3 problems found
\(A\) and \(B\) both toss the same biased coin. The probability that the coin shows heads is \(p\), where \(0 < p < 1\), and the probability that it shows tails is \(q = 1 - p\). Let \(X\) be the number of times \(A\) tosses the coin until it shows heads. Let \(Y\) be the number of times \(B\) tosses the coin until it shows heads.
A candidate finishes examination questions in time \(T\), where \(T\) has probability density function \[ \mathrm{f}(t)=t\mathrm{e}^{-t}\qquad t\geqslant0, \] the probabilities for the various questions being independent. Find the moment generating function of \(T\) and hence find the moment generating function for the total time \(U\) taken to finish two such questions. Show that the probability density function for \(U\) is \[ \mathrm{g}(u)=\frac{1}{6}u^{3}\mathrm{e}^{-u}\qquad u\geqslant0. \] Find the probability density function for the total time taken to answer \(n\) such questions.
Solution: \begin{align*} && M_T(x) &= \mathbb{E}[e^{xT}] \\ &&&= \int_0^{\infty} e^{xt}te^{-t} \d t \\ &&&= \int_0^{\infty}te^{(x-1)t} \d t \\ &&&= \left [ \frac{t}{x-1} e^{(x-1)t} \right]_0^{\infty} - \int_0^\infty \frac{e^{(x-1)t}}{x-1} \d t \\ &&&= \left [ \frac{e^{(x-1)t}}{(x-1)^2} \right]_0^{\infty} \\ &&&= \frac{1}{(x-1)^2} \\ \\ && M_U(x) &= M_{T_1+T_2}(x) \\ &&&= \frac1{(x-1)^4} \\ \\ && I_n &= \int_0^{\infty} t^ne^{(x-1)t} \d t \\ &&&= \left[ \frac{1}{(x-1)}t^ne^{(x-1)t} \right]_0^{\infty} - \frac{n}{(x-1)} \int_0^{\infty}t^{n-1}e^{(x-1)t} \d t \\ &&&= -\frac{n}{(x-1)}I_{n-1} \\ \Rightarrow && I_n &= \frac{n!}{(1-x)^{n+1}} \\ \\ \Rightarrow && \int_0^{\infty} e^{xt} \frac16u^3e^{-u} \d u &= \int_0^{\infty} \frac16u^3e^{(x-1)u} \d u \\ &&&= \frac{1}{(1-x)^4} \\ \Rightarrow && f_U(u) &= \frac16u^3e^{-u} \\ \\ && M_{X_1+\cdots+X_n}(x) &= \frac{1}{(x-1)^{2n}} \\ \Rightarrow && f_{X_1+\cdots+X_n}(t) &= \frac1{(2n-1)!} t^{2n-1}e^{-t} \end{align*} (NB: This is the gamma distribution)
An examination consists of several papers, which are marked independently. The mark given for each paper can be an integer from \(0\) to \(m\) inclusive, and the total mark for the examination is the sum of the marks on the individual papers. In order to make the examination completely fair, the examiners decide to allocate the mark for each paper at random, so that the probability that any given candidate will be allocated \(k\) marks \((0\leqslant k\leqslant m)\) for a given paper is \((m+1)^{-1}\). If there are just two papers, show that the probability that a given candidate will receive a total of \(n\) marks is \[ \frac{2m-n+1}{\left(m+1\right)^{2}} \] for \(m< n\leqslant2m\), and find the corresponding result for \(0\leqslant n\leqslant m\). If the examination consists of three papers, show that the probability that a given candidate will receive a total of \(n\) marks is \[ \frac{6mn-4m^{2}-2n^{2}+3m+2}{2\left(m+1\right)^{2}} \] in the case \(m< n\leqslant2m\). Find the corresponding result for \(0\leqslant n\leqslant m\), and deduce the result for \(2m< n\leqslant3m\).
Solution: In order to receive \(n\) marks over the two papers, where \(m < n \leq 2m\) the student must receive \(k\) and \(n-k\) marks in each paper. Since \(n > m\), \(n-k\) is a valid mark when \(n-k \leq m\) ie when \(n-m\leq k\), therefore the probability is: \begin{align*} \sum_{k = n-m}^m \mathbb{P}(\text{scores }k\text{ and }n-k) &= \sum_{k=n-m}^m \frac{1}{(m+1)^2} \\ &= \frac{m-(n-m-1)}{(m+1)^2} \\ &= \frac{2m-n+1}{(m+1)^2} \end{align*} If \(0 \leq n \leq m\) then we need \(n-k\) marks in the second paper to be positive, ie \(n-k \geq 0 \Rightarrow n \geq k\), so \begin{align*} \sum_{k = 0}^n \mathbb{P}(\text{scores }k\text{ and }n-k) &= \sum_{k = 0}^n \frac{1}{(m+1)^2} \\ &= \frac{n+1}{(m+1)^2} \end{align*} On the first paper, they can score any number of marks, since \(n > m\), so we must have: \begin{align*} \sum_{k=0}^m \mathbb{P}(\text{scores }k\text{ and }n-k) &= \frac{1}{m+1} \sum_{k=0}^m \mathbb{P}(\text{scores }n-k\text{ on second papers}) \\ &= \frac{1}{m+1}\l \sum_{k=0}^{n-m} \frac{2m-(n-k)+1}{(m+1)^2} +\sum_{k=n-m+1}^m \frac{n-k+1}{(m+1)^2}\r \end{align*}