Change the sample space.

P(A | B) considers only outcomes where B occurred. Divide the probability of both A and B by the probability of B, provided $P(B)$ > 0. This restriction changes the denominator.

Independent events satisfy $P(A\cap B)$ = P(A) $P(B)$ . Mutually exclusive events cannot happen together, which is a different idea. To find P(A ∪ B), add the two probabilities then subtract the overlap that was counted twice.

INTERACTIVE MODEL

Try it. Watch it change.

Counts in a two-way table
Event	B	not B	Total
A	8	12	20
not A	15	10	25
Total	23	22	45

A and B count8A and not B count12not A and B count15

P(A) = 0.4444 · P(B) = 0.5111 · P(A | B) = 0.3478
Not independent (or conditioning event has zero probability)

Conditioning on B restricts the denominator to the B column. Cell counts represent an illustrative population.

Explore: Find cell counts where P(A | B) equals $P(A)$ . Then change one cell and retest independence.

Conditional probability

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

WORKED EXAMPLE

$P(A)$ = 0.4, $P(B)$ = 0.5 and $P(A\cap B)$ = 0.1. Find P(A | B) and test independence.

P(A | B) = $\frac{0.1}{0.5}$ = 0.2.
P(A) $P(B)$ = 0.2, which differs from the intersection 0.1.
The events are not independent.

Assumed knowledge

Two-way tables, complements and fractions.

Learning checkpoints & source

YOUR LEARNING CHECKPOINT

Use conditional probability and the addition rule.
Test independence using probabilities.

QCAA Mathematical Methods 2025 v1.3 · p. 18 ↗

READY TO TRY IT?

Change the sample space.

P(A)P(A)P(A) = 0.4, P(B)P(B)P(B) = 0.5 and P(A∩B)P(A\cap B)P(A∩B) = 0.1. Find P(A | B) and test independence.

Put the idea to work.

$P(A)$ = 0.4, $P(B)$ = 0.5 and $P(A\cap B)$ = 0.1. Find P(A | B) and test independence.