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  1. #1
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    Probability Help

    I got this 2 question that I have being trying to solve for hours but can't seem to figure out.

    1- P(A)= 0.71 ; P(B|A)= 0.17 ; P(B|not A)= 0.59 ; P(B)= ?
    2- P(B)= 0.27 ; P(A|B)= 0.84 : P(A|not B)= 0.14 ; P(B|A)= ?


    Also can explain how to get P(A|not B) and P(B|not A) for future reference?
    thank you very much
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  2. #2
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    Quote Originally Posted by Panzz View Post
    I got this 2 question that I have being trying to solve for hours but can't seem to figure out.

    1- P(A)= 0.71 ; P(B|A)= 0.17 ; P(B|not A)= 0.59 ; P(B)= ?
    2- P(B)= 0.27 ; P(A|B)= 0.84 : P(A|not B)= 0.14 ; P(B|A)= ?


    Also can explain how to get P(A|not B) and P(B|not A) for future reference?
    thank you very much
    In both cases drawing a tree diagram makes things simple. I'll do 1.

    The first two branches are A (0.71) and not A (0.29). The second two branches going from each of the first are B and not B:

    Going from A it's B (0.17) and not B (0.83).

    Going from not A it's B (0.59) and not B (0.41).

    Then Pr(B) = (0.71)(0.17) + (0.29)(0.59) = ....

    Pr(A | not B) = (0.71)(0.83)/[(0.71)(0.83) + (0.29)(0.41)] = ....

    Pr(not A | not B) = (0.29)(0.41)/[(0.71)(0.83) + (0.29)(0.41)] = ....

    The tree diagram makes it all simple and obvious.


    Note that for question 2, the first two branches are B and not B. The seocnd two branches are A and not A.
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