LaTeX is giving me some serious parsing issues today, so I'll try to use it sparsely.
An airplane is reported as missing. Investigators assume that the airplane has disappeared in one of three given zones. Denote the probability that the airplane is found in zone k, if it is there, with 1- . Different values of may depend on geographical reasons, differences in vegetation, et cetera.
Calculate the probability that the airplane is in zone k (k=1,2,3) if a search for the airplane in zone 1 didn't result in the airplane being found.
First off, I'm not really sure how one failed attempt to look for an airplane would have an impact on where it actually is. Of course, if a large number of searches are being made for it in zone 1 without it being found, it would seem more likely for it to be in zone 2 or 3, but the more I think about it, the less sure I am that there's an explicit connection.
Assuming that there is, it may be sensible to assume an a priori probability of 1/3 for the airplane being in a given zone (before considering the effects of how influences the conditional probability). Then again, the wording "Investigators assume that the airplane has disappeared in one of three given zones" makes it sound like a bad idea to not take an "outside zones 1-3" region into account.
So, let (1 <= k <= 3) be the event that the airplane is in zone k. (We may also let be the event that the plane is not in either of the three zones.) Furthermote, let (1 <= j <= 3) be the event that the airplane is found in zone j.
What we're looking for, then, I think, is
P(A_1 | F_1^c), P(A_2 | F_1^c) and P(A_3 | F_1^c), respectively.
P(A_1 | F_1^c) = (Bayes' theorem) = P(F_1^c | A_1)*P(A_1) / P(F_1^c) = *P(A_1) / P(F_1^c) = / (3*P(F_1^c)).
The expressions for P(A_2 | F_1^c) and P(A_3 | F_1^c) don't simplify as nicely.
I've drawn probability trees and venn diagrams, but I haven't managed to get further than what's seen above. Any hints?