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)

However, the P(a) = .20, P(w) = .10, and P(d) = .60. What went wrong?

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- January 27th 2011, 03:19 PMdwsmithMed 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)

However, the P(a) = .20, P(w) = .10, and P(d) = .60. What went wrong? - January 27th 2011, 03:40 PMPlato
Say twelve people apply. According to that two will be admitted, one will be wait-listed, and nine will be denied.

- January 27th 2011, 06:40 PMdwsmith
- January 27th 2011, 10:53 PMIthaka
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, 02:31 PMdwsmith