Conditional Probability and Independence

≈ 20 min

Conditional probability is the chance of $A$ given that $B$ has happened:

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

Knowing $B$ shrinks the world to just the outcomes where $B$ is true, and we ask what fraction of those also have $A$ .

Two events are independent when one gives no information about the other. Formally,

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

Worked example. If $P(A\cap B)=0.2$ and $P(B)=0.5$ , then $P(A\mid B) = \dfrac{0.2}{0.5} = 0.4$ .

For dependent events, like drawing marbles without replacement, the second probability changes because the first draw altered what is left.

Two events $A$ and $B$ are independent when:

$P(A\cap B) = P(A) + P(B)$

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

$P(A\mid B) = 0$

$A$ and $B$ cannot both happen

If $P(A\cap B)=0.2$ and $P(B)=0.5$ , find $P(A\mid B)$ .

Events $A$ and $B$ are independent with $P(A)=0.3$ and $P(B)=0.6$ . Find $P(A\cap B)$ .

A bag has 4 red and 2 blue marbles. Two are drawn without replacement. What is the probability that both are red? Give a decimal.