1 problem found
A group of biologists attempts to estimate the magnitude, \(N\), of an island population of voles ({\it Microtus agrestis}). Accordingly, the biologists capture a random sample of 200 voles, mark them and release them. A second random sample of 200 voles is then taken of which 11 are found to be marked. Show that the probability, \(p_N\), of this occurrence is given by $$ p_N = k{{{\big((N-200)!\big)}^2} \over {N!(N-389)!}}, $$ where \(k\) is independent of \(N\). The biologists then estimate \(N\) by calculating the value of \(N\) for which \(p_N\) is a maximum. Find this estimate. All unmarked voles in the second sample are marked and then the entire sample is released. Subsequently a third random sample of 200 voles is taken. Write down the probability that this sample contains exactly \(j\) marked voles, leaving your answer in terms of binomial coefficients. Deduce that $$ \sum_{j=0}^{200}{389 \choose j}{3247 \choose {200-j}} = {3636 \choose 200}. $$
Solution: There will be \(200\) marked vols out of \(N\), and we are finding \(11\) of them. There are \(\binom{200}{11}\) ways to chose the \(11\) marked voles and \(\binom{N - 200}{200-11}\) ways to choose the unmarked voles. The total number of ways to choose \(200\) voles is \(\binom{N}{200}\). Therefore the probability is \begin{align*} p_N &= \frac{\binom{200}{11} \cdot \binom{N - 200}{200-11}}{\binom{N}{200}} \\ &= \binom{200}{11} \cdot \frac{ \frac{(N-200)!}{(189)!(N - 389)!} }{\frac{N!}{(N-200)!(200)!}} \\ &= \binom{200}{11} \frac{200!}{189!} \frac{\big((N-200)!\big)^2}{N!(N-389)!} \end{align*} As required and \(k = \binom{200}{11} \frac{200!}{189!}\). We want to maximise \(\frac{(N-200)!^2}{N!(N-389)!}\), we will do this by comparing consecutive \(p_N\). \begin{align*} \frac{p_{N+1}}{p_N} &= \frac{\frac{(N+1-200)!^2}{(N+1)!(N+1-389)!}}{\frac{(N-200)!^2}{N!(N-389)!}} \\ &= \frac{(N-199)!^2 \cdot N! \cdot (N-389)!}{(N+1)!(N-388)!(N-200)!^2} \\ &= \frac{(N-199)^2 \cdot 1 \cdot 1}{(N+1) \cdot (N-388)\cdot 1} \\ \end{align*} \begin{align*} && \frac{p_{N+1}}{p_N} &> 1 \\ \Leftrightarrow && \frac{(N-199)^2 \cdot 1 \cdot 1}{(N+1) \cdot (N-388)\cdot 1} & > 1 \\ \Leftrightarrow && (N-199)^2 & > (N+1) \cdot (N-388) \\ \Leftrightarrow && N^2-2\cdot199N+199^2 & > N^2 - 387N -388 \\ \Leftrightarrow && -398N+199^2 & > - 387N -388 \\ \Leftrightarrow && 199^2+388 & > 11N\\ \Leftrightarrow && \frac{199^2+388}{11} & > N\\ \Leftrightarrow && 3635\frac{4}{11} & > N\\ \end{align*} Therefore \(p_N\) is increasing if \(N \leq 3635\), so we should take \(N = 3636\). \[ \P(\text{exactly } j \text{ marked voles}) = \frac{\binom{389}{j} \cdot \binom{3636 - 389}{200-j}}{\binom{3636}{200}}\] Since \begin{align*} && 1 &= \sum_{j=0}^{200} \P(\text{exactly } j \text{ marked voles}) \\ && &= \sum_{j=0}^{200} \frac{\binom{389}{j} \cdot \binom{3247}{200-j}}{\binom{3636}{200}} \\ \Leftrightarrow&& \binom{3636}{200} &= \sum_{j=0}^{200} \binom{389}{j} \cdot \binom{3247}{200-j} \end{align*}