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Thread: Med School probabilities

  1. #1
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    Med School probabilities

    The probability of being wait-listed, P(w), is half that of being admitted and 3 times as many applicants are denied than are admitted.

    P(a), P(w) = 1/2P(a), P(d) = 3P(a)

    P(a) + .5P(a) + 3P(a) = 1\Rightarrow 4.5P(a)=1\Rightarrow P(a)=\frac{2}{9}

    However, the P(a) = .20, P(w) = .10, and P(d) = .60. What went wrong?
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    Say twelve people apply. According to that two will be admitted, one will be wait-listed, and nine will be denied.
    Last edited by Plato; Jan 27th 2011 at 03:54 PM.
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    Quote Originally Posted by Plato View Post
    Say twelve people apply. According to that two will be admitted, one will be wait-listed, and nine will be denied.
    What I don't understand is why the approach I took didn't work.
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    Quote Originally Posted by dwsmith View Post
    However, the P(a) = .20, P(w) = .10, and P(d) = .60. What went wrong?
    Do you mean this should be the answer? (P(a)=0.2)? What seems wrong here is that the sum is not 1.

    Your answer, on the other hand, makes sense:
    p(a)=2/9
    p(w)=1/2 of p(a)=1/9
    p(d)=3*p(a)=6/9

    The sum is 1 and it can be verified provided the number of candidates is a multiple of 9 - for example if there are 9 candidates, then 2 are admitted, 1 is wait-listed and 6 are denied admission.
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    Quote Originally Posted by Ithaka View Post
    Do you mean this should be the answer? (P(a)=0.2)? What seems wrong here is that the sum is not 1.

    Your answer, on the other hand, makes sense:
    p(a)=2/9
    p(w)=1/2 of p(a)=1/9
    p(d)=3*p(a)=6/9

    The sum is 1 and it can be verified provided the number of candidates is a multiple of 9 - for example if there are 9 candidates, then 2 are admitted, 1 is wait-listed and 6 are denied admission.
    I had class today and the book is definitely wrong.
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