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Math Help - Mixed Probability

  1. #1
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    Mixed Probability

    Events A & B are such that P(AB) = 1/3, P(B|A)= 1/4, and P(B'|A') = 4/5.

    Hence find a) P(B'|A) b)P(AnB) c) P(B) d) P(AUB)

    I really do not know where to start with this :S
    I've been staring at it for 1/2 hour and still haven't got anywhere
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  2. #2
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    Do you know that \displaystyle P(B|A) = \frac{P(A\cap B)}{P(A)} ?
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  3. #3
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    P(1/4) = P(AnB)/P(1/2)

    So P(AnB) = 1/8 ( I think)

    I'm slowly getting there...
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  4. #4
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    Quote Originally Posted by Wildsurf View Post
    P(1/4) = P(AnB)/P(1/2)

    So P(AnB) = 1/8 ( I think)
    No indeed. In fact part of the given is P(AB)=P(A\cap B)=\frac{1}{3}
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  5. #5
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    P(AnB) = P(A)P(B)

    1/8 = (1/3)P(B)

    1/8 = P(B)
    ----
    1/3

    1/8 X 3/1 = P(B)

    3/8 = P(B)
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  6. #6
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    Quote Originally Posted by Plato View Post
    No indeed. In fact part of the given is P(AB)=P(A\cap B)=\frac{1}{3}
    My bad P(AB) was in fact meant to be P(A)

    Really bad typo
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  7. #7
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    For a) P(B^c|A)=1-P(B|A).

    BUT P(B|A^c)\ne 1-P(B|A).
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