Proof of this probability equation

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SUMMARY

The proof of the conditional probability equation states that for disjoint events B1, B2, ..., the equation P(B1, B2, ... | A) = P(B1 | A) + P(B2 | A) + ... holds true when P(A) > 0. The key to this proof lies in recognizing that if A and B are disjoint, then P(A ∪ B) = P(A) + P(B). The proof demonstrates that P(B1, B2, ... | A) can be expressed as the sum of the individual conditional probabilities P(Bi | A) by manipulating the intersection of events with A.

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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?
 
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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 [itex]P(A\cup B)= P(A)+ P(B)[/itex].
 
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) + ... ?
 

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