1. conditional probability proof...

I am trying to prove this question, but I am not sure if what I did is correct..

q: Suppose that A1, A2, ..., An are events in a random experiment whose intersection has positive probability. Prove the multiplication rule of probability.

$\displaystyle P({A_1} \cap {A_2} \cap....{A_n})$ $\displaystyle = P(A_1) P({A_2} | {A_1}) P({A_3} | {A_1} \cap A2) ...... P({A_n} | {A_1} {A_2}......{A_{n-1})$

This is what I have done:

$\displaystyle P(A_1) = P(A_1)$

$\displaystyle P({A_2}\cap{A_1}) = P({A_2}|{A_1}) P({A_1})$ [by the definition of conditional prob]

$\displaystyle P[{A_3} \cap ({A_2}\cap{A_1})]$ $\displaystyle = P({A_3} |{A_2}\cap{A_1}) P({A_2}\cap{A_1})$ $\displaystyle =P({A_3} |{A_2}\cap{A_1}) P({A_2}|{A_1}) P({A_1})$

now assuming this is true for all n, we have

$\displaystyle P[{A_{n+1}} \cap ({A_1} \cap ...\cap{A_n})]$ $\displaystyle = P({A_1})P({A_2}|{A_1}) P({A_3} |{A_2}\cap{A_1}).......P({A_{n+1}}|{A_1} \cap.......\cap{A_n})$

thus it is true for all n..

Is it path I have followed correct? Any inputs will be appreciated!

Thanks

2. It's fine, you can think about picking cards from a deck.
If you want to select three aces, then you want an Ace on the first, next another Ace on the second pick GIVEN an Ace on the first pick.....
You can use induction if that will convince you that this is ok.

3. (yes)