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

Mutually exclusive events cannot occur together, so P(A∩ B)=0 and P(A B)=P(A)+P(B). Independent events satisfy P(A∩ B)=P(A)P(B), equivalently P(A| B)=P(A) when P(B)>0.

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.

Draft — not yet indexed

Mutually exclusive and independent events

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

Explanation

Mutually exclusive events cannot occur together, so $P(A\cap B)=0$ and $P(A\cup B)=P(A)+P(B)$ . Independent events satisfy $P(A\cap B)=P(A)P(B)$ , equivalently $P(A\mid B)=P(A)$ when $P(B)>0$ .
Use $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ for any two events, subtracting the overlap to avoid double-counting. Events of positive probability cannot be both mutually exclusive and independent; the same rules apply to events defined from discrete or continuous random variables.
Independence gives $0.18=P(A)P(B)=0.45P(B)$ , so $P(B)=0.18/0.45=0.40$ . Then $P(A\cup B)=0.45+0.40-0.18=0.67$ .
State explicitly which event property justifies each probability equation in the solution.
For a continuous random variable, probability is represented by area under its density curve, so interval probabilities are integrals and the total area is $1$ .

Worked example

Events $A$ and $B$ are independent. Given $P(A)=0.45$ and $P(A\cap B)=0.18$ , find $P(B)$ and $P(A\cup B)$ .

1.Independence gives $0.18=P(A)P(B)=0.45P(B)$ , so $P(B)=0.18/0.45=0.40$ .
2.Then $P(A\cup B)=0.45+0.40-0.18=0.67$ .

Answer: $P(B)=0.40$ .; $P(A\cup B)=0.67$ .

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.

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

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Other points in S3 Probability

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