An automated mobile dummy target for gunnery practice is moving anti-clockwise around the circumference of a large circle of radius \(R\) in a horizontal plane at a constant angular speed \(\omega\). A shell is fired from \(O\), the centre of this circle, with initial speed \(V\) and angle of elevation \(\alpha\). Show that if \(V^2 < gR\), then no matter what the value of \(\alpha\), or what vertical plane the shell is fired in, the shell cannot hit the target. Assume now that \(V^2 > gR\) and that the shell hits the target, and let \(\beta\) be the angle through which the target rotates between the time at which the shell is fired and the time of impact. Show that \(\beta\) satisfies the equation $$ g^2{{\beta}^4} - 4{{\omega}^2}{V^2}{{\beta}^2} +4{R^2}{{\omega}^4}=0. $$ Deduce that there are exactly two possible values of \(\beta\). Let \(\beta_1\) and \(\beta_2\) be the possible values of \(\beta\) and let \(P_1\) and \(P_2\) be the corresponding points of impact. By considering the quantities \((\beta_1^2 +\beta_2^2) \) and \(\beta_1^2\beta_2^2\,\), or otherwise, show that the linear distance between \(P_1\) and \(P_2\) is \[ 2R \sin\Big( \frac\omega g \sqrt{V^2-Rg}\Big) \;. \]
Solution: \begin{align*} && 0 &= V\sin \alpha t-\frac12 gt^2 \\ \Rightarrow && t &= \frac{2V \sin \alpha}{g} \\ && R &= V \cos \alpha \, t \\ &&&= \frac{2V^2 \sin \alpha \cos \alpha}{g} \\ &&&= \frac{V^2 \sin 2 \alpha}{g} \end{align*} Therefore the max distance is \(\frac{V^2}{g}\), therefore we cannot hit the target if \(R > \frac{V^2}{g} \Rightarrow gR > V^2\). We have \(\beta = \omega t \Rightarrow t = \frac{\beta}{\omega}\) \begin{align*} && \sin \alpha &= \frac{gt}{2V} \\ && \cos \alpha &= \frac{R}{Vt} \\ \Rightarrow && 1 &= \left (\frac{gt}{2V} \right)^2 + \left ( \frac{R}{Vt} \right)^2 \\ &&&= \left (\frac{g\beta}{2V \omega} \right)^2 + \left ( \frac{R\omega}{V\beta} \right)^2 \\ &&&= \frac{g^2 \beta^2}{4 V^2 \omega^2} + \frac{R^2 \omega^2}{V^2 \beta ^2} \\ \Rightarrow && 4V^2 \omega^2 \beta^2 &= g^2 \beta^4 + 4R^2 \omega^4 \\ \Rightarrow && 0 &= g^2 \beta^4 - 4\omega^2 V^2 \beta^2+4R^2\omega^4 \end{align*} This (quadratic) equation in terms of \(\beta^2\) has two solution if \(\Delta = 16\omega^4V^4-16g^2R^2\omega^4 =16\omega^4(V^4-g^2R^2) > 0\) since \(V^2 > gR\). Since \(\beta > 0\) there are exactly two solutions, once we have values for \(\beta\). First notice, \begin{align*} && \beta_1^2 + \beta_2^2 &= \frac{4\omega^2V^2}{g^2} \\ && \beta_1^2\beta_2^2 &= \frac{4R^2\omega^4}{g^2} \end{align*} Then notice the positions of \(P_1\) and \(P_2\) are \((R\cos \beta_1 , R\sin \beta_1)\) and \((R\cos \beta_2, R\sin \beta_2)\). \begin{align*} && d^2 &= R^2\left ( \cos \beta_1 - \cos \beta_2 \right)^2 + R^2 \left ( \sin \beta_1 - \sin \beta_2 \right)^2 \\ &&&= 2R^2 - 2R^2(\cos \beta_1 \cos \beta_2 + \sin \beta_1 \sin \beta_2) \\ &&&= 2R^2-2R^2\cos(\beta_1 - \beta_2) \\ &&&= 2R^2 \left (1-\cos(\sqrt{(\beta_1-\beta_2)^2} \right ) \\ &&&= 2R^2 \left (1 - \cos\left ( \sqrt{\frac{4\omega^2 V^2}{g^2} - \frac{4R\omega^2}{g}} \right) \right) \\ &&&= 2R^2 \left (1 - \cos\left (\frac{2\omega}{g} \sqrt{V^2 - Rg} \right) \right) \\ &&&= 4 R^2 \sin^2 \left (\frac{\omega}{g} \sqrt{V^2 - Rg} \right) \end{align*} which gives us the required result.
It is known that there are three manufacturers \(A, B, C,\) who can produce micro chip MB666. The probability that a randomly selected MB666 is produced by \(A\) is \(2p\), and the corresponding probabilities for \(B\) and \(C\) are \(p\) and \(1 - 3p\), respectively, where \({{0} \le p \le {1 \over 3}}.\) It is also known that \(70\%\) of MB666 micro chips from \(A\) are sound and that the corresponding percentages for \(B\) and \(C\) are \(80\%\) and \(90\%\), respectively. Find in terms of \(p\), the conditional probability, \(\P(A {\vert} S)\), that if a randomly selected MB666 chip is found to be sound then it came from \(A\), and also the conditional probability, \(\P(C {\vert} S)\), that if it is sound then it came from \(C\). A quality inspector took a random sample of one MB666 micro chip and found it to be sound. She then traced its place of manufacture to be \(A\), and so estimated \(p\) by calculating the value of \(p\) that corresponds to the greatest value of \(\P(A {\vert} S)\). A second quality inspector also a took random sample of one MB666 chip and found it to be sound. Later he traced its place of manufacture to be \(C\) and so estimated \(p\) by applying the procedure of his colleague to \(\P(C {\vert} S)\). Determine the values of the two estimates and comment briefly on the results obtained.
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.