Multiplication rule

Statement

Let $(\Omega, \mathcal{F}, P)$ be a probability space and let $A_1, \dots, A_n \in \mathcal{F}$ with $n \ge 2$ and $P(A_1 \cap \dots \cap A_{n-1}) > 0$. Then $$P(A_1 \cap \dots \cap A_n) = P(A_1) \, P(A_2 \mid A_1) \, P(A_3 \mid A_1 \cap A_2) \cdots P(A_n \mid A_1 \cap \dots \cap A_{n-1}).$$ Here $P(\cdot \mid \cdot)$ is the conditional probability.

Remarks

The case $n = 2$ is $P(A \cap B) = P(A) \, P(B \mid A)$ for $P(A) > 0$. It is the definition of $P(B \mid A)$ multiplied by $P(A)$. The hypothesis makes each conditional probability in the formula defined, because each conditioning event contains $A_1 \cap \dots \cap A_{n-1}$. Example: draw two cards from a deck without replacement. The probability of two aces is $\frac{4}{52} \cdot \frac{3}{51} = \frac{1}{221}$.

Explanations

L1 human

To find the chance that several things all happen, take them one at a time. Multiply the chance of the first by the chance of the second given the first. Then multiply by the chance of the third given the first two, and so on.