Probability

Section S3

Years 1 and 2

3 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 9MA0 section S3

Checked against Edexcel 9MA0 section S3. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Mathematics (9MA0) specification; registry verification recorded 11 July 2026.

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S3.1

Year 1

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

Notes

Worked answers & exam appearances →

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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.

Tier 1 · Easy

ORIGINAL

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)

Tier 2 · Standard

ORIGINAL

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 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

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 1 is 5 marks)

Your progress and exam materials

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Answer conventions

Follow the wording on the question and its mark scheme. awrt means an appropriately rounded value is accepted; an exact answer must stay as a fraction, surd, logarithm or multiple of π when required, and a rounded decimal may be disallowed. Include requested units and forms. A cso tag protects that accuracy mark, while earlier method marks follow the question-specific dependencies.

S3.2

Year 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).

Notes

Worked answers & exam appearances →

Evidence from your answers: none yet

Your confidence:

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.

Tier 1 · Easy

ORIGINAL

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)

Tier 2 · Standard

ORIGINAL

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 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

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 1 is 5 marks)

S3.3

Year 2

Modelling with probability, including critiquing assumptions made and the likely effect of more realistic assumptions.

Notes

Worked answers & exam appearances →

Evidence from your answers: none yet

Your confidence:

Explanation

A probability model simplifies a real process by specifying possible outcomes and assigning probabilities to them.
State assumptions explicitly, such as independence, constant probabilities, identical trials or equally likely outcomes, and judge them in context.
More realistic dependence or changing probabilities can alter both central probabilities and tail risks, so identify the likely direction of the effect where possible.
Validate a model by comparing its predictions with observed data; a close fit in one sample does not prove that its assumptions are true.

Worked example

A factory model treats defects in two items from the same batch as independent, each with probability $0.03$ . It therefore predicts probability $0.03^2$ that both are defective. Explain how batch-to-batch variation is likely to affect this prediction.

1.The model gives $P(\text{both})=0.03^2=0.0009$ .
2.If an unobserved batch condition raises the defect probability for both items, learning that one is defective increases the chance that the other is defective.
3.This positive dependence makes the simple independent model likely to underestimate the joint probability.

Answer: Items from the same poor-quality batch are positively associated rather than independent.; Therefore two defects together are likely to occur more often than the modelled probability $0.0009$ .

Common mistakes

Don't assume repeated trials are independent although the first outcome changes the conditions for later trials.
Don't accept independence because it simplifies calculation without considering shared batch conditions.

Exam tip

Critique a probability model by naming the dependence mechanism and stating the likely direction of bias.

Tier 1 · Easy

ORIGINAL

A model assigns probability

1/200

to each ticket winning a draw. State one assumption behind this model and one reason it might fail.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

A weather model assumes that rain on successive days is independent and that the probability of rain each day is

0.2

. Find its probability of five consecutive dry days, then give one reason the independence assumption may be unrealistic and one possible refinement.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

A transport model assumes that each of

5

commuters independently chooses route A with probability

0.4

. Calculate the modelled probability that all

5

choose route A. A closure on route B can influence every commuter on the same day. Critique the independence assumption and state the likely effect on the probability just calculated.

(5)

(Total for Question 1 is 5 marks)

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