Prove the Multiplication Rule (Conditional Form) with more than two events.
For events $A_1, A_2,\ldots, A_n$ prove that
$$
P(A_1 \cap A_2 \cap\ldots\cap A_n)=
P(A_1)\ P(A_2|A_1)\ P(A_3|A_1 \cap A_2)\ \ldots\ P(A_n|A_1 \cap A_2 \cap … \cap\ A_{n-1}).
$$
My first attempt was to try induction and I do get through the first two induction steps but I am not getting the answer when trying to prove for all $n$
Any help would be highly appreciated
Thanks
Best Answer
You know that the definition of conditional probability is
$$P(B | A) = \frac{P(A\cap B)}{P(A)},$$ so just apply the definition to every term in the right hand side of your equation. Starting with $$\quad P(A_1)\ P(A_2|A_1)\ P(A_3|A_1 \cap A_2)\ \cdots\ P(A_n|A_1 \cap A_2 \cap ... \cap\ A_{n-1})$$ and applying the definition of conditional probability to each term we get $$ \ \color{blue}{P(A_1)} \frac{\color{red}{P(A_1 \cap A_2)}}{\color{blue}{P(A_1)}}\frac{\color{green}{P(A_1 \cap A_2 \cap A_3)}}{\color{red}{P(A_1 \cap A_2)}} \cdots \frac{P(A_1 \cap A_2 \cap\cdots \cap A_n) }{\color{purple}{P(A_1 \cap A_2 \cap\cdots \cap A_{n-1})}} $$ And almost every term will cancel out, except $P(A_1 \cap A_2 \cap \cdots \cap A_n)$. Hence
$ P(A_1 \cap A_2 \cap \cdots \cap A_n)= P(A_1)\ P(A_2|A_1)\ P(A_3|A_1 \cap A_2)\ \cdots\ P(A_n|A_1 \cap A_2 \cap \cdots \cap\ A_{n-1}). $