Tree Diagrams and Conditional Probability

Multiply along the branches, add between branches. "Given that" shrinks the sample space to the outcomes you know have happened, and $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$ does the shrinking.

Cambridge 9709 Paper 5Edexcel YMA01 WST01Edexcel 9MA0 Paper 3AQA 7357 Paper 3≈ 40 minRevision 5 · updated 2026-10-27

Section 1 of 14

Learning objectives

Draw and use tree diagrams for two or three stages, with or without replacement.
Understand conditional probability as probability within a reduced sample space.
Use $P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}$ .
Reverse a tree: find the probability of an earlier event given a later one.

Pause and think

$2\%$ of people have a disease. A test catches $95\%$ of cases but also gives a positive result for $10\%$ of healthy people. You test positive. Are you probably ill?

Section 2 of 14

Core idea

Start hereA positive test: how worried should you be?

A disease affects 1% of people. A test catches 95% of cases but also wrongly flags 10% of healthy people. You test positive. The chance you're ill is not 95%: it's under 9%. Conditional probability explains the surprise.

The big idea

$P(B\mid A)=\dfrac{P(A\cap B)}{P(A)}$ : probability of $B$ given $A$ has happened. Equivalently $P(A\cap B)=P(A)P(B\mid A)$ , which is what tree diagrams use.

1. Shrinking the sample space

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

$\mid$: "given that".
$P(A)$: must be non-zero.

Knowing $A$ happened, only outcomes in $A$ remain possible. Among those, the fraction also in $B$ is $\frac{n(A\cap B)}{n(A)}$ . With the 30 students: P(Physics | Maths) $=\frac7{18}$ .

2. Tree diagrams

Tree diagram: two sweets without replacement

Multiply along a path for "this and then that". Add between paths for "either this way or that way". The four totals add to

1

Worked example5 red, 3 blue; two taken without replacement

Find P(both red), P(second red), P(first red | second red).

Goal: Three probabilities.
Strategy: Tree; multiply along, add between.

$P(RR)=\tfrac58\cdot\tfrac47=\tfrac{20}{56}$
Along a path.
$P(R_2)=\tfrac{20}{56}+\tfrac{15}{56}=\tfrac58$
RR or BR.
$P(R_1\mid R_2)=\frac{20/56}{35/56}=\tfrac47$
Conditional formula.

Answer

\frac5{14}

\frac58

\frac47

✓ CheckP(second red) equals P(first red) by symmetry ✓.

Worked exampleThe medical test

P(disease)=0.01; P(+ | disease)=0.95; P(+ | healthy)=0.10. Find P(disease | +).

Goal: Reverse conditional.
Strategy: Tree; positive paths.

$P(D\cap+)=0.01\times0.95=0.0095$
Ill and positive.
$P(+)=0.0095+0.99\times0.10=0.1085$
All positives.
$P(D\mid+)=\tfrac{0.0095}{0.1085}=0.0876$
Conditional.

AnswerAbout

8.8\%

✓ CheckOf 1000 people: 10 ill (9.5 test +), 990 healthy (99 test +):

\frac{9.5}{108.5}\approx9\%

✓.

Common misconception $P(A\mid B)\neq P(B\mid A)$ P(positive | ill)

=0.95

but P(ill | positive)

\approx0.09

Pause and think

$P(A)=0.6$ , $P(A\cap B)=0.24$ . $P(B\mid A)$ ?

Summary

$P(B|A)=P(A\cap B)/P(A)$ .
Trees: × along, + between.
Reverse with the formula.

Conditional probability: Probability given extra information.

HistoryThe Reverend Bayes. Thomas Bayes, an English minister, worked out how to "reverse" conditional probabilities. He never published it; his friend Richard Price found the essay among his papers and sent it to the Royal Society in 1763. Bayes' theorem now underpins spam filters, medical diagnosis and machine learning.

Fun factThe Monty Hall problem. On a game show you pick one of three doors; one hides a car. The host, who knows where the car is, opens another door to show a goat and offers you a switch. Switching wins

\tfrac23

of the time. When Marilyn vos Savant explained this in a 1990 magazine column, thousands of readers, including mathematicians, wrote to tell her she was wrong. She wasn't.

Keep reading for free

Create a free account to read the rest of this lesson: key formulae, worked examples, explore the graph, common mistakes, exam tip, guided practice and more. Your answers are checked and your progress is saved.

Create free account Sign in