Notes · StatisticsUK · A-Levels

Probability

This topic develops the language and laws of probability: sample spaces and events, the addition and multiplication laws, conditional probability and independence. It uses Venn diagrams, two-way tables and probability trees as tools, and builds to the law of total probability, which underpins later work on Bayes' theorem and the distributions.

5 sections~12 min reading time3 competenciesLevel Foundation 1 · Standard 3 · Advanced 1

T·0333 / 18

Exam profile

AO1 · Apply the probability laws, conditional probability and set operations correctlyAO2 · Represent situations with Venn diagrams, tables and trees and interpret themAO3 · Solve multi-stage and conditional probability problems in context

Operators:calculatefinddetermineshow thatexplaininterpret

basic level

The AS foundation expects the addition and multiplication laws, simple conditional probability, and Venn and tree diagrams.

higher level

The full A-Level expects fluent conditional reasoning, independence tests, and the law of total probability across multi-stage problems.

Depth

Text

Contents · 5 sections

Probability

Sample spaces, events and the addition law#

FoundationPearson Edexcel 9ST0, Topic 2 (Paper 1)

The sample space

S

is the set of all possible outcomes of an experiment, and an event is a subset of

S

. When outcomes are equally likely, the probability of an event is

\frac{\text{number of favourable outcomes}}{\text{total number of outcomes}}

. Every probability satisfies

0 \leq P(A) \leq 1

, the whole sample space has probability

P(S) = 1

, and the complement satisfies

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

, a relationship that often turns an awkward calculation ('at least one') into an easy one.

For events that are not necessarily exclusive, the addition law is

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

. The subtraction corrects for the overlap

A \cap B

, which would otherwise be counted twice; forgetting it is the single most common error in this topic. The union

A \cup B

is the event that

A

B

(or both) occurs, and the intersection

A \cap B

is the event that both occur.

Two events are mutually exclusive when they cannot occur together, so

A \cap B = \varnothing

and

P(A \cap B) = 0

; for such events the addition law reduces to

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

. Recognising exclusivity is what licenses the simpler form, and wrongly assuming it is a frequent slip.

When outcomes are not equally likely, probabilities are estimated by relative frequency,

\frac{\text{number of times the event occurred}}{\text{number of trials}}

, which stabilises towards the true probability as the number of trials grows. This empirical view links probability to the data and diagrams of the previous topic.

P(A \cup B) = P(A) + P(B) - P(A \cap B), \qquad P(A') = 1 - P(A)

Addition law and complement

Subtract the overlap once; the complement often simplifies 'at least one' problems.

Worked example

Applying the addition law

Given $P(A) = 0.45$ , $P(B) = 0.30$ and $P(A \cap B) = 0.15$ , find $P(A \cup B)$ and $P(A' \cap B')$ .

Union
$P(A \cup B) = 0.45 + 0.30 - 0.15 = 0.60$ .
$P(A \cup B) = 0.45 + 0.30 - 0.15 = 0.60$
Complement of the union
$A' \cap B'$ is 'neither $A$ nor $B$ ', which is the complement of $A \cup B$ .
Compute
$P(A' \cap B') = 1 - P(A \cup B) = 1 - 0.60 = 0.40$ .

Result: $P(A \cup B) = 0.60$ and $P(A' \cap B') = 0.40$ .

Typical mistakes

Omitting the $-P(A \cap B)$ term in the addition law for non-exclusive events.
Assuming events are mutually exclusive without justification.

Active revision

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

Active recall

Recall the key points — then reveal.

Sources: Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification (Pearson Edexcel)

Conditional probability and the multiplication law#

StandardPearson Edexcel 9ST0, Topic 2 (Paper 1)

Two-way table of sex and handedness

Fig. 1Conditional probabilities are read using row or column totals, e.g. P(left-handed given male) = 9/52.

The conditional probability

P(A \mid B)

is the probability of

A

given that

B

has occurred, defined by

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

provided

P(B) > 0

. Conditioning on

B

restricts attention to the outcomes in

B

and asks what proportion of them also lie in

A

; this is why a two-way table, where you simply read along the relevant row or column, is such a natural tool.

Rearranging the definition gives the multiplication law

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

. This is the engine of tree diagrams: the probability of a path is the product of the conditional probabilities along its branches, and each branching set sums to 1. The two forms make clear that

P(A \mid B)

and

P(B \mid A)

are generally different, a distinction that becomes decisive in Bayes' theorem.

In the two-way table below, of 100 people classified by sex and handedness, the conditional probability that a person is left-handed given that they are male is read directly:

P(\text{left} \mid \text{male}) = \frac{9}{52}

, using the male row total as the new denominator. Contrast this with

P(\text{male} \mid \text{left}) = \frac{9}{13}

, which conditions on the left-handed column instead.

The order of conditioning must match the wording exactly. 'The probability that a randomly chosen left-hander is male' conditions on being left-handed; 'the probability that a randomly chosen male is left-handed' conditions on being male. Reading the condition carefully, and choosing the correct denominator, is where most marks are won or lost in this topic.

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

Conditional probability and the multiplication law

Conditioning on $B$ rescales by $P(B)$ ; the multiplication law gives path probabilities on a tree.

Worked example

Conditional probability from a table

Using the sex-and-handedness table, find the probability that a randomly chosen person is left-handed given that they are male, and comment on whether it differs from the overall left-handed rate.

Restrict to males
There are 52 males, of whom 9 are left-handed.
Conditional probability
$P(\text{left} \mid \text{male}) = \frac{9}{52} = 0.173$ (3 s.f.).
$P(\text{left} \mid \text{male}) = \frac{9}{52} = 0.173$
Overall rate
The overall left-handed rate is $\frac{13}{100} = 0.13$ .
Comment
The male left-handed rate (0.173) exceeds the overall rate (0.13), so sex and handedness are not independent in this sample.

Result: $P(\text{left} \mid \text{male}) = \frac{9}{52} = 0.173$ , which exceeds the overall rate 0.13, suggesting dependence.

Typical mistakes

Confusing $P(A \mid B)$ with $P(B \mid A)$ — they use different denominators.
Dividing by the grand total instead of the conditioning total when finding a conditional probability.

Active revision

Using the table above, find $P(\text{male} \mid \text{left-handed})$ and $P(\text{right-handed} \mid \text{female})$ .

Active recall

Recall the key points — then reveal.

Sources: Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification (Pearson Edexcel)

Independence and mutually exclusive events#

StandardPearson Edexcel 9ST0, Topic 2 (Paper 1)

Two events are independent when the occurrence of one does not change the probability of the other, so

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

. Substituting into the multiplication law gives the working test for independence:

A

and

B

are independent if and only if

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

. This product form is the one to check, because it is symmetric in

A

and

B

and needs only the three probabilities.

Independence and mutual exclusivity are opposite ideas and must never be confused. Mutually exclusive events cannot occur together (

P(A \cap B) = 0

), so knowing that one happened tells you the other did not — they are strongly dependent. Independent events (with non-zero probabilities) CAN occur together, and in fact must, since

P(A \cap B) = P(A)P(B) > 0

. Two events with non-zero probability cannot be both mutually exclusive and independent.

To test independence in practice, compute

P(A)P(B)

and compare it with

P(A \cap B)

: equality means independent, inequality means dependent. This is a frequent exam task, often set from a Venn diagram or a two-way table, and the conclusion must be stated explicitly with the comparison shown.

For a sequence of independent events the probabilities simply multiply, which is what makes repeated-trial models such as the binomial possible. Sampling with replacement gives independence between draws; sampling without replacement does not, because each draw changes the composition of what remains — a distinction that decides whether a binomial model is appropriate later.

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

Test for independence

Equivalently $P(A \mid B) = P(A)$ ; contrast with mutually exclusive, where $P(A \cap B) = 0$ .

Worked example

Testing independence

For events with $P(A) = 0.45$ , $P(B) = 0.30$ and $P(A \cap B) = 0.15$ , determine whether $A$ and $B$ are independent.

Compute the product
$P(A)\,P(B) = 0.45 \times 0.30 = 0.135$ .
Compare
$P(A \cap B) = 0.15 \neq 0.135$ .
$0.15 \neq 0.45 \times 0.30 = 0.135$
Conclude
Since the intersection probability differs from the product, $A$ and $B$ are not independent.

Result: Because $P(A \cap B) = 0.15 \neq 0.135 = P(A)P(B)$ , the events are dependent.

Typical mistakes

Treating mutually exclusive events as independent; they are the opposite (strongly dependent).
Multiplying probabilities that are not independent, especially when sampling without replacement.

Active revision

Events $A$ and $B$ satisfy $P(A) = 0.5$ , $P(B) = 0.3$ and $P(A \cap B) = 0.2$ . Determine whether $A$ and $B$ are independent.

Active recall

Recall the key points — then reveal.

Sources: Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification (Pearson Edexcel)

Venn diagrams and set operations#

StandardPearson Edexcel 9ST0, Topic 2 (Paper 1)

Venn diagram of two events

Fig. 2Survey of 100 students: 30 study only Mathematics, 15 study only Physics, 15 study both; 40 study neither (outside both circles).

A Venn diagram shows the sample space as a rectangle and events as overlapping regions, so that set operations become areas: the union

A \cup B

is everything in either circle, the intersection

A \cap B

is the overlap, and the complement

A'

is everything outside

A

. Filling the diagram from the inside out — the intersection first, then the parts of each circle outside it, then the region outside both — keeps the numbers consistent.

The diagram below records a survey of 100 students who may study Mathematics (

A

) or Physics (

B

): 15 study both, 30 study only Mathematics, 15 study only Physics, and the remaining 40 study neither. From these region counts every probability follows by division by 100, and the addition law can be checked directly against the diagram.

Probabilities read straight from the regions:

P(A) = \frac{45}{100} = 0.45

(the whole Maths circle),

P(A \cap B) = \frac{15}{100} = 0.15

, and

P(A \cup B) = \frac{60}{100} = 0.60

, which agrees with

0.45 + 0.30 - 0.15

. Conditional probabilities also come straight from the diagram by re-basing on the conditioning region, for example

P(B \mid A) = \frac{15}{45} = \frac{1}{3}

For three events the diagram has eight regions and the same inside-out method applies, starting from the central triple intersection. Venn diagrams are especially powerful for 'exactly one', 'at least two' and complement questions, which are cumbersome algebraically but immediate once the regions are filled.

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

Addition law on a Venn diagram

Each region is counted once; the overlap is subtracted to avoid double counting.

Worked example

Reading a Venn diagram

Using the survey of 100 students (only Maths 30, only Physics 15, both 15, neither 40), find $P(\text{exactly one subject})$ and $P(A \cup B)$ .

Exactly one
Exactly one means only Maths or only Physics: $30 + 15 = 45$ students.
Probability
$P(\text{exactly one}) = \frac{45}{100} = 0.45$ .
Union
$A \cup B$ is anyone studying at least one subject: $30 + 15 + 15 = 60$ , so $P(A \cup B) = 0.60$ .
$P(A \cup B) = \frac{30+15+15}{100} = 0.60$

Result: $P(\text{exactly one}) = 0.45$ and $P(A \cup B) = 0.60$ .

Typical mistakes

Placing a total in a circle instead of splitting it into the 'only' and 'both' regions.
Forgetting the outside region (neither event), so the region counts do not sum to the total.

Active revision

In the survey above, find the probability that a student chosen at random studies exactly one of the two subjects.

Active recall

Recall the key points — then reveal.

Sources: Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification (Pearson Edexcel)

Tree diagrams and the law of total probability#

AdvancedPearson Edexcel 9ST0, Topic 2 (Paper 1)

Probability tree for two machines

Fig. 3Path products give P(A and defective) = 0.03 and P(B and defective) = 0.032; the total defective rate is 0.062.

A probability tree lays out a multi-stage experiment as branches. The probabilities on the branches leaving any node sum to 1, and the probability of a complete path from root to leaf is the product of the branch probabilities along it — a direct application of the multiplication law. Trees are the clearest way to organise problems with conditional stages, such as drawing without replacement or a two-machine production process.

The tree below models components made by two machines: machine A makes 60% of output with a 5% defective rate, machine B makes 40% with an 8% defective rate. Multiplying along each path gives

P(A \cap \text{def}) = 0.6 \times 0.05 = 0.03

and

P(B \cap \text{def}) = 0.4 \times 0.08 = 0.032

, and the four path probabilities sum to 1, a useful check.

The law of total probability collects the paths leading to an event. If the events

A

and

A'

partition the sample space, then

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

; more generally the sum runs over any set of mutually exclusive, exhaustive events. Here the total defective rate is

0.03 + 0.032 = 0.062

, obtained by adding the probabilities of every path that ends in a defect.

This decomposition is the foundation of Bayes' theorem in the next topic: once the total probability of the observed event is known, a reverse conditional such as

P(A \mid \text{def})

can be found. Mastery here is therefore not an end in itself but the machinery on which the whole of statistical inference by conditioning is built.

P(B) = P(B \mid A)\,P(A) + P(B \mid A')\,P(A')

Law of total probability

Sum the probabilities of every path (through a partition of the sample space) that reaches $B$ .

Worked example

Total probability of a defect

Machine A makes 60% of components (5% defective) and machine B makes 40% (8% defective). Find the probability that a randomly chosen component is defective.

Path through A
$P(A \cap \text{def}) = 0.6 \times 0.05 = 0.03$ .
Path through B
$P(B \cap \text{def}) = 0.4 \times 0.08 = 0.032$ .
Total probability
$P(\text{def}) = 0.03 + 0.032 = 0.062$ .
$P(\text{def}) = 0.6(0.05) + 0.4(0.08) = 0.062$
Interpret
About 6.2% of all components are defective across the two machines.

Result: The total probability that a component is defective is 0.062 (6.2%).

Typical mistakes

Adding branch probabilities along a path instead of multiplying them.
For sampling without replacement, reusing the first-stage probabilities on the second stage.

Active revision

Using the two-machine tree, find the total probability that a randomly chosen component is defective.

Active recall

Recall the key points — then reveal.

Sources: Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification (Pearson Edexcel)

References & sources

Sources

Pearson Edexcel

Pearson Edexcel Level 3 Advanced GCE in Statistics (9ST0) Specification

Probability

5 sections~12 min reading time3 competenciesLevel Foundation 1 · Standard 3 · Advanced 1

Probability

Key points

Applying the addition law

Key points

Conditional probability from a table

Key points

Testing independence

Key points

Reading a Venn diagram

Key points

Total probability of a defect

Sources

Probability

Key points

Applying the addition law

Key points

Conditional probability from a table

Key points

Testing independence

Key points

Reading a Venn diagram

Key points

Total probability of a defect

Sources