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Math Help - Conditional Probabilty question

  1. #1
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    Conditional Probabilty question

    okay the problem statement is as follows
    A gene can be either type A or type B (ie note type A). A gene can also be either dominant or regressive.
    If a gene is type B, then there is a 0.31 probability that it is dominant. When a gene is dominant, then there is also a 0.22 probability that it is type B.

    Let A be the event that the gene is of 'type A' and let D be the even that the gene is 'dominant'. What is the probability that the gene is type A?

    so what is given is P(dominant|B)=0.31 and P(B|dominant)=0.22 and that A and B are mutually exclusive

    Now what I'm not entirely sure is correct is, P(regressive|B)=0.69 and P(A|dominant)=0.78

    Breaking down P(dominant|B) = P(dom intersect B)/P(B) and similarly for the others. I've tried every combination I can think of for solving for P(A) and I always end up with more unknowns than equations. What should I do?

    I also have P(A) = (A int R + A int D)/(A int R + A int D + B int R + B int D)
    similarly for P(B) = (B int R + B int D)/(A int R + A int D + B int R + B int D)
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  2. #2
    Senior Member
    Joined
    Feb 2008
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    I don't think you can do this without additional information. This is what I'd start with:

    P(B|D)=\frac{P(D|B)P(B)}{P(D|B)P(B)+P(D|A)P(A)}

    There are a couple of variables here that we don't know; namely:

    P(B); P(A); P(D|A)

    However, the question implies the following:

    P(A)+P(B)=1

    Which means:

    P(B)=1-P(A)

    Which we can plug in to the original equation:

    P(B|D)=\frac{P(D|B)[1-P(A)]}{P(D|B)[1-P(A)]+P(D|A)P(A)}

    But what about P(D|A)? Well, let's explore the Bayesian equations for this. The first one that comes to mind is:

    P(A|D)=\frac{P(D|A)P(A)}{P(D|A)P(A)+P(D|B)P(B)}

    This might help, because, remember:

    P(B)=1-P(A)

    Which means that:

    P(A|D)=1-P(B|D)

    So let's plug those in:

    1-P(B|D)=\frac{P(D|A)P(A)}{P(D|A)P(A)+P(D|B)[1-P(A)]}

    And now let's rearrange it:

    P(D|A)=\frac{P(D|B)P(A)+P(B|D)P(D|B)-P(B|D)P(D|B)P(A)-P(D|B)}{P(A)-P(B|D)P(A)-P(A)}

    Now you should be able to plug that in and solve for P(A)
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