Probability • Topic 1 of 3

Conditional Probability & Multiplication Rule

Conditional probability $P(A\mid B)$ is the probability of $A$ given that $B$ has occurred:

$$P(A\mid B)=\frac{P(A\cap B)}{P(B)},\qquad P(B)>0.$$

Rearranging gives the probability that both events happen:

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

Events $A$ and $B$ are independent if one does not affect the other:

$$P(A\cap B)=P(A)\,P(B)\quad\Longleftrightarrow\quad P(A\mid B)=P(A).$$

Do not confuse independent with mutually exclusive: mutually exclusive events ($P(A\cap B)=0$) with non-zero probabilities are in fact dependent.

Solved Examples

Worked Example

A die is rolled. Find $P(\text{number}>3 \mid \text{number is even})$.

Solution

Even outcomes: $\{2,4,6\}$, so $P(B)=\tfrac36$. Even and $>3$: $\{4,6\}$, so $P(A\cap B)=\tfrac26$. Thus $P(A\mid B)=\dfrac{2/6}{3/6}=\dfrac{2}{3}.$

Worked Example

If $P(A)=0.5,\ P(B)=0.4,\ P(A\cap B)=0.2$, are $A,B$ independent?

Solution

$P(A)P(B)=0.5\times0.4=0.2=P(A\cap B)$. Yes, they are independent.

Worked Example

Two cards are drawn without replacement from a deck. Find $P(\text{both kings})$.

Solution

$P=\dfrac{4}{52}\times\dfrac{3}{51}=\dfrac{12}{2652}=\dfrac{1}{221}.$

Worked Example

If $P(A)=0.6,\ P(B\mid A)=0.5$, find $P(A\cap B)$.

Solution

$P(A\cap B)=P(A)P(B\mid A)=0.6\times0.5=0.3.$

Quick Check — 4 MCQs

Tap an option to check your answer0 / 4

Q1.$P(A\mid B)$ equals:

Explanation: Definition of conditional probability.

Q2.$A,B$ independent means $P(A\cap B)=$

Explanation: Product of probabilities.

Q3.On a die, $P(>3 \mid \text{even})=$

Explanation: $\{4,6\}$ out of $\{2,4,6\}$.

Q4.If $P(A)=0.6,P(B\mid A)=0.5$, then $P(A\cap B)=$

Explanation: $0.6\times0.5=0.3$.