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Proof of this probability equation

  1. Sep 25, 2011 #1
    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. jcsd
  3. Sep 25, 2011 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    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].
     
  4. Sep 25, 2011 #3
    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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