Notes · MathematicsUK · A-Levels

Probability

Probability quantifies uncertainty. The A-Level covers the probability of events, mutually exclusive and independent events, the addition and multiplication rules, Venn diagrams and set notation, conditional probability and its formula, and tree diagrams. These tools support the distributions and hypothesis-testing topics that follow.

4 sections~10 min reading time3 competenciesLevel Foundation 1 · Standard 2 · Advanced 1

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Exam profile

AO1 · Calculate probabilities using the addition, multiplication and conditional rulesAO2 · Reason about independence and conditional probabilityAO3 · Model situations with Venn and tree diagrams

Operators:findcalculateshow thatdetermineexplain

basic level

AS-Level covers probability of events, mutually exclusive and independent events, Venn and tree diagrams.

higher level

The full A-Level adds conditional probability, the conditional-probability formula and set notation for events.

Depth

Text

Contents · 4 sections

Probability

Events, sample spaces and the addition rule#

FoundationDfE GCE Mathematics — probabilityAQA 7357 (Statistics)

Sample space for two dice

Fig. 1The 36 equally likely outcomes; the total 7 occurs 6 times, so

P(7) = \tfrac{6}{36} = \tfrac{1}{6}

The sample space is the set of all possible outcomes of an experiment, and an event is a subset of the sample space. For equally likely outcomes, the probability of an event is

P(A) = \dfrac{\text{number of outcomes in } A}{\text{total number of outcomes}}

. Every probability lies between 0 and 1, the probabilities of all outcomes sum to 1, and the complement satisfies

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

, a relation that often turns a hard 'at least one' calculation into an easy one.

Two events are mutually exclusive if they cannot both occur, so

P(A \cap B) = 0

. For such events the addition rule simplifies to

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

. This is the case of rolling a 2 or a 5 on a die, where the outcomes do not overlap and their probabilities simply add.

In general, events can overlap, and the general addition rule is

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

. The subtraction corrects for the overlap, which would otherwise be counted twice — the probability that a card is a heart or a king is

P(\text{heart}) + P(\text{king}) - P(\text{heart and king})

, subtracting the king of hearts once. Recognising whether events overlap decides which form of the rule you use.

A sample space diagram (a grid of outcomes) is a reliable way to list equally likely outcomes for two combined experiments, such as rolling two dice. Counting favourable outcomes from the grid then gives probabilities directly, and it makes clear which events are mutually exclusive and which overlap.

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

The general addition rule

Subtract the overlap so it is not counted twice; for mutually exclusive events $P(A \cap B) = 0$ .

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

The complement

Useful for 'at least one' problems, where the complement 'none' is easier to compute.

Worked example

The addition rule with overlap

In a group, $P(\text{studies French}) = 0.4$ , $P(\text{studies German}) = 0.3$ and $P(\text{both}) = 0.1$ . Find the probability that a student studies French or German.

Identify the overlap
The events overlap, with $P(F \cap G) = 0.1$ .
Apply the general addition rule
$P(F \cup G) = 0.4 + 0.3 - 0.1$ .
$P(F \cup G) = 0.4 + 0.3 - 0.1 = 0.6$
State
$P(F \cup G) = 0.6$ .

Result: $P(\text{French or German}) = 0.6$ .

Typical mistakes

Adding $P(A)$ and $P(B)$ for overlapping events without subtracting $P(A \cap B)$ .
Giving a probability outside the range 0 to 1.

Active revision

A card is drawn from a standard pack. Find the probability that it is a face card (jack, queen or king) or a spade.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Venn diagrams and set notation#

StandardDfE GCE Mathematics — probabilityAQA 7357 (Statistics)

A two-event Venn diagram

Fig. 2Of a group, 12 are in A only, 8 in B only, 5 in both; the 5 in the overlap is placed first.

A Venn diagram represents events as overlapping regions inside a rectangle (the sample space), and it is the clearest way to organise information about how events combine. The notation is standard set theory:

A \cap B

(intersection, 'A and B'),

A \cup B

(union, 'A or B'),

A'

(complement, 'not A'), and the empty set for impossible events. Filling in a Venn diagram from given probabilities, working from the middle out, is a core technique.

The reliable method is to place the probability of the intersection first, then subtract to find the parts of each event that do not overlap, and finally deduce the region outside all events so that the total is 1. Once the diagram is complete, any required probability is read off by adding the appropriate regions, which removes almost all the risk of double-counting.

Set notation also expresses conditional and combined events precisely. For example

P(A \cap B')

is the probability of A but not B, and

P((A \cup B)')

is the probability of neither. Translating a wordy description ('passed maths but not physics') into set notation, and back, is examined and underpins accurate calculation.

Venn diagrams extend to three events, dividing the space into eight regions, and are especially useful for problems involving three overlapping categories. They also make independence and conditional probability visual: a conditional probability

P(A \mid B)

restricts attention to the part of the diagram inside

B

, which is the intuition behind its formula.

A \cap B\ (\text{and}), \quad A \cup B\ (\text{or}), \quad A'\ (\text{not } A)

Set notation for events

Intersection, union and complement describe how events combine on a Venn diagram.

Worked example

Completing a Venn diagram

For events with $P(A) = 0.5$ , $P(B) = 0.4$ and $P(A \cap B) = 0.2$ , find $P(A' \cap B')$ .

Place the intersection
$P(A \cap B) = 0.2$ in the overlap.
Find the 'only' regions
A only $= 0.5 - 0.2 = 0.3$ ; B only $= 0.4 - 0.2 = 0.2$ .
Deduce the outside
Total inside $= 0.3 + 0.2 + 0.2 = 0.7$ , so $P(A' \cap B') = 1 - 0.7 = 0.3$ .
$P(A' \cap B') = 1 - 0.7 = 0.3$

Result: $P(A' \cap B') = 0.3$ .

Typical mistakes

Placing the full $P(A)$ in the 'A only' region instead of subtracting the overlap.
Forgetting the region outside all events when checking the total is 1.

Active revision

In a class of 30, 18 study maths, 15 study physics and 10 study both. Draw a Venn diagram and find the probability that a randomly chosen student studies neither.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Conditional probability#

AdvancedDfE GCE Mathematics — probabilityAQA 7357 (Statistics)

A two-way table

Fig. 3From the table,

P(\text{passed} \mid \text{female}) = \tfrac{35}{50} = 0.7

, restricting to the 50 females.

Conditional probability is the probability of one event given that another has occurred, written

P(A \mid B)

('the probability of A given B'). Conditioning on

B

restricts attention to the outcomes in

B

, so the definition is

P(A \mid B) = \dfrac{P(A \cap B)}{P(B)}

, provided

P(B) \neq 0

. It is the fraction of

B

's probability that also lies in

A

Rearranging the definition gives the multiplication rule

P(A \cap B) = P(B)\,P(A \mid B)

, which is exactly what the branches of a tree diagram multiply. This is the general form; only when the events are independent does it reduce to

P(A \cap B) = P(A)\,P(B)

. Keeping the general form in mind prevents the common error of assuming independence.

Two events are independent when the occurrence of one does not change the probability of the other, i.e.

P(A \mid B) = P(A)

, equivalently

P(A \cap B) = P(A)\,P(B)

. Testing this equality is the standard way to decide, in a calculation, whether two events are independent — you compute both sides and compare.

Two-way tables and Venn diagrams are the natural tools for conditional problems, because you can read the restricted totals directly. Given a table of frequencies,

P(A \mid B)

is simply the number in both

A

and

B

divided by the total in

B

— a concrete version of the formula that many candidates find clearer.

P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \qquad P(A \cap B) = P(B)\,P(A \mid B)

Conditional probability and the multiplication rule

Conditioning restricts to $B$ ; rearranging gives the multiplication rule used on tree diagrams.

A, B \text{ independent} \iff P(A \cap B) = P(A)\,P(B)

Test for independence

Compute both sides and compare; equality means the events are independent.

Worked example

Conditional probability from a table

Using the two-way table (40, 10 for males; 35, 15 for females), find the probability that a person passed given they are male, and test whether passing and being male are independent.

Conditional probability
$P(\text{pass} \mid \text{male}) = \dfrac{40}{50} = 0.8$ .
Marginal probabilities
$P(\text{pass}) = \dfrac{75}{100} = 0.75$ and $P(\text{male}) = 0.5$ .
Test independence
$P(\text{pass} \mid \text{male}) = 0.8 \neq 0.75 = P(\text{pass})$ , so the events are not independent.
$0.8 \neq 0.75$

Result: $P(\text{pass} \mid \text{male}) = 0.8$ ; since this differs from $P(\text{pass}) = 0.75$ , passing and being male are not independent.

Typical mistakes

Confusing $P(A \mid B)$ with $P(B \mid A)$ — the condition matters.
Assuming events are independent and multiplying probabilities without checking.

Active revision

In a survey, $P(\text{owns a car}) = 0.6$ , $P(\text{owns a bike}) = 0.5$ and $P(\text{owns both}) = 0.3$ . Find $P(\text{owns a bike} \mid \text{owns a car})$ and determine whether the two events are independent.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

Tree diagrams and independence#

StandardDfE GCE Mathematics — probabilityAQA 7357 (Statistics)

Tree diagram for two draws without replacement

Fig. 4Bag of 3 red, 2 blue: drawing two without replacement; the leaf products give each outcome's probability.

A tree diagram lays out a sequence of events as branches, with each branch labelled by a probability. The probabilities on the branches leaving any node must sum to 1, and the probability of a particular sequence of outcomes is the product of the probabilities along its path from the root. This 'multiply along the branches' rule is a direct application of the multiplication rule

P(A \cap B) = P(A)\,P(B \mid A)

To find the probability of an event that can happen in several ways, you add the path probabilities of all the routes that give it. So

P(\text{one red and one blue})

from two draws is the sum of the 'red then blue' and 'blue then red' paths. The combination of multiplying along paths and adding across paths handles the great majority of multi-stage probability problems.

The branch probabilities depend on whether the draws are with or without replacement. With replacement, the situation resets and successive events are independent, so the second-stage probabilities equal the first. Without replacement, the second-stage probabilities are conditional on the first outcome (the counts change), and the tree's later branches must be adjusted accordingly.

Tree diagrams also make conditional questions tractable: to find

P(\text{first was red} \mid \text{second was blue})

you form the conditional-probability quotient using path probabilities from the tree. Drawing the tree, labelling every branch, and being careful about replacement is the safest route through these questions.

P(\text{path}) = \prod (\text{branch probabilities along the path})

Multiplying along a tree

Multiply the branch probabilities from root to leaf; add across paths for combined events.

Worked example

Tree diagram without replacement

A bag has 3 red and 2 blue counters. Two are drawn without replacement. Find the probability of one of each colour.

First-stage probabilities
$P(R) = \tfrac{3}{5} = 0.6$ , $P(B) = \tfrac{2}{5} = 0.4$ .
Path probabilities
$P(RB) = 0.6 \times \tfrac{2}{4} = 0.3$ ; $P(BR) = 0.4 \times \tfrac{3}{4} = 0.3$ .
$P(RB) = 0.6 \times 0.5 = 0.3$
Add the two paths
$P(\text{one of each}) = P(RB) + P(BR) = 0.3 + 0.3 = 0.6$ .

Result: $P(\text{one of each colour}) = 0.6$ .

Typical mistakes

Keeping the first-stage probabilities on the second stage when sampling without replacement.
Multiplying when you should add (for alternative paths) or vice versa.

Active revision

A bag contains 5 red and 3 green counters. Two are drawn without replacement. Draw a tree diagram and find the probability that the two counters are different colours.

Active recall

Recall the key points — then reveal.

Sources: Mathematics: AS and A level content (GCE subject content) (Department for Education) · AQA A-level Mathematics 7357 specification (AQA)

References & sources

Sources

Department for Education

Mathematics: AS and A level content (GCE subject content)

AQA

AQA A-level Mathematics 7357 specification

Probability

4 sections~10 min reading time3 competenciesLevel Foundation 1 · Standard 2 · Advanced 1

Probability

Key points

The addition rule with overlap

Key points

Completing a Venn diagram

Key points

Conditional probability from a table

Key points

Tree diagram without replacement

Sources

Probability

Key points

The addition rule with overlap

Key points

Completing a Venn diagram

Key points

Conditional probability from a table

Key points

Tree diagram without replacement

Sources