqrt{57})/20$ . The plus sign gives $P(B)>1$ , so $P(A)=(15-\sqrt{57})/20$ . Substituting into $2x^2$ , or using $3x-21/25$ , gives $(141-15\sqrt{57})/100$ .

(5 marks)

Tier 2 · Standard

Events

A

and

B

are independent, with

P(A)=0.35

and

P(B)=0.60

. Show that

A'

and

B

are independent, and find

P(A\cup B')

(4)

(Total for Question 7 is 4 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$P(A'\cap B)=0.39=P(A')P(B)$ , so $A'$ and $B$ are independent. $P(A\cup B')=0.61$	4
Notes
Independence gives $P(A\cap B)=0.35(0.60)=0.21$ . Therefore $P(A'\cap B)=P(B)-P(A\cap B)=0.60-0.21=0.39$ . Also $P(A')P(B)=0.65(0.60)=0.39$ , proving independence. The complement of $A\cup B'$ is $A'\cap B$ , so $P(A\cup B')=1-0.39=0.61$ .

(4 marks)

Tier 3 · Hard

A fair coin is tossed twice, so

HH,HT,TH,TT

are equally likely. Let

A

be the event that the first toss is a head,

B

the event that the second toss is a head, and

C

the event that both tosses give the same result. Show that each pair of events is independent, but that the three events are not mutually independent.

(6)

(Total for Question 8 is 6 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$P(A)=P(B)=P(C)=\frac12$ . $P(A\cap B)=P(A\cap C)=P(B\cap C)=\frac14$ , equal to the relevant products, so every pair is independent. $P(A\cap B\cap C)=\frac14$ but $P(A)P(B)P(C)=\frac18$ . Therefore $A$ , $B$ and $C$ are pairwise independent but not mutually independent.	6
Notes
Each event contains two of the four equally likely outcomes. Every pair intersects only in $HH$ , so each pair has intersection probability $1/4=(1/2)(1/2)$ . The intersection of all three events is also $\{HH\}$ and has probability $1/4$ , whereas mutual independence would require $(1/2)^3=1/8$ . Pairwise checks alone therefore do not establish mutual independence.

(6 marks)

Tier 3 · Hard

A discrete random variable

X

has probabilities

P(X=1)=k

P(X=2)=0.2

P(X=3)=0.3

and

P(X=4)=0.5-k

, where all probabilities are non-negative. Let

A

be the event

X\leq2

and

B

the event that

X

is even. Find all values of

k

for which

A

and

B

are independent. For these values, find

P(A\cup B)

and state whether

A

and

B

are mutually exclusive.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$k=0.2$ or $k=0.3$ $P(A\cup B)=0.7$ for either value $A$ and $B$ are not mutually exclusive because $P(A\cap B)=0.2>0$ .	6
Notes
$P(A)=k+0.2$ , $P(B)=0.7-k$ and $P(A\cap B)=P(X=2)=0.2$ . Independence requires $(k+0.2)(0.7-k)=0.2$ , giving $k^2-0.5k+0.06=0=(k-0.2)(k-0.3)$ . Both roots make every listed probability non-negative. The union probability is $(k+0.2)+(0.7-k)-0.2=0.7$ . The positive intersection rules out mutual exclusivity.

(6 marks)

Q10

Tier 3 · Hard

10.

The continuous random variable

X

is uniformly distributed on

-2\leq x\leq3

. For

-1<t<1

, let

A

be the event

X>t

and

B

the event

|X|<1

. Find the value of

t

for which

A

and

B

are independent. For this value of

t

, find

P(A\cup B)

(5)

(Total for Question 10 is 5 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	$t=-\dfrac13$ $P(A\cup B)=\dfrac45$	5
Notes
The uniform interval has length $5$ . For $-1<t<1$ , $P(A)=(3-t)/5$ , $P(B)=2/5$ and $P(A\cap B)=(1-t)/5$ . Independence gives $(1-t)/5=[(3-t)/5](2/5)$ , so $5-5t=6-2t$ and $t=-1/3$ . Then $P(A)=2/3$ , $P(B)=2/5$ and $P(A\cap B)=4/15$ , hence $P(A\cup B)=2/3+2/5-4/15=4/5$ .

(5 marks)

Verified exam appearances

We have not yet indexed a verified real-paper appearance for S3.1. Browse the Edexcel A-level Maths 9MA0 past papers directly.

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