S3 Probability — notes on one page

Every Edexcel A-level Maths 9MA0 specification point in S3, read straight through: 3 points with definitions, explanations, mark-scheme wording and worked examples. Practice questions are in the printable question pack.

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.

How this checking works

S3.1 · Understand and use mutually exclusive and independent events when calculating probabilities; link to discrete and continuous distributions.

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 $P(A\cap B)=P(A)P(B)$ .

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)

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.

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.

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).

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.

S3.3 · Modelling with probability, including critiquing assumptions made and the likely effect of more realistic assumptions.

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.