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Probability proof

  1. Sep 21, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove if P(A|B) = P(A|B') then A and B are independent.

    where B' is the complement of B

    2. Relevant equations

    if independent, P(A|B) = P(A)
    also, P(A∩B) = P(A)P(B)

    for conditional probability,
    P(A|B) = P(A∩B) / P(B)

    3. The attempt at a solution

    P(A|B) = P(A∩B) / P(B) = P(B|A)P(A) / P(B)
    P(A|B') = P(A∩B') / P(B') = P(B'|A)P(A) / P(B')

    I'm not really sure how to go from here... What do I do?
     
  2. jcsd
  3. Sep 21, 2009 #2
    If you know that P(A|B) = P(A|B') and

    P(A|B) = P(AB)/P(B) and P(A|B') = P(AB')/P(B')

    (where AB = A intersect B)

    then why not set these right-hand sides equal and see what happens?

    --Elucidus
     
  4. Sep 21, 2009 #3
    Yeah, I did that. Then P(A∩B) / P(B) = P(A∩B') / P(B') where ∩ means intersection

    Sorry, I still don't see what should follow.
     
    Last edited: Sep 22, 2009
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