EnglandMathsSyllabus dot point

How do you calculate the likelihood of events, including when events depend on or exclude one another?

Probability of events, mutually exclusive and independent events, the addition and multiplication rules, Venn diagrams and tree diagrams, and conditional probability.

A focused answer to the AQA A-Level Mathematics probability content, covering single and combined events, mutually exclusive and independent events, the addition and multiplication rules, Venn and tree diagrams, and conditional probability.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Basic probability
Mutually exclusive and independent events
Combining events
Conditional probability and tree diagrams
Testing for independence

What this dot point is asking

AQA wants you to calculate probabilities of single and combined events, distinguish mutually exclusive from independent events, use the addition and multiplication rules, represent situations with Venn and tree diagrams, and calculate conditional probabilities. Probability appears on Paper 3 and underpins both the binomial and the hypothesis-testing topics, so the foundations here matter beyond their own marks.

Basic probability

For equally likely outcomes, $P(\text{event}) = \dfrac{\text{number of favourable outcomes}}{\text{total number of outcomes}}$ . Every probability satisfies $0 \le P \le 1$ , and the probabilities of all distinct outcomes of an experiment sum to $1$ . The complement rule, $P(\text{not } A) = 1 - P(A)$ , is often the quickest route, especially for "at least one" questions where the complement is "none".

Mutually exclusive and independent events

These two ideas are frequently confused but are entirely different.

Combining events

Venn diagrams make the addition rule visual: the union is everything inside either circle, and the intersection is the overlap. A reliable method is to fill the intersection first, then subtract it from each event total to find the "only" regions, then place "neither" outside so all four regions sum to one.

Conditional probability and tree diagrams

Without-replacement tree diagram

A bag has $3$ red and $2$ blue counters. Two are drawn without replacement. Find the probability that both are red, and the probability that exactly one is red.

Step 1: First-draw probabilities

$P(\text{red first}) = \frac{3}{5}$ and $P(\text{blue first}) = \frac{2}{5}$ .

Step 2: Second-draw probabilities (reduced totals)

After a red is removed, $4$ counters remain with $2$ red: $P(\text{red second} \mid \text{red first}) = \frac{2}{4}$ . After a blue, $3$ red remain in $4$ : $P(\text{red second} \mid \text{blue first}) = \frac{3}{4}$ .

Step 3: Multiply along the both-red branch

$P(\text{both red}) = \frac{3}{5} \times \frac{2}{4} = \frac{6}{20} = \frac{3}{10}$ .

Step 4: Add the two exactly-one-red paths

Red then blue is $\frac{3}{5}\times\frac{2}{4} = \frac{6}{20}$ ; blue then red is $\frac{2}{5}\times\frac{3}{4} = \frac{6}{20}$ . So $P(\text{exactly one red}) = \frac{12}{20} = \frac{3}{5}$ .

Testing for independence

To test whether $A$ and $B$ are independent, check whether $P(A \cap B) = P(A)\,P(B)$ , or equivalently whether $P(A \mid B) = P(A)$ . If the two sides differ, the events are dependent. This is a common exam instruction ("determine whether the events are independent") and requires you to show both quantities and compare them explicitly.

Exam-style practice questions

Practice questions written in the style of AQA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AQA 20186 marksPaper 3, Section A. In a sixth form,

60

percent of students study Mathematics,

35

percent study Physics, and

25

percent study both. A student is chosen at random. (a) Draw a Venn diagram to represent this information. (b) Find the probability that the student studies Mathematics or Physics or both. (c) Given that the student studies Mathematics, find the probability they also study Physics.

Show worked answer →

Convert to probabilities: $P(M) = 0.6$ , $P(P) = 0.35$ , $P(M \cap P) = 0.25$ . In the Venn diagram the intersection is $0.25$ , Mathematics only is $0.6 - 0.25 = 0.35$ , Physics only is $0.35 - 0.25 = 0.10$ , and neither is $1 - (0.35 + 0.25 + 0.10) = 0.30$ . For (b), $P(M \cup P) = 0.6 + 0.35 - 0.25 = 0.70$ . For (c), $P(P \mid M) = \frac{P(M \cap P)}{P(M)} = \frac{0.25}{0.6} \approx 0.417$ . Markers reward a fully labelled Venn diagram whose regions sum to one, correct use of the addition rule, and the conditional probability formula.

AQA 20225 marksPaper 3, Section B. A box contains

7

red and

3

green counters. Two counters are drawn at random without replacement. (a) Draw a tree diagram for the two draws. (b) Calculate the probability that the two counters are different colours. (c) Given that the second counter is green, calculate the probability that the first was red.

Show worked answer →

For (a) the first-draw branches are $\frac{7}{10}$ red and $\frac{3}{10}$ green; second-draw branches use the reduced denominator $9$ . For (b), different colours means red then green or green then red: $\frac{7}{10}\times\frac{3}{9} + \frac{3}{10}\times\frac{7}{9} = \frac{21}{90} + \frac{21}{90} = \frac{42}{90} = \frac{7}{15}$ . For (c), $P(R_1 \mid G_2) = \frac{P(R_1 \cap G_2)}{P(G_2)}$ . Here $P(R_1 \cap G_2) = \frac{7}{10}\times\frac{3}{9} = \frac{21}{90}$ and $P(G_2) = \frac{21}{90} + \frac{3}{10}\times\frac{2}{9} = \frac{21}{90} + \frac{6}{90} = \frac{27}{90}$ , so $P(R_1 \mid G_2) = \frac{21}{27} = \frac{7}{9}$ . Markers reward correct without-replacement denominators and the conditional formula.

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Sources & how we know this

AQA A-level Mathematics (7357) specification — AQA (2017)