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Conditional Probabilities Complementary Proof

  1. Dec 3, 2007 #1
    I'm having trouble seeing how this works out. It's blatantly obvious that this is true, but somehow I can't seem to get anywhere on paper with it to simplify it down to anything. Any help would be greatly appreciated!

    [tex]P\left(A\right|B)=1-P\left(not A\right|B)[/tex]
  2. jcsd
  3. Dec 4, 2007 #2


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    P(A|B) = P(A&B)/P(B)
    P(~A|B) = P(~A&B)/P(B)

    What is P(A&B) + P(~A&B) = ?
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