# Proof of this probability equation

1. Sep 25, 2011

### zeion

1. The problem statement, all variables and given/known data

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..

2. Relevant equations

3. The attempt at a solution

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

How do I get +'s out?

2. Sep 25, 2011

### 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 $P(A\cup B)= P(A)+ P(B)$.

3. Sep 25, 2011

### 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) + ... ?