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Learning objectives
Core idea
Key formulae
Worked examples
Explore the graph
Common mistakes
Exam tip
Guided practice
Practice questions
Challenge
Summary
Quick quiz
Same topic, different board
Past paper questions

LearnYMA01WST01 · Statistics 1ProbabilityCombined events and Venn diagrams

Probability · Subtopic 1 of 3

Combined Events and Venn Diagrams

Probability starts with counting equally likely outcomes. A sample space lists them; a Venn diagram shows how events overlap; and the addition rule stops the overlap being counted twice.

Cambridge 9709 Paper 5Edexcel YMA01 WST01Edexcel 9MA0 Paper 3AQA 7357 Paper 3≈ 30 minRevision 5 · updated 2026-10-27

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Section 1 of 14

Learning objectives

Find probabilities by counting equally likely outcomes in a sample space.
Use the notation $A\cup B$ , $A\cap B$ and $A'$ .
Use the addition rule $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ .
Fill in and read Venn diagrams, using "at least one $=1-$ none".

Pause and think

One card is drawn from a standard pack. There are $13$ hearts and $4$ kings. Is the probability of "a heart or a king" equal to $\dfrac{13+4}{52}$ ?

Section 2 of 14

Core idea

Start hereMeasuring uncertainty

Roll two dice. Is a total of 7 more likely than 12? There's one way to get 12 (6+6) but six ways to get 7. Probability turns "more likely" into numbers between $0$ and $1$ that can be added and multiplied.

The big idea

For equally likely outcomes, $P(A)=\frac{n(A)}{n(S)}$ . $0\le P(A)\le1$ ; $P(A')=1-P(A)$ ; $P(A\cup B)=P(A)+P(B)-P(A\cap B)$ .

1. Sample spaces

Sum of two dice: 6 of the 36 outcomes make 7

P(\text{sum}=7)=\dfrac{6}{36}=\dfrac16

. The sevens lie on a diagonal.

P(A)=\frac{\text{number of outcomes in }A}{\text{total number of equally likely outcomes}}

$S$: the sample space: all possible outcomes.
$A'$: "not $A$ ".

Worked exampleTwo dice

Find P(total 7) and P(at least one six).

Goal: Two probabilities.
Strategy: Count in the 36-cell grid; use the complement for "at least".

$P(7)=\tfrac6{36}=\tfrac16$
Six cells on the diagonal.
$P(\text{no six})=\tfrac56\times\tfrac56=\tfrac{25}{36}$
Neither die shows six.
$P(\text{at least one six})=1-\tfrac{25}{36}=\tfrac{11}{36}$
Complement.

Answer

\frac16

;

\frac{11}{36}

✓ CheckCount directly: 6 in the six-row + 6 in the six-column − 1 double-counted = 11 ✓.

2. The addition rule

30 students: 18 do Maths, 12 do Physics, 7 do both

Fill the overlap first (7), then "Maths only"

=18-7=11

, "Physics only"

=12-7=5

, neither

=30-23=7

P(A\cup B)=P(A)+P(B)-P(A\cap B)

$n(A)+n(B)\text{ counts the overlap twice}$
Outcomes in both are in each count.
$n(A\cup B)=n(A)+n(B)-n(A\cap B)$
Remove one copy.
$\div n(S)$
Convert to probabilities.

Worked example30 students: 18 Maths, 12 Physics, 7 both

P(Maths or Physics)? P(exactly one)?

Goal: Two probabilities.
Strategy: Venn diagram, overlap first.

$M\text{ only}=11,\ \text{both}=7,\ P\text{ only}=5,\ \text{neither}=7$
Fill from the middle.
$P(M\cup P)=\tfrac{23}{30}$
$\frac{18+12-7}{30}$ .
$P(\text{exactly one})=\tfrac{16}{30}=\tfrac8{15}$
$11+5$ .

Answer

\frac{23}{30}

;

\frac8{15}

✓ Check

11+7+5+7=30

✓.

Pause and think

$P(A)=0.5$ , $P(B)=0.4$ , $P(A\cup B)=0.7$ . $P(A\cap B)$ ?

Summary

Count equally likely outcomes.
Complement for "at least".
Addition rule subtracts the overlap.

Sample space: All possible outcomes.
Complement: $A'$ : not $A$ .

HistoryA gambler's problem. Probability theory began with gambling. Gerolamo Cardano wrote a book on dice games around 1564, and in 1654 Blaise Pascal and Pierre de Fermat exchanged letters about the "problem of points": how to split the stakes fairly when a game is interrupted. Their solution is usually taken as the birth of modern probability.

Fun factThe birthday surprise. In a room of just

23

people, the probability that two share a birthday is over

50\%

; with

70

people it is over

99.9\%

. It feels wrong because we think of matching our own birthday, but there are

253

different pairs among

23

people, each one a chance of a match.

Section 3 of 14

Key formulae

Equally likely outcomes

P(A)=\frac{n(A)}{n(\text{total})}

Complement

P(A')=1-P(A)

Addition rule

P(A\cup B)=P(A)+P(B)-P(A\cap B)

At least one

P(\text{at least one})=1-P(\text{none})

TrickFill the Venn diagram from the middle. Put

P(A\cap B)

in the overlap first. Then "

A

only" is

P(A)-P(A\cap B)

, "

B

only" is

P(B)-P(A\cap B)

, and the outside is whatever is left to make

1

. Almost every Venn question becomes reading off regions.

Section 4 of 14

Worked examples

Example 1 · Sample space0 / 2 steps

Two fair dice are thrown. Find the probability that the product of the scores is even.

Example 2 · Venn diagram0 / 2 steps

30

students,

18

study French,

12

study Spanish and

5

study both. A student is chosen at random. Find the probability that they study neither.

Example 3 · Using the addition rule0 / 2 steps

P(A)=0.5

P(B)=0.35

and

P(A\cup B)=0.7

. Find

P(A\cap B)

and

P(A\cap B')

Section 5 of 14

Explore the graph

At least one six

Loading the interactive graph…

Try this

The curve is $P(\text{at least one six in }n\text{ throws})=1-\left(\tfrac56\right)^n$ .
How many throws are needed before a six is more likely than not? (Where does the curve cross $0.5$ ?)
The 17th-century gambler the Chevalier de Méré won money betting on "at least one six in $4$ throws". Check the point at $k=4$ .

"At least one" is easiest through the complement: $1-P(\text{none})$ .

In the real worldWhat "70% chance of rain" means. A forecast of a

70\%

chance of rain means that, on days when the weather conditions look like today's, rain fell on about

70\%

of them. Forecasters check themselves this way: over many "

70\%

days", it should rain on about seven in ten.

Section 6 of 14

Common mistakes

Incorrect

P(A\cup B)=P(A)+P(B)

always

CorrectSubtract

P(A\cap B)

Otherwise the overlap is counted twice.

IncorrectPutting

P(A)

in the "

A

only" region

Correct

P(A)-P(A\cap B)

P(A)

is the whole circle, overlap included.

IncorrectAt least one six in

4

throws

=4\times\tfrac16

Correct

1-\left(\tfrac56\right)^4

Adding counts the outcomes with two or more sixes several times.

IncorrectProbabilities greater than

1

CorrectCheck every answer is in

[0,1]

A quick sanity check catches many slips.

Expectation

$P(A\cup B)=P(A)+P(B)$

Reality

The overlap, counted twice, getting a probability of $1.3$

Section 7 of 14

Exam tip

Exam tipDraw the diagram even when not asked. A quick Venn diagram or sample-space grid turns most early probability questions into reading off numbers. Fractions or exact decimals are fine; if you give a decimal that is not exact, give it to

3

significant figures.

Section 8 of 14

Guided practice

SA-9709-P5-PR-G012 marksL1 FoundationCalculatorBasic rules

P(A)=0.4

P(B)=0.5

and

P(A\cap B)=0.15

. Find

P(A\cup B)

Salam Academy Original Practice · Proposed mark scheme

Section 9 of 14

Practice questions

SA-9709-P5-PR1-0012 marksL1 FoundationCalculatorBasic rules

One card is drawn at random from a standard pack of

52

. Find the probability that it is a heart or a king.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P5-PR-0081 markL1 FoundationCalculatorBasic rules

Two fair dice are thrown. Find the probability that the total is

8

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P5-PR1-0041 markL1 FoundationCalculatorBasic rules

Two fair dice are thrown. Find the probability that the total is

7

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P5-PR1-0023 marksL1 FoundationCalculatorVenn diagrams

30

students,

18

study French,

12

study Spanish and

5

study both. A student is chosen at random. Find the probability that the student studies neither language.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P5-PR1-0033 marksL2 StandardCalculatorVenn diagrams

P(A)=0.5

P(B)=0.35

and

P(A\cup B)=0.7

. Find

P(A\cap B)

and

P(A\cap B')

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P5-PR1-0063 marksL2 StandardCalculatorVenn diagrams

P(A')=0.6

P(B)=0.3

and

P(A\cap B)=0.1

. Find

P(A\cup B)

and

P(A'\cap B')

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P5-PR1-0072 marksL2 StandardCalculatorBasic rules

Two fair dice are thrown. Find the probability that the product of the scores is even.

Salam Academy Original Practice · Proposed mark scheme

SA-9709-P5-PR-0132 marksL2 StandardCalculatorBasic rules

A fair die is thrown

4

times. Find the probability of at least one six, correct to 3 significant figures.

Salam Academy Original Practice · Proposed mark scheme

More questions on every skill in this subtopic are in the question bank under Practise.

Section 10 of 14

Challenge

SA-9709-P5-PR1-0083 marksL3 ExamCalculatorVenn diagrams

In a class of

50

20

play football,

15

play tennis and

8

play both. A pupil is chosen at random. Find the probability that the pupil plays exactly one of the two sports.

Salam Academy Original Practice · Proposed mark scheme

Section 11 of 14

Summary

Equally likely outcomes: count and divide.
$P(A')=1-P(A)$ ; at least one $=1-$ none.
$P(A\cup B)=P(A)+P(B)-P(A\cap B)$ .
Venn diagrams: fill the overlap first.

Section 12 of 14

Quick quiz

Q1 · SA-9709-P5-PR-Q031 markL1 FoundationCalculatorBasic rules

A

and

B

are mutually exclusive, then

Salam Academy Original Practice · Proposed mark scheme

Q2 · SA-9709-P5-PR1-0051 markL1 FoundationCalculatorBasic rules

A fair coin is tossed and a fair die is thrown. Find the probability of a head and an even number.

Salam Academy Original Practice · Proposed mark scheme

Section 13 of 14

Same topic, different board

Specification: IAL Mathematics
Where: WST01 (Statistics 1) · Probability
Expected: Venn diagrams (up to three events), the addition rule and set notation; probabilities from two-way tables.
Calculator: Permitted. Statistical tables are provided in the formula booklet.
Typical style: Venn diagrams with three events, and set notation such as P(A′ ∩ B), are common.
Open in: YMA01 · WST01 9MA0 · Paper 3

Section 14 of 14

Past paper questions

4 questions (32 marks) from real exams test this subtopic. Try each one under timed conditions, then mark it with the official mark scheme.

WST01/01 · June 2025Question 44 marks
Union of events and conditional probability.
Question paper, page 10 Mark scheme
WST01/01 · January 2025Question 49 marks
Three-set Venn diagram, a test for independence and conditional probability.
Question paper, page 12 Mark scheme
9709/52 · Oct/Nov 2024Question 16 marks
Two-way table: conditional probability, independence and a binomial probability.
Question paper, page 3 Mark scheme
WST01/01 · June 2024Question 613 marks
Venn diagrams, independence and conditional probability.
Question paper, page 22 Mark scheme

All YMA01 past paper questions by topic All 9709 past paper questions by topic

Salam Academy original content · Content status: reviewed and published. Syllabus mapping: Pearson Edexcel YMA01, 2018 specification (official specification).