9 problems found
In this question, the notation \(\lfloor x \rfloor\) denotes the greatest integer less than or equal to \(x\), so for example \(\lfloor \pi\rfloor = 3\) and \(\lfloor 3 \rfloor =3\).
Solution:
Bag \(P\) and bag \(Q\) each contain \(n\) counters, where \(n\ge2\). The counters are identical in shape and size, but coloured either black or white. First, \(k\) counters (\(0\le k\le n\)) are drawn at random from bag \(P\) and placed in bag \(Q\). Then, \(k\) counters are drawn at random from bag \(Q\) and placed in bag \(P\).
Solution:
Solution:
A pack of cards consists of \(n+1\) cards, which are printed with the integers from \(0\) to \(n\). A~game consists of drawing cards repeatedly at random from the pack until the card printed with 0 is drawn, at which point the game ends. After each draw, the player receives \(\pounds 1\) if the card drawn shows any of the integers from \(1\) to \(w\) inclusive but receives nothing if the card drawn shows any of the integers from \(w+1\) to \(n\) inclusive.
A bag contains \(b\) black balls and \(w\) white balls. Balls are drawn at random from the bag and when a white ball is drawn it is put aside.
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*}
An examiner has to assign a mark between 1 and \(m\) inclusive to each of \(n\) examination scripts (\(n\leqslant m\)). He does this randomly, but never assigns the same mark twice. If \(K\) is the highest mark that he assigns, explain why \[ \mathrm{P}(K=k)=\left.\binom{k-1}{n-1}\right/\binom{m}{n} \] for \(n\leqslant k\leqslant m,\) and deduce that \[ \sum_{k=n}^{m}\binom{k-1}{n-1}=\binom{m}{n}\,. \] Find the expected value of \(K\).
Solution: If the highest mark is \(k\), then there are \(n-1\) remaining marks to give, and they have to be chosen from the numbers \(1, 2, \ldots, k-1\), ie in \(\binom{k-1}{n-1}\) ways. There are \(n\) numbers to be chosen from \(1, 2, \ldots, m\) in total, therefore \(\displaystyle \mathbb{P}(K=k) = \left.\binom{k-1}{n-1} \right/ \binom{m}{n}\) Since \(K\) can take any of the values \(n, \cdots, m\), we must have \begin{align*} && 1 &= \sum_{k=n}^m \mathbb{P}(K=k) \\ &&&= \sum_{k=n}^m \left.\binom{k-1}{n-1} \right/ \binom{m}{n} \\ \Rightarrow && \binom{m}{n} &= \sum_{k=n}^m \binom{k-1}{n-1} \\ \\ && \mathbb{E}(K) &= \sum_{k=n}^m k \cdot \mathbb{P}(K=k) \\ &&&= \sum_{k=n}^m k \cdot \left.\binom{k-1}{n-1} \right/ \binom{m}{n} \\ &&&= n\binom{m}{n}^{-1} \sum_{k=n}^m \frac{k}{n} \cdot \binom{k-1}{n-1} \\ &&&= n\binom{m}{n}^{-1} \sum_{k=n}^m \binom{k}{n} \\ &&&= n\binom{m}{n}^{-1} \sum_{k=n+1}^{m+1} \binom{k-1}{n+1-1} \\ &&&= n\binom{m}{n}^{-1} \binom{m+1}{n+1} \\ &&&= n \cdot \frac{m+1}{n+1} \end{align*}
I have a bag containing \(M\) tokens, \(m\) of which are red. I remove \(n\) tokens from the bag at random without replacement. Let \[ X_{i}=\begin{cases} 1 & \mbox{ if the ith token I remove is red;}\\ 0 & \mbox{ otherwise.} \end{cases} \] Let \(X\) be the total number of red tokens I remove.
Solution:
Balls are chosen at random without replacement from an urn originally containing \(m\) red balls and \(M-m\) green balls. Find the probability that exactly \(k\) red balls will be chosen in \(n\) choices \((0\leqslant k\leqslant m,0\leqslant n\leqslant M).\) The random variables \(X_{i}\) \((i=1,2,\ldots,n)\) are defined for \(n\leqslant M\) by \[ X_{i}=\begin{cases} 0 & \mbox{ if the \(i\)th ball chosen is green}\\ 1 & \mbox{ if the \(i\)th ball chosen is red. } \end{cases} \] Show that
Solution: There are \(\displaystyle \binom{m}{k} \binom{M-m}{n-k}\) ways to choose \(k\) red and and \(n-k\) green balls out of a total \(\displaystyle \binom{M}{n}\) ways to choose balls. Therefore the probability is: \[ \mathbb{P}(\text{exactly }k\text{ red balls in }n\text{ choices}) = \frac{\binom{m}{k} \binom{M-m}{n-k}}{ \binom{M}{n}}\]