183 problems found
Tabulated values of \({\Phi}(\cdot)\), the cumulative distribution function of a standard normal variable, should not be used in this question. Henry the commuter lives in Cambridge and his working day starts at his office in London at 0900. He catches the 0715 train to King's Cross with probability \(p\), or the 0720 to Liverpool Street with probability \(1-p\). Measured in minutes, journey times for the first train are \(N(55,25)\) and for the second are \(N(65,16)\). Journey times from King's Cross and Liverpool Street to his office are \(N(30,144)\) and \(N(25,9)\), respectively. Show that Henry is more likely to be late for work if he catches the first train. Henry makes \(M\) journeys, where \(M\) is large. Writing \(A\) for \(1-{\Phi}(20/13)\) and \(B\) for \(1-{\Phi}(2)\), find, in terms of \(A\), \(B\), \(M\) and \(p\), the expected number, \(L\), of times that Henry will be late and show that for all possible values of \(p\), $$BM \le L \le AM.$$ Henry noted that in 3/5 of the occasions when he was late, he had caught the King's Cross train. Obtain an estimate of \(p\) in terms of \(A\) and \(B\). [A random variable is said to be \(N\left({{\mu}, {\sigma}^2}\right)\) if it has a normal distribution with mean \({\mu}\) and variance \({\sigma}^2\).]
Solution: If Henry catches the first train, his journey time is \(N(55+30,25+144) = N(85,13^2)\). He is on time if the journey takes less than \(105\) minutes, \(\frac{20}{13}\) std above the mean. If he catches the second train, his journey times is \(N(65+25, 16+9) = N(90, 5^2)\). He is on time if his journey takes less than \(80\) minutes, ie \(\frac{10}{5} = 2\) standard deviations above the mean. This is more likely than from the first train. \(A = 1 - \Phi(20/13)\) is the probability he is late from the first train. \(B = 1 - \Phi(2)\) is the probability he is late from the second train. The expected number of lates is \(L = M \cdot p \cdot A + M \cdot (1-p) \cdot B\), since \(B \leq A\) we must have \(BM \leq L \leq AM\) \begin{align*} && \frac35 &= \frac{pA}{pA + (1-p)B} \\ \Rightarrow && 3(1-p)B &= 2pA \\ \Rightarrow && p(2A+3B) &= 3B \\ \Rightarrow && p &= \frac{3B}{2A+3B} \end{align*}
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*}
In a lottery, any one of \(N\) numbers, where \(N\) is large, is chosen at random and independently for each player by machine. Each week there are \(2N\) players and one winning number is drawn. Write down an exact expression for the probability that there are three or fewer winners in a week, given that you hold a winning ticket that week. Using the fact that $$ {\biggl( 1 - {a \over n} \biggr) ^n \approx \e^{-a}}$$ for \(n\) much larger than \(a\), or otherwise, show that this probability is approximately \({2 \over 3}\) . Discuss briefly whether this probability would increase or decrease if the numbers were chosen by the players. Show that the expected number of winners in a week, given that you hold a winning ticket that week, is \( 3-N^{-1}\).
The random variable \(X\) takes only the values \(x_1\) and \(x_2\) (where \( x_1 \not= x_2 \)), and the random variable \(Y\) takes only the values \(y_1\) and \(y_2\) (where \(y_1 \not= y_2\)). Their joint distribution is given by $$ \P ( X = x_1 , Y = y_1 ) = a \ ; \ \ \P ( X = x_1 , Y = y_2 ) = q - a \ ; \ \ \P ( X = x_2 , Y = y_1 ) = p - a \ . $$ Show that if \(\E(X Y) = \E(X)\E(Y)\) then $$ (a - p q ) ( x_1 - x_2 ) ( y_1 - y_2 ) = 0 . $$ Hence show that two random variables each taking only two distinct values are independent if \(\E(X Y) = \E(X) \E(Y)\). Give a joint distribution for two random variables \(A\) and \(B\), each taking the three values \(- 1\), \(0\) and \(1\) with probability \({1 \over 3}\), which have \(\E(A B) = \E( A)\E (B)\), but which are not independent.
Solution: \begin{align*} \mathbb{P}(X = x_1) &= a + q - a = q \\ \mathbb{P}(X = x_2) &= 1 - q \\ \mathbb{P}(Y = y_1) & = a + p - a = p \\ \mathbb{P}(Y = y_2) & = 1 - p \end{align*} \begin{align*} \mathbb{E}(X)\mathbb{E}(Y) &= \l qx_1 + (1-q)x_2 \r \l p y_1 + (1-p)y_2\r \\ &= qpx_1y_1 + q(1-p)x_1y_2 + (1-q)px_2y_1 + (1-q)(1-p)x_2y_2 \\ \mathbb{E}(XY) &= ax_1y_1 + (q-a)x_1y_2 + (p-a)x_2y_1 + (1 + a - p - q)x_2y_2 &= \end{align*} Therefore \(\mathbb{E}(XY) - \mathbb{E}(X)\mathbb{E}(Y)\) is a degree 2 polynomial in the \(x_i, y_i\). If \(x_1 = x_2\) then we have: \begin{align*} \mathbb{E}(X)\mathbb{E}(Y) &=x_1 \l p y_1 + (1-p)y_2\r \\ \mathbb{E}(XY) &= x_1(ay_1 + (q-a)y_2 + (p-a)y_1 + (1 + a - p - q)y_2) \\ &= x_1 (py_1 + (1-p)y_2) \end{align*} Therefore \(x_1 - x_2\) is a root and by symmetry \(y_1 - y_2\) is a root. Therefore it remains to check the coefficient of \(x_1y_1\) which is \(a - pq\) to complete the factorisation. For any two random variables taking two distinct values, we can find \(a, q, p\) satisfying the relations above. We also note that \(X\) and \(Y\) are independent if \(\mathbb{P}(X = x_i, Y = y_i) = \mathbb{P}(X = x_i)\mathbb{P}(Y = y_i)\). Since \(x_1 \neq x_2\) and \(y_1 \neq y_2\) and \(\E(A B) = \E( A)\E (B) \Rightarrow a = pq\). But if \(a = pq\), we have \(\mathbb{P}(X = x_1, Y = y_1) = \mathbb{P}(X = x_1)\mathbb{P}(Y = y_1)\) and all the other relations drop out similarly. Consider \begin{align*} \mathbb{P}(A = -1, B = 1) &= \frac{1}{6} \\ \mathbb{P}(A = -1, B = -1) &= \frac{1}{6} \\ \mathbb{P}(A = 0, B = 0) &= \frac{1}{3} \\ \mathbb{P}(A = 1, B = -1) &= \frac{1}{6} \\ \mathbb{P}(A = -1, B = -1) &= \frac{1}{6} \end{align*}
Solution:
Bar magnets are placed randomly end-to-end in a straight line. If adjacent magnets have ends of opposite polarities facing each other, they join together to form a single unit. If they have ends of the same polarity facing each other, they stand apart. Find the expectation and variance of the number of separate units in terms of the total number \(N\) of magnets.
Solution: There are \(N-1\) gaps between the magnets which are independently gaps or not gaps. Therefore the total number of gaps is \(X \sim Binomial(N-1, \frac12)\) and \begin{align*} \mathbb{E}(X) &= \frac{N-1}{2} \\ \textrm{Var}(X) &= \frac{N-1}{4} \end{align*}
When I throw a dart at a target, the probability that it lands a distance \(X\) from the centre is a random variable with density function \[ \mathrm{f}(x)=\begin{cases} 2x & \text{ if }0\leqslant x\leqslant1;\\ 0 & \text{ otherwise.} \end{cases} \] I score points according to the position of the dart as follows: %
A stick is broken at a point, chosen at random, along its length. Find the probability that the ratio, \(R\), of the length of the shorter piece to the length of the longer piece is less than \(r\). Find the probability density function for \(R\), and calculate the mean and variance of \(R\).
Solution: Let \(X \sim U[0, \tfrac12]\) be the shorter piece, so \(R = \frac{X}{1-X}\), and \begin{align*} && \mathbb{P}(R \leq r) &= \mathbb{P}(\tfrac{X}{1-X} \leq r) \\ &&&= \mathbb{P}(X \leq r - rX) \\ &&&= \mathbb{P}((1+r)X \leq r) \\ &&&= \mathbb{P}(X \leq \tfrac{r}{1+r} ) \\ &&&= \begin{cases} 0 & r < 0 \\ \frac{2r}{1+r} & 0 \leq r \leq 1 \\ 1 & r > 1 \end{cases} \\ \\ && f_R(r) &= \begin{cases} \frac{2}{(1+r)^2} & 0 \leq r \leq 1 \\ 0 & \text{otherwise} \end{cases} \end{align*} Let \(Y \sim U[\tfrac12, 1]\) be the longer piece, then \(R = \frac{1-Y}{Y} = Y^{-1} - 1\) and \begin{align*} \E[R] &= \int_{\frac12}^1 (y^{-1}-1) 2 \d y \\ &= 2\left [\ln y - y \right]_{\frac12}^1 \\ &= -2 + 2\ln2 +2\frac12 \\ &= 2\ln2 -1 \\ \\ \E[R^2] &= \int_{\frac12}^1 (y^{-1}-1)^2 2 \d y\\ &= 2\left [-y^{-1} -2\ln y + 1 \right]_{\frac12}^1 \\ &= 2 \left ( 2 - 2\ln 2+\frac12\right) \\ &= 3-4\ln 2 \\ \var[R] &= 3 - 4 \ln 2 -(2\ln 2-1)^2 \\ &= 2 - 4(\ln 2)^2 \end{align*}
You play the following game. You throw a six-sided fair die repeatedly. You may choose to stop after any throw, except that you must stop if you throw a 1. Your score is the number obtained on your last throw. Determine the strategy that you should adopt in order to maximize your expected score, explaining your reasoning carefully.
Solution: Once you have thrown, all previous throws are irrelevant so the only thing which can affect your decision is the current throw. Therefore the strategy must consist of a list of states we re-throw from, and a list of states we stick on. It must also be the case that if we stick on \(k\) we stick on \(k+1\) (otherwise we can improve our strategy by switching those two values around). Therefore we can form a table of our expected score: \begin{array}{c|c|c} \text{stop on} & \text{possible outcomes} & \E[\text{score}] \\ \hline \geq 2 & \{1,2,3,4,5,6\} & \frac{21}{6} = 3.5 \\ \geq 3 & \{1,3,4,5,6\} & \frac{19}{5} = 3.8 \\ \geq 4 & \{1,4,5,6\} & \frac{16}{4} = 4 \\ \geq 5 & \{1,5,6\} & \frac{12}{3} = 4 \\ =6 & \{1,6\} & \frac{7}{2} = 3.5 \end{array} Therefore the optimal strategy is to stop on \(4\) or higher. If we cared about variance we might look at the variance of the two best strategies, \(4\) or higher has a variance of \(\frac{1+16+25+36}{4} - 16 = 3.5\) and \(5\) or higher has a variance of \(\frac{1+25+36}3 - 16 = \frac{14}3 > 3.5\) so \(4\) or higher is probably better in most scenarios.
In the game of endless cricket the scores \(X\) and \(Y\) of the two sides are such that \[ \P (X=j,\ Y=k)=\e^{-1}\frac{(j+k)\lambda^{j+k}}{j!k!},\] for some positive constant \(\lambda\), where \(j,k = 0\), \(1\), \(2\), \(\ldots\).
Solution:
The cakes in our canteen each contain exactly four currants, each currant being randomly placed in the cake. I take a proportion \(X\) of a cake where \(X\) is a random variable with density function \[{\mathrm f}(x)=Ax\] for \(0\leqslant x\leqslant 1\) where \(A\) is a constant.
In the basic version of Horizons (H1) the player has a maximum of \(n\) turns, where \(n \ge 1\). At each turn, she has a probability \(p\) of success, where \(0 < p < 1\). If her first success is at the \(r\)th turn, where \(1 \le r \le n\), she collects \(r\) pounds and then withdraws from the game. Otherwise, her winnings are nil. Show that in H1, her expected winnings are $$ p^{-1}\left[1+nq^{n+1}-(n+1)q^n\right]\quad\hbox{pounds}, $$ where \(q=1-p\). The rules of H2 are the same as those of H1, except that \(n\) is randomly selected from a Poisson distribution with parameter \(\lambda\). If \(n=0\) her winnings are nil. Otherwise she plays H1 with the selected \(n\). Show that in H2, her expected winnings are $$ {1 \over p}{\left(1-{\e^{-{\lambda}p}}\right)} -{{\lambda}q}{\e^{-{\lambda}p}} \quad\hbox{pounds}. $$
Solution: \begin{align*} && \E[H1] &= \sum_{r=1}^n r \cdot \mathbb{P}(\text{first success on }r\text{th turn}) \\ &&&= \sum_{r=1}^n r \cdot q^{r-1}p \\ &&&= p\sum_{r=1}^n r q^{r-1} \\ \\ && \frac{1-x^{n+1}}{1-x} &= \sum_{r=0}^n x^r \\ \Rightarrow && \sum_{r=1}^n r x^{r-1} &= \frac{-(n+1)x^n(1-x) +(1-x^{n+1})}{(1-x)^2} \\ &&&= \frac{1-(n+1)x^n+nx^{n+1}}{(1-x)^2} \\ \\ && \E[H1] &= p\sum_{r=1}^n r q^{r-1} \\ &&&= p\frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2} \\ &&&= p^{-1}(1-(n+1)q^{n} + nq^{n+1}) \end{align*} Not that if \(n =0\) , the formula for \(\E[H1] = 0\). So \begin{align*} && \E[H2] &= \E[\E[H1|n=N]] \\ &&&= p^{-1}\E \left [ 1-(N+1)q^{N} + Nq^{N+1}\right] \\ &&&= p^{-1}\E \left [ 1-((1-q)N+1)q^{N} \right] \\ &&&= p^{-1}\left (1 - p\E[Nq^N] - G_{Po(\lambda)}(q) \right) \\ &&&= p^{-1}(1-e^{-\lambda(1-q)}) - \E[Nq^N] \\ &&&= p^{-1}(1-e^{-\lambda(1-q)}) - q\lambda e^{-\lambda(1-q)} \\ &&&= p^{-1}(1-e^{-\lambda p}) - q\lambda e^{-\lambda p} \end{align*}
Suppose that a solution \((X,Y,Z)\) of the equation \[X+Y+Z=20,\] with \(X\), \(Y\) and \(Z\) non-negative integers, is chosen at random (each such solution being equally likely). Are \(X\) and \(Y\) independent? Justify your answer. Show that the probability that \(X\) is divisible by \(5\) is \(5/21\). What is the probability that \(XYZ\) is divisible by 5?
Solution: They are not independent: \begin{align*} && \mathbb{P}(X = 20 \,\, \cap Y = 20) = 0 \\ && \mathbb{P}(X = 20 )\mathbb{P}(Y = 20) \neq 0 \\ \end{align*} \begin{align*} X = 0: && 21 \text{ solutions} \\ X = 5: && 16 \text{ solutions} \\ X = 10: && 11 \text{ solutions} \\ X = 15: && 6 \text{ solutions} \\ X = 20: && 1 \text{ solutions} \\ 5 \mid X: && 55 \text{ solutions} \\ \\ && \binom{20+2}{2} = 11 \cdot 21 \text{ total solutions} \\ \Rightarrow && \mathbb{P}(5 \mid X) = \frac{55}{11 \cdot 21} = \frac{5}{21} \end{align*} \begin{align*} \mathbb{P}(5 \mid XYZ) &= 3\cdot \mathbb{P}(5 \mid X) - 2\mathbb{P}(5 \mid X, Y, Z) \\ &= \frac{3 \cdot 55 - 2 \cdot \binom{4+2}{2}}{11 \cdot 21} = \frac{35}{77} \end{align*}
The diagnostic test AL has a probability 0.9 of giving a positive result when applied to a person suffering from the rare disease mathematitis. It also has a probability 1/11 of giving a false positive result when applied to a non-sufferer. It is known that only \(1\%\) of the population suffer from the disease. Given that the test AL is positive when applied to Frankie, who is chosen at random from the population, what is the probability that Frankie is a sufferer? In an attempt to identify sufferers more accurately, a second diagnostic test STEP is given to those for whom the test AL gave a positive result. The probablility of STEP giving a positive result on a sufferer is 0.9, and the probability that it gives a false positive result on a non-sufferer is \(p\). Half of those for whom AL was positive and on whom STEP then also gives a positive result are sufferers. Find \(p\).
Solution: \begin{align*} \mathbb{P}(M | P_{AL}) &= \frac{\mathbb{P}(M \cap P_{AL})}{\mathbb{P}(P_{AL})} \\ &= \frac{\frac{1}{100} \frac{9}{10}}{\frac{1}{100} \frac{9}{10} + \frac{99}{100} \frac{1}{11}} \\ &= \frac{\frac{9}{10}}{\frac{9}{10} + \frac{9}{1}} \\ &= \frac{9}{99} = \frac{1}{11} \\ \end{align*} \begin{align*} && \frac12 &= \mathbb{P}(M | P_{STEP}, P_{AL}) \\ &&&= \frac{\frac{1}{100} \frac{9}{10} \frac{9}{10}}{\frac{1}{100} \frac{9}{10} \frac{9}{10} + \frac{99}{100} \frac{1}{11}p} \\ &&&= \frac{81}{81+900p} \\ \Rightarrow && p &= \frac{81}{900} = \frac{9}{100} \end{align*} Therefore \(p = 9\%\)
The staff of Catastrophe College are paid a salary of \(A\) pounds per year. With a Teaching Assessment Exercise impending it is decided to try to lower the student failure rate by offering each lecturer an alternative salary of \(B/(1+X)\) pounds, where \(X\) is the number of his or her students who fail the end of year examination. Dr Doom has \(N\) students, each with independent probability \(p\) of failure. Show that she should accept the new salary scheme if $$A(N+1)p < B(1-(1-p)^{N+1}).$$ Under what circumstances could \(X\), for Dr Doom, be modelled by a Poisson random variable? What would Dr Doom's expected salary be under this model?
Solution: \begin{align*} && \E[\text{salary}] &= B\sum_{k=0}^N \frac{1}{1+k}\binom{N}{k}p^k(1-p)^{N-k} \\ \\ && (q+x)^N &= \sum_{k=0}^N \binom{N}{k}x^kq^{N-k} \\ \Rightarrow && \int_0^p(q+x)^N \d x &= \sum_{k=0}^N \binom{N}{k} \frac{p^{k+1}}{k+1}q^{N-k} \\ && \frac{(q+p)^{N+1}-q^{N+1}}{N+1} &= \frac{p}{B} \E[\text{salary}] \\ \Rightarrow && \E[\text{salary}] &= B\frac{1-q^{N+1}}{p(N+1)} \end{align*} Therefore if \(Ap(N+1) < B(1-(1-p)^{N+1})\) the expected value of the new salary is higher. (Whether or not the new salary is worth it in a risk adjusted sense is for the birds). We could model \(X\) by a Poisson random variable if \(N\) is large and \(Np = \lambda \) is small. Suppose \(X \approx Po(\lambda)\) then \begin{align*} \E \left [\frac{B}{1+X} \right] &= B\sum_{k=0}^\infty \frac{1}{1+k}\frac{e^{-\lambda}\lambda^k}{k!} \\ &= \frac{B}{\lambda} \sum_{k=0}^\infty e^{-\lambda} \frac{\lambda^{k+1}}{(k+1)!} \\ &= \frac{B}{\lambda}e^{-\lambda}(e^{\lambda}-1) \\ &= \frac{B(1-e^{-\lambda})}{\lambda} = B \frac{1-e^{-Np}}{Np} \end{align*}