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Baye's Theorem

  1. Aug 6, 2013 #1
    P(B/A) = P(A/B).P(B) / P(A)

    Later we expand P(A) as P(A/B).P(B) + P(A/B).P(B) ... B is complement of B

    I don't understand how we can expand P(A) like that. Doesn't that assume that A ℂ B?
  2. jcsd
  3. Aug 6, 2013 #2


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    Think about what it is saying. The probability that A happens is the probability that A happens given that B happens plus the probability that A happens given that B doesn't happen. Both cases are needed to cover all possibilities.
  4. Aug 6, 2013 #3
    Well basically what my book says is that : -
    P(A) = P(A/B1).P(B1) + P(A/B2).P(B2) + ... + P(A/Bn).P(Bn)

    Doesn't this assume that B1 U B2 ... U Bn is a super-set of A?
  5. Aug 6, 2013 #4

    D H

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    Of course.

    Your text should have specified that B1, B2, ···, Bn are a set of mutually disjoint subsets of the universe U of possible outcomes and that B1B2 ∪ ··· ∪ Bn=U. The set A must be a subset of this universe of outcomes U; otherwise it doesn't even make sense to talk about P(B1|A).
  6. Aug 6, 2013 #5
    It didn't. Anyways, thank you. This clears my doubt.
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