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Year 11 Methods (Unit 1 & 2) Probability

Conditional Probability

20 practice questions 1 video lesson Theory + worked examples

Understand conditional probability for Queensland Year 11 Mathematical Methods (QCAA). Conditional probability measures the chance of one event given that another has already happened, working within a reduced sample space.

You will learn to read conditionals from two-way tables, Venn diagrams and tree diagrams, apply the multiplication rule linking joint and conditional probabilities, and find the complement of a conditional — an essential stepping stone into QCAA statistics.

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Theory

In Year 11 Mathematical Methods (QCAA, Unit 1 Topic 5), a conditional probability \(P(A\mid B)\) is the probability of \(A\) given that \(B\) has happened. It uses the reduced sample space of \(B\): \(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\). This page reads conditionals from tables, Venn diagrams and trees.

A conditional probability \(P(A\mid B)\) is read ‘the probability of \(A\) given \(B\)’. Knowing \(B\) has occurred restricts attention to the outcomes in \(B\) — the reduced sample space.

We divide the overlap by the condition: \(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}\). With counts this is \(\dfrac{n(A\cap B)}{n(B)}\) — the favourable count over the size of \(B\), not the whole sample space.

Rearranging gives the multiplication rule \(P(A\cap B)=P(A\mid B)\,P(B)\). The complement of a conditional keeps the same condition: \(P(A'\mid B)=1-P(A\mid B)\).

Condition = new denominator. ‘Given \(B\)’ means divide by \(P(B)\) (or count only within \(B\)), never by the full total.
100 students by handedness and glasses
GlassesNo glassesTotal
Left-handed121830
Right-handed284270
Total4060100
Given a student is left-handed, restrict to that row of \(30\): \(P(\text{glasses}\mid\text{left})=\dfrac{12}{30}=\dfrac{2}{5}\).
Conditional probability restricts to event BOnly the shaded circle B counts; within it the A part is zero point two four out of zero point four. A B 0.24 0.16
Only circle \(B\) counts once \(B\) is given. Inside it, \(P(A\cap B)=0.24\) out of \(P(B)=0.40\), so \(P(A\mid B)=0.6\).

The conditional probability formula:

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

With counts (reduced sample space):

\[P(A\mid B)=\dfrac{n(A\cap B)}{n(B)}\]
P(AB)=n(AB)n(B)

The multiplication rule (rearranged):

\[P(A\cap B)=P(A\mid B)\,P(B)\]
P(AB)=P(AB)P(B)
Complement of a conditional: \(P(A'\mid B)=1-P(A\mid B)\) — the condition \(B\) stays the same.

How to find a conditional probability

  1. Spot the condition: the event after ‘given’ is \(B\); it becomes the new denominator.
  2. Find the overlap: \(P(A\cap B)\) or the count \(n(A\cap B)\) inside \(B\).
  3. Divide: \(P(A\mid B)=\dfrac{P(A\cap B)}{P(B)}=\dfrac{n(A\cap B)}{n(B)}\).
  4. Rearrange if needed: use \(P(A\cap B)=P(A\mid B)P(B)\), or the complement \(P(A'\mid B)=1-P(A\mid B)\).
Example 1 — Using the formula
Events \(A\) and \(B\) satisfy \(P(A\cap B)=0.24\) and \(P(B)=0.4\). Find \(P(A\mid B)\).
Solution

Apply the conditional formula:

\(P(A\mid B)\)\(=\)\(\dfrac{P(A\cap B)}{P(B)}\)
\(=\)\(\dfrac{0.24}{0.4}\)

Divide:

\(=\)\(0.6\)

\(P(A\mid B)=0.6\).

P of A given B from probabilitiesInside B, zero point two four is also in A out of the total zero point four in B. A B 0.24 0.16
Example 2 — From a two-way table
Using the survey table, find the probability a person owns a pet given that they are female.
Survey of 200 people
Owns a petNo petTotal
Male6040100
Female7030100
Total13070200
Solution

‘Given female’ restricts to the female row: \(n(\text{female})=100\).

\(n(\text{female}\cap\text{pet})\)\(=\)\(70\)
\(n(\text{female})\)\(=\)\(100\)

Divide within the reduced sample space:

\(P(\text{pet}\mid\text{female})\)\(=\)\(\dfrac{70}{100}\)
\(=\)\(\dfrac{7}{10}\)

\(P(\text{pet}\mid\text{female})=\dfrac{7}{10}\).

Example 3 — From a tree (without replacement)
A bag holds \(4\) red and \(6\) blue balls. Two are drawn without replacement. Find the probability the second is red given the first is red.
Solution

Given the first is red, one red is removed: \(3\) red remain of \(9\).

\(P(\text{2nd red}\mid\text{1st red})\)\(=\)\(\dfrac{3}{9}\)
\(=\)\(\dfrac{1}{3}\)

\(P(\text{2nd red}\mid\text{1st red})=\dfrac{1}{3}\).

Two draws without replacement from 4 red and 6 blueGiven the first ball is red, the second red branch has probability three ninths. 4/10 R 3/9 R RR 6/9 B RB 6/10 B 4/9 R BR 5/9 B BB
Example 4 — Multiplication rule and complement
For events \(A\) and \(B\), \(P(B)=0.5\) and \(P(A\mid B)=0.7\). Find \(P(A\cap B)\) and \(P(A'\mid B)\).
Solution

Rearrange for the overlap:

\(P(A\cap B)\)\(=\)\(P(A\mid B)\,P(B)\)
\(=\)\(0.7\times 0.5\)
\(=\)\(0.35\)

Complement of the conditional (same condition \(B\)):

\(P(A'\mid B)\)\(=\)\(1-P(A\mid B)\)
\(=\)\(1-0.7\)
\(=\)\(0.3\)

\(P(A\cap B)=0.35\) and \(P(A'\mid B)=0.3\).

Intersection from a conditional probabilityWithin B of total zero point five, the A part is zero point three five. A B 0.35 0.15

Common pitfalls

Dividing by the whole total. ‘Given \(B\)’ means divide by \(P(B)\) (or \(n(B)\)), not by the full sample space.
Swapping the condition. \(P(A\mid B)\) and \(P(B\mid A)\) are usually different — they have different denominators.
Complementing the wrong part. \(P(A'\mid B)=1-P(A\mid B)\) keeps \(B\); it is not \(1-P(A\mid B')\).
Forgetting to reduce for no replacement. On a tree, the ‘given’ branch already uses the reduced counts — read the second-stage probability directly.

Frequently asked questions

What is conditional probability?

\(P(A\mid B)\) is the probability of \(A\) given that \(B\) has occurred, found by \(\dfrac{P(A\cap B)}{P(B)}\).

What does ‘reduced sample space’ mean?

Once \(B\) is given, only outcomes in \(B\) are possible, so \(B\) becomes the new total (denominator).

How do you read a conditional probability from a two-way table?

Restrict to the row or column for the condition, then divide the overlap cell by that total, e.g. \(\dfrac{n(A\cap B)}{n(B)}\).

Is P(A given B) the same as P(B given A)?

No — they usually differ because they divide by different totals, \(P(B)\) versus \(P(A)\).

What is the multiplication rule?

\(P(A\cap B)=P(A\mid B)\,P(B)\), found by rearranging the conditional formula.

What is the complement of a conditional probability?

\(P(A'\mid B)=1-P(A\mid B)\); the condition \(B\) is unchanged.