Question

Problem 5 5.a Consider the following identity. For all positive integers n and k with n 2k, (n choose k) + (n choose k-1) = (n+1 choose k). This can be demonstrated either algebraically or via a story proof. To prove the identity algebraically, we can write (n choose k) + (n choose k-1) = n!/[k!(n-k)!] + n!/[(k-1)!(n-k+1)!] = [(n-k+1)n! + (k)n!]/[k!(n-k+1)!] [n!(n+1)/k!(n-k+1)!] = (n+1 choose k). Which of the following is a story proof of the identity? Consider a group of n-k+1 peacocks and k toucans. The right-hand side is the number of ways to choose k of these birds, with order not mattering. The left-hand side counts the same thing in a different way, breaking it into two cases: all k toucans are chosen (first term) or at least one toucan is not chosen (second term). Consider n+1 people, with one of them pre-designated as the "president" of the group. The right-hand side is the number of ways to choose k out of these n+1 people, with order not mattering. The left-hand side counts the same thing in a different way, by considering two disjoint cases: the president is not in the chosen subset (first term) or is in the chosen subset (second term). Consider a group of n-1 peacocks and 2 toucans. The right-hand side is the number of ways to choose k of these birds, with order not mattering. The left-hand side counts the same thing in a different way, breaking it into two cases: no toucans are chosen (first term) or at least one toucan is chosen (second term). Consider n+1 people, with one of them pre-designated as the "president" of the group. The right-hand side is the number of committees of size k that can be formed from the people. The left-hand side is the number of committees of size k or k-1 that can be formed, if the president is not allowed to be on the committee. 5.b Elk dwell in a certain forest. There are N elk, of which a simple random sample of size n are captured and tagged ("simple random sample" means that all (Nn) sets of n elk are equally likely). The captured elk are returned to the population, and then a new sample is drawn, this time with size m. This is an important method that is widely used in ecology, known as capture- recapture. What is the probability that exactly k of the melk in the new sample were previously tagged? (Assume that an elk that was captured before doesn't become more or less likely to be captured again.) (n choose k)*(m/N)^{"{{N-m}/N)^{n-k} [(n choose m)*(N-n choose k)]/(N choose m) (m choose k)*(n/N)^k"({N-ny/N)^{m-k} [(n choose k)*(N-n choose m-k)]/(N choose m)

Answer 1

5.a) Option 4

5.b) Option 4 because of the n tagged elks , k are to be chosen, this can be done in (n choose k) ways. The remaining (m-k) elks can be chosen from (N-n) untagged elks in ((N-n) choose (m-k)) ways. In general m elks can be selected from N elks in (N choose m) ways. Thus the required probability is

(n choose k)*((N-n) choose(m-k))/(N choose m)

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