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There are b animals in a forest of which a are marked. Let X be the number of animals to be captured to obtain m marked animals. Find P(X=n). The answer given in the book ((n,m) is the binomial coeff) is (a/b) (a-1,m-1)(b-a,n-m)/(b-1,n-1)However, when n=b, and m=a, this does not give P(X=n) = 1 as it should. It seems to me the correct answer is(a,m) (b-a,n-m)/(b,n) ?What am i missing here ? Thanks for all answers !

I object to using animals. The ones you captured before are going to be more or less likely to be caught twice, compared to animals that have never been caught. I imagine you will just keep catching the same subset over and over.

However, when n=b, and m=a, this does not give P(X=n) = 1 as it should.I don't think so. If m=a, means that you want to capture all marked animals. P(X=n) is the probability that you have to catch every one from the forest in order to do so. If you are lucky m catches will be enough, so P(X=n) < 1, unless m=n also.

I got a different answer to the book I think:m/n * (a choose m)/(b choose n) * ( (b-a) choose (n-m) )When n = b and m = a, this simplifies to a/bSanity check: 1 marked animal out of b. What's the chance we have to catch all of them to find it ... 1/b ... OK works.2 marked animals out of 5 ... what's the chance you have to catch all 5 ... MM, MUM, MUUM, MUUUM, UMM, UMUM, UMUUM, UUMM, UUMUM, UUUMMso it's 2/5, again my formula works

I don't think so. If m=a, means that you want to capture all marked animals. P(X=n) is the probability that you have to catch every one from the forest in order to do so. If you are lucky m catches will be enough, so P(X=n) < 1, unless m=n also.Thanks. I see where my mistake is. I was thinking that P(X=n) is the prob. that a collection of n animals will have m marked ones.

In fact my answer is the same as the book since:a * ( (a-1) choose (m-1) ) = (a choose m) * m, and b * ( (b-1) choose (n-1) ) = (b choose n) * n so the book's formula collapses to mine.My representation is much more natural and arose from how I solved it rather than all those silly '-1's coming in.

How do you explain this answer?

required prob= prob of getting one marked animal * prob of getting (m-1) marked out of (n-1)=(a/b) * (pick (m-1) marked animals from (a-1) * pick (n-m) unmarked animals from (b-a) / pick (n-1) from (b-1))=stated book answer

QuoteOriginally posted by: MattFMy representation is much more natural and arose from how I solved it rather than all those silly '-1's coming in.The book answer is natural if you think inductively.