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Math Help - P(A|B)

  1. #1
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    P(A|B)

    I hate Statistics.

    Anyway.

    Two independant events are such that P(A)=0.2, P(B)=0.25, P(AuB)=0.4, P(AnB)=0.05

    All I have to figure out is:

    Given that exactly one event has occured, find the probablity that A occurs.

    I know the formula for given is P(X|Y)=P(XnY)/P(Y) and the answer is 0.2*0.75/0.35

    I just want to know where each of the numbers come from. Mainly the 0.75. Thanks.
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  2. #2
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    Quote Originally Posted by StephenPoco View Post
    I hate Statistics.

    Anyway.

    Two independant events are such that P(A)=0.2, P(B)=0.25, P(AuB)=0.4, P(AnB)=0.05

    All I have to figure out is:

    Given that exactly one event has occured, find the probablity that A occurs.

    I know the formula for given is P(X|Y)=P(XnY)/P(Y) and the answer is 0.2*0.75/0.35

    I just want to know where each of the numbers come from. Mainly the 0.75. Thanks.
    Honestly I don't know where the 0.75 comes from, but I get the same answer.

    We are looking for the P(only A occurred | (only A occurred) or (only B occurred)).

    In other words we want P(A and not B | (A and not B) or (B and not A)).

    In other words, P(A and not B | (A or B) - (A and B)).

    With Venn diagrams it looks really simple.

    So we get (0.2-0.05) / (0.4-0.05) = 0.15/0.35 = 3/7.
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  3. #3
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    Yeah, I know that it's 0.15/0.35. It's just I can't seem to figure out where the 0.75 came from which really threw me off track.
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  4. #4
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    Quote Originally Posted by StephenPoco View Post
    Yeah, I know that it's 0.15/0.35. It's just I can't seem to figure out where the 0.75 came from which really threw me off track.
    Well P(A and not B) is 75% of P(A).

    But why anyone would try to solve it that way is beyond me.
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