Proof of this probability equation

[zeion]

Homework Statement

Hi,

I don't understand the proof of this for conditional probability:

If B1, B2, ... are disjoint,
then P(B1, B2, B3, ... | A) = P(B1 | A) + P(B2 | A) + ...

From my notes I only have P(A) > 0 , P(B ∩ A) ≥ 0
but I don't see how those help me prove it..

Homework Equations

The Attempt at a Solution

P(B1, B2, ... | A) = P(B1, B2, B3, ... ∩ A) / P(A)

How do I get +'s out?

 

[HallsofIvy]

Well, you are right- if you have only those two rules to work with, you cannot possibly do it. But it seems to me that whoever gave you this exercise expected you to know more- specifically, that if A and B are disjoint, then $P(A\cup B)= P(A)+ P(B)$.

 

[zeion]

Ok so I have:

P(B1, B2, ... | A)
= P(B1, B2, ... ∩ A) / P(A)
= (1/P(A)) (P(B1, B2, ... ∩ A)
= ?
= (1/P(A)) (P(B1 ∩ A) + P(B2 ∩ A) + ... )
= [P(B1 ∩ A) / P(A)] + [P(B2 ∩ A) / P(A)] + ...
= P(B1 | A) + P(B2 | A) + ...


how can I show that P(B1, B2, ... ∩ A) = P(B1 ∩ A) + P(B2 ∩ A) + ... ?