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P(A|B') why?

  1. Apr 21, 2015 #1


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    I know the equations that P(A|B') = P(AnB) / P(B')

    But why isn't it P(A) - P(A n B)

    See photo attachment

    We know b didn't happen so isn't it just A minus the middle?

    Attached Files:

  2. jcsd
  3. Apr 21, 2015 #2


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

    Your version will hold only of P(B')=1.
  4. Apr 22, 2015 #3
    Specifically, [itex]P(A | B^{C}) := P(A) - P(A \cap B)[/itex] would still include B in the sample space. Given that B hasn't happen, you don't want B in the sample space.
  5. Apr 23, 2015 #4

    Stephen Tashi

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    In an expression for probability P(S), there are more things involved that the set S. The expression for a probability involves (perhaps implicitly) a particular "probability space". The expressions [itex] P(A|B') [/itex] and [itex] P(A \cap B') [/itex] both refer to the same set. However, they refer to different probability spaces. In the probability space for [itex] P(A|B') [/itex] no events in [itex] A \cap B [/itex] exist. In the probability space for [itex]A \cap B' [/itex] , events in [itex] A \cap B [/itex] may exist and may be assigned nonzero probabilities.

    Your reasoning with the Venn diagram doesn't include the information about what sets are in the two different probability spaces.
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