Conditional probability, the conditional probability formula, the multiplication rule, independence, and probability from two-way tables and tree diagrams.

What this dot point is asking

WJEC wants you to calculate conditional probability (the probability of one event given another has occurred), to use the multiplication rule in its general form, to test independence, and to extract conditional probabilities from two-way tables and tree diagrams. This A2 topic builds on the AS probability work and underpins the distribution and hypothesis-testing material in the unit.

The answer

The conditional probability formula

When you know event $B$ has occurred, the sample space shrinks to $B$, so probabilities are rescaled by $P(B)$.

The multiplication rule

Rearranging the conditional formula gives the general multiplication rule for the probability that both events occur.

For independent events this simplifies to $P(A \cap B) = P(A)\,P(B)$, because then $P(A \mid B) = P(A)$.

Independence

Two-way tables and tree diagrams

A two-way table records frequencies for combinations of two categories; condition by looking only at the relevant row or column total. A tree diagram shows sequences; the second set of branches carries conditional probabilities, which is exactly why multiplying along a branch gives the joint probability.

Examples in context

Example 1. Medical testing. A test is positive for $98\%$ of people with a condition and for $5\%$ without it; $1\%$ of people have the condition. The probability of having the condition given a positive test is found by conditioning on "positive", and turns out surprisingly low because positives among the healthy majority dominate. Conditional probability explains why a positive screening result is not a diagnosis.

Example 2. Drawing without replacement. Two cards are drawn from a pack. The probability the second is an ace given the first was an ace is $\dfrac{3}{51}$, because one ace and one card are gone. The conditioning changes both the favourable count and the total, exactly as the formula requires.

Try this

Q1. $P(A \cap B) = 0.2$ and $P(B) = 0.4$. Find $P(A \mid B)$. [2 marks]

• Cue. $P(A \mid B) = \dfrac{0.2}{0.4} = 0.5$.

Q2. Events $A$ and $B$ have $P(A) = 0.3$, $P(B) = 0.5$, $P(A \cap B) = 0.15$. Are they independent? [2 marks]

• Cue. $P(A)P(B) = 0.15 = P(A \cap B)$, so yes, independent.

Q3. From a tree diagram, $P(\text{rain}) = 0.3$ and $P(\text{late} \mid \text{rain}) = 0.6$. Find $P(\text{rain and late})$. [2 marks]

• Cue. Multiply along the branch: $0.3 \times 0.6 = 0.18$.