# Med School probabilities

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• January 27th 2011, 02:19 PM
dwsmith
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?
• January 27th 2011, 02:40 PM
Plato
Say twelve people apply. According to that two will be admitted, one will be wait-listed, and nine will be denied.
• January 27th 2011, 05:40 PM
dwsmith
Quote:

Originally Posted by Plato
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.
• January 27th 2011, 09:53 PM
Ithaka
Quote:

Originally Posted by dwsmith
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.
• January 28th 2011, 01:31 PM
dwsmith
Quote:

Originally Posted by Ithaka
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.