Notes · StatisticsUK · A-Levels

Bayes' theorem

This topic uses the law of total probability and Bayes' theorem to reverse a conditional probability, updating a prior belief in the light of evidence to a posterior. It applies the method to diagnostic and screening problems, where the interplay of a low base rate with test accuracy produces the counter-intuitive but important base-rate effect.

4 sections~10 min reading time3 competenciesLevel Standard 1 · Advanced 3

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Exam profile

AO1 · Apply the law of total probability and Bayes' theoremAO2 · Interpret prior and posterior probabilities correctlyAO3 · Solve real diagnostic and screening problems and comment on the results

Operators:calculatefinddetermineshow thatinterpretcomment

basic level

The AS foundation expects the law of total probability and simple conditional reversals via a tree or table.

higher level

The full A-Level expects the formal statement of Bayes' theorem, prior/posterior language and the base-rate interpretation.

Depth

Text

Contents · 4 sections

Bayes' theorem

The law of total probability#

StandardPearson Edexcel 9ST0, Topic 8 (Paper 1)

Bayes' theorem is built on the law of total probability, which computes the overall probability of an event by summing over the ways it can occur. If the events

A_1, A_2, \ldots

are mutually exclusive and exhaustive (they partition the sample space), then

P(B) = \sum_i P(B \mid A_i)\,P(A_i)

. Each term is the probability of one route to

B

, and the routes are added.

The simplest and most common form uses a two-part partition,

A

and its complement

A'

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

. This is exactly the sum of the two tree paths that reach

B

, and it converts a conditional description of a process into the unconditional probability of the observed event.

In the diagnostic context that runs through this topic, let

D

be 'has the disease' with prevalence

P(D) = 0.01

, and let

+

be 'tests positive', with

P(+ \mid D) = 0.99

(the test's sensitivity) and

P(+ \mid D') = 0.05

(the false-positive rate). The total probability of a positive test is

P(+) = 0.99(0.01) + 0.05(0.99) = 0.0099 + 0.0495 = 0.0594

This total is the denominator of Bayes' theorem. Notice how the large group of healthy people (

P(D') = 0.99

) contributes many false positives (0.0495) despite the low individual false-positive rate, so that most positive results come from healthy people. Recognising which paths dominate the total is the key to the base-rate interpretation later.

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

Law of total probability (two-part partition)

Sum the probabilities of every route to $B$ through a partition of the sample space.

Worked example

Total probability of a positive test

A disease affects 1% of a population. A test gives a positive result for 99% of sufferers and 5% of non-sufferers. Find $P(+)$ .

Route through D
$P(+ \mid D)P(D) = 0.99 \times 0.01 = 0.0099$ .
Route through D'
$P(+ \mid D')P(D') = 0.05 \times 0.99 = 0.0495$ .
Total
$P(+) = 0.0099 + 0.0495 = 0.0594$ .
$P(+) = 0.99(0.01) + 0.05(0.99) = 0.0594$

Result: $P(+) = 0.0594$ , of which most (0.0495) comes from healthy people.

Typical mistakes

Using a partition whose parts are not exhaustive, so the routes do not cover every case.
Adding conditional probabilities without weighting them by the probabilities of the partition.

Active revision

A disease has prevalence 2%. A test is positive for 95% of sufferers and for 3% of non-sufferers. Find the overall probability of a positive test.

Active recall

Recall the key points — then reveal.

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

Bayes' theorem#

AdvancedPearson Edexcel 9ST0, Topic 8 (Paper 1)

Probability tree for a diagnostic test

Fig. 1Path products: P(D and +) = 0.0099 and P(D' and +) = 0.0495; so P(D given +) = 0.0099 / 0.0594 = 0.167.

Bayes' theorem reverses a conditional probability. Starting from the definition

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

and writing the numerator as

P(B \mid A)P(A)

and the denominator by the law of total probability, gives

P(A \mid B) = \dfrac{P(B \mid A)P(A)}{P(B \mid A)P(A) + P(B \mid A')P(A')}

. It converts the known direction

P(B \mid A)

into the wanted direction

P(A \mid B)

The tree below organises the diagnostic problem for direct use of the formula. The numerator is the single path

D \to +

, with probability

0.99 \times 0.01 = 0.0099

, and the denominator is the total probability of a positive test,

0.0594

, the sum of both positive paths. The posterior is

P(D \mid +) = \dfrac{0.0099}{0.0594} = 0.167

(3 s.f.).

Reading a Bayes problem from a tree is the reliable method: the numerator is the target path's product, and the denominator is the sum of ALL paths that produce the observed evidence. This makes the arithmetic mechanical once the tree is drawn, and it exposes exactly which competing explanations of the evidence are being weighed.

The result

P(D \mid +) = 0.167

is strikingly low given a 99% sensitive test: even after a positive result, the probability of disease is only about 1 in 6. This is not a paradox but a consequence of the low prevalence, and it is the reason Bayes' theorem is essential for interpreting screening tests honestly.

P(A \mid B) = \frac{P(B \mid A)P(A)}{P(B \mid A)P(A) + P(B \mid A')P(A')}

Bayes' theorem

The wanted conditional equals the target path over the total probability of the evidence.

Worked example

Reversing the conditional with Bayes

With $P(D) = 0.01$ , $P(+ \mid D) = 0.99$ and $P(+ \mid D') = 0.05$ , find the probability that a person who tests positive actually has the disease.

Target path
$P(D \cap +) = P(+ \mid D)P(D) = 0.99 \times 0.01 = 0.0099$ .
Total evidence
$P(+) = 0.0099 + 0.05 \times 0.99 = 0.0594$ .
Apply Bayes
$P(D \mid +) = \dfrac{0.0099}{0.0594} = 0.167$ (3 s.f.).
$P(D \mid +) = \frac{0.0099}{0.0594} = 0.167$
Interpret
Despite the positive result, there is only about a 16.7% chance the person has the disease.

Result: $P(D \mid +) = 0.167$ (3 s.f.), so a positive result implies only about a 1-in-6 chance of disease.

Typical mistakes

Putting only one positive path in the denominator instead of the total probability of the evidence.
Confusing $P(D \mid +)$ with $P(+ \mid D)$ — Bayes' theorem exists precisely to distinguish them.

Active revision

Using the diagnostic tree, find $P(D \mid +)$ from the path products.

Active recall

Recall the key points — then reveal.

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

Priors, posteriors and updating#

AdvancedPearson Edexcel 9ST0, Topic 8 (Paper 1)

Bayes' theorem is naturally read as belief updating. The prior probability

P(A)

is the belief before the evidence, the likelihood

P(B \mid A)

is how probable the evidence is under

A

, and the posterior

P(A \mid B)

is the revised belief after the evidence. In the diagnostic example the prior 0.01 is updated by the positive test to the posterior 0.167 — a large relative increase, from 1% to about 17%, even though it remains below one half.

The strength of the update depends on how much more probable the evidence is under

A

than under

A'

, a ratio known as the likelihood ratio. When the evidence is far more likely under

A

, the posterior rises sharply; when it is only slightly more likely, the belief barely moves. This explains why a single positive on a test with a high false-positive rate is only weakly informative.

Updating can be repeated: the posterior after one piece of evidence becomes the prior for the next, provided the pieces of evidence are conditionally independent given the state. Two independent positive tests, for instance, multiply the evidence and can push a modest posterior much higher, which is why doctors confirm a screening result with a second, independent test.

The framework insists that a conclusion depends on the prior as much as on the evidence. The same positive test yields a very different posterior in a high-prevalence population than in a low-prevalence one. Stating the prior, and acknowledging its influence, is part of an honest Bayesian interpretation and a frequent discussion point in examination answers.

\text{posterior} \;\propto\; \text{likelihood} \times \text{prior}: \quad P(A \mid B) \propto P(B \mid A)\,P(A)

Bayesian updating

The posterior is proportional to the likelihood times the prior; normalise by $P(B)$ .

Worked example

The prior changes the posterior

With the same test ( $P(+ \mid D) = 0.99$ , $P(+ \mid D') = 0.05$ ), recompute $P(D \mid +)$ when the prevalence is $P(D) = 0.20$ .

Target path
$P(D \cap +) = 0.99 \times 0.20 = 0.198$ .
Total evidence
$P(+) = 0.198 + 0.05 \times 0.80 = 0.198 + 0.04 = 0.238$ .
Posterior
$P(D \mid +) = \dfrac{0.198}{0.238} = 0.832$ (3 s.f.).
$P(D \mid +) = \frac{0.198}{0.238} = 0.832$
Compare
The same test gives a posterior of 0.832 at 20% prevalence versus 0.167 at 1% — the prior dominates the interpretation.

Result: At 20% prevalence the posterior is 0.832, far above the 0.167 at 1%, showing the decisive role of the prior.

Typical mistakes

Ignoring the prior and quoting the likelihood $P(B \mid A)$ as if it were the posterior.
Updating with a second piece of evidence that is not conditionally independent of the first.

Active revision

Explain how the posterior probability of disease after a positive test would change if the disease were much more common (say 20% prevalence).

Active recall

Recall the key points — then reveal.

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

Diagnostic testing and the base-rate effect#

AdvancedPearson Edexcel 9ST0, Topic 8 (Paper 1)

Frequency table for 10 000 tested people

Fig. 2Of 594 positive tests only 99 are true cases, so P(D given +) = 99/594 = 0.167.

The base-rate effect is the tendency to neglect the underlying prevalence (the base rate) when interpreting a test, focusing instead on the test's accuracy. The diagnostic example makes it concrete: with a 1% base rate, a 99%-sensitive test still leaves a positive result only 16.7% likely to be a true case, because the healthy majority generates so many false positives.

A frequency (natural-numbers) table often communicates this more clearly than probabilities. Imagining 10 000 people, 100 have the disease and 9 900 do not; of the 100, 99 test positive, while of the 9 900, 495 test positive as false alarms. Of the

99 + 495 = 594

positives, only 99 are true cases, so

P(D \mid +) = \dfrac{99}{594} = 0.167

— the same answer, now visible in whole people.

The table also names the four cells of any diagnostic test: true positives (99), false positives (495), false negatives (1) and true negatives (9 405). Sensitivity is the true-positive rate among the diseased, and specificity the true-negative rate among the healthy; the positive predictive value,

P(D \mid +)

, is what Bayes' theorem computes and is what actually matters to a patient.

The practical lesson is that a good test is not enough when the condition is rare: screening programmes must weigh the harm of many false positives against the benefit of detecting rare true cases, and confirmatory testing is standard. Communicating this honestly, with the base rate made explicit, is exactly the AO3 skill the topic assesses.

P(D \mid +) = \frac{\text{true positives}}{\text{true positives} + \text{false positives}} = \frac{99}{99 + 495} = 0.167

Positive predictive value

The proportion of positive tests that are true cases; low when the base rate is low.

Worked example

The base-rate effect in whole numbers

Among 10 000 people, 100 have a disease. A test detects 99 of them but also flags 495 of the 9 900 healthy people. Find the probability that a positive test is a true case.

Count the positives
Positives total $99 + 495 = 594$ .
True cases among them
Only 99 of these are genuinely diseased.
Predictive value
$P(D \mid +) = \dfrac{99}{594} = 0.167$ (3 s.f.).
$P(D \mid +) = \frac{99}{594} = 0.167$
Interpret
Because the disease is rare, over 83% of positive results are false alarms, so a confirmatory test is warranted.

Result: $P(D \mid +) = 0.167$ ; most positives are false alarms because of the low base rate.

Typical mistakes

Equating the sensitivity $P(+ \mid D)$ with the predictive value $P(D \mid +)$ .
Ignoring the base rate when judging how informative a positive result is.

Active revision

For a condition with 0.5% prevalence and a test with 98% sensitivity and 96% specificity, build a table for 10 000 people and find the positive predictive value.

Active recall

Recall the key points — then reveal.

Sources: GCE AS and A level subject content (Statistics) (Department for Education / Ofqual)

References & sources

Sources

Pearson Edexcel

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

Department for Education / Ofqual

GCE AS and A level subject content (Statistics)

Bayes' theorem

4 sections~10 min reading time3 competenciesLevel Standard 1 · Advanced 3

Bayes' theorem

Key points

Total probability of a positive test

Key points

Reversing the conditional with Bayes

Key points

The prior changes the posterior

Key points

The base-rate effect in whole numbers

Sources

Bayes' theorem

Key points

Total probability of a positive test

Key points

Reversing the conditional with Bayes

Key points

The prior changes the posterior

Key points

The base-rate effect in whole numbers

Sources