Proof of this probability equation

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The discussion focuses on proving the equation for conditional probability involving disjoint events. The user struggles to understand how to derive the equation P(B1, B2, ... | A) = P(B1 | A) + P(B2 | A) + ... using only basic probability rules. They attempt to express the joint probability in terms of the intersection with A and recognize that additional knowledge about disjoint events is necessary. The key point is that the proof hinges on the property that if events are disjoint, their probabilities can be summed. The conversation emphasizes the need for a deeper understanding of probability concepts to complete the proof.
zeion
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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 P(A\cup B)= P(A)+ P(B).
 
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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