2.8 Multiplication rule

By the definition of conditional probability

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

Multiply \(\text{P}(B)\) on both sides, we have

\[\text{P}(A \cap B)=\text{P}(B) \cdot \text{P}(A \mid B)\]

The above is the multiplication rule.

If we swap \(A\) and \(B\) in the above formula, we have

\[\text{P}(B \cap A)=\text{P}(A) \cdot \text{P}(B \mid A)\]

Note \(\text{P}(B \cap A)=\text{P}(A \cap B)\), thus we have

\[\text{P}(A \cap B)=\text{P}(A) \cdot \text{P}(B \mid A)\]

The multiplication rule can be described as:

“The probability that both events occur is the probability that first event occurs multiply by the (conditional) probability that the second event occurs, given that the first event has occurred.”

Multiplication rule for three events

\[ \text{P}(A \cap B \cap C)=\text{P}(A) \cdot \text{P}(B \mid A) \cdot \text{P}(C \mid A \cap B) \]

General form of multiplication rule

For any \(n\) events \(A_1, A_2, \cdots, A_n\) with positive probability, we have

\[ \begin{aligned} \text{P}(A_1 \cap A_2 \cap \cdots \cap A_n)=& \; \text{P}(A_1) \cdot \\ &\; \text{P}(A_2 \mid A_1) \cdot \\ &\; \text{P}(A_3 \mid A_1 \cap A_2) \cdot \\ &\; \cdots \\ &\; \text{P}(A_n \mid A_1 \cap A_2 \cap \cdots \cap A_{n-1}) \\ \end{aligned} \]

Exercise

We deal from a well-shuffled 52-card deck.

What is the chance none of the first 3 cards is a heart?

Can you solve it using

• combinatorics?
• multiplication rule?