Prove the Multiplication Rule (Conditional Form) with more than two events. 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
 A: 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}). $
A: Starting with the required result
$$
P(A_1 \cap A_2 \cap\ldots\cap A_n)=
P(A_1)\ P(A_2|A_1)\ P(A_3|A_1 \cap A_2)\ \ldots\ P(A_n|A_1 \cap A_2 \cap ...  \cap\ A_{n-1}),
$$
just look at the last term and use the definition of conditional probability: $P(E|F)=P(E\cap F)/P(F)$. You can rewrite the last term as
$$
P(A_n|A_1 \cap A_2 \cap ...  \cap\ A_{n-1})=
\frac{P(A_n\cap A_1 \cap A_2 \cap ...  \cap\ A_{n-1})}
{P(A_1 \cap A_2 \cap ...  \cap\ A_{n-1})}
=\frac{P(A_1 \cap A_2 \cap ...  \cap\ A_{n})}
{P(A_1 \cap A_2 \cap ...  \cap\ A_{n-1})}.
$$
Therefore the result is true if
$$
P(A_1 \cap A_2 \cap\ldots\cap A_n)=
P(A_1)\ P(A_2|A_1)\ P(A_3|A_1 \cap A_2)\ \ldots\ P(A_{n-1}|A_1 \cap A_2 \cap ...  \cap\ A_{n-2})\
\frac{P(A_1 \cap A_2 \cap ...  \cap\ A_{n})}
{P(A_1 \cap A_2 \cap ...  \cap\ A_{n-1})}.
$$
Moving terms around, this is equivalent to
$$
P(A_1 \cap A_2 \cap\ldots\cap A_{n-1})=
P(A_1)\ P(A_2|A_1)\ P(A_3|A_1 \cap A_2)\ \ldots\ P(A_{n-1}|A_1 \cap A_2 \cap ...  \cap\ A_{n-2})
$$
which is precisely the required property when $n$ is replaced by $n-1$. This establishes the basis for induction on $n$. All that remains to show is that the identity holds for $n=2$, which as you know it does.
A: Using your notation,
$P(($intersection of all $A_i$'s from $i=1$ to $n)=$ $P(($intersection of all $A_i$'s from $i=1$ to $n-1,$ intersected with $An)= $probability of intersection up to n-1, times probability of An given interaction up to n-1 (this is just using the base case)=$P(A1)P(A2|A1)P(A3|A1 \A2) ..... P(An-1|A1 \A2 \ ....  \ An-2) $(by induction step) times $P(An|A1 \A2 \ ....  \ An-1)$.
