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It goes like this: There are two stables on a farm, one that houses 20 horses and 13 mules, the other with 25 horses and 8 mules. Without any pattern, animals occasionally leave their stables and then return to their stables. Suppose that during a period when all the animals are in their stables, a horse comes out of a stable and then returns. What is the probability that the next animal coming out of the same stable will also be a horse?

I'm thinking Bayes' Rule will come into play here some how but I'm not sure.

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It is Bayes' rule, not Baye's rule, and there might not be enough information here to solve the problem without making some additional assumptions. – Dilip Sarwate Oct 12 '12 at 3:08
@Dilip SarwateThats what I thought too. What kind of assumptions are you thinking? Thanks for the correction. – TheHopefulActuary Oct 12 '12 at 3:10
I think if we interpret "without any pattern" to mean that any one of the 66 animals is as likely to leave its stable as any other, then there is enough information to solve the problem. – Gerry Myerson Oct 12 '12 at 3:12
up vote 1 down vote accepted

The probability that the next animal out of the same stable will be a horse, is (probability it was stable 1 times probability next animal out of stable 1 is a horse) plus (probability it was stable 2 times probability next animal out of stable 2 is a horse).

Now, probability next animal out of 1 is a horse is (on the simplest assumption) 20/33, and probability next animal out of 2 is a horse is 25/33, so we just have to find probability it was stable 1, given that a horse came out of it (and probability it was stable 2, but that's just one minus probability it was stable 1).

Let's write it as $P(1|H)$. Presumably you have a formula for $P(A|B)$ in terms of $P(A\cap B)$ and $P(B)$. You will need to work out a few things like $P(1)$ and $P(H|1)$ and $P(H\cap1)$ --- can you see your way through this now?

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Yes I can, thank you for the help! – TheHopefulActuary Oct 12 '12 at 5:10

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