The diagrams below show two separate systems of particles, strings and pulleys. In both systems, the pulleys are smooth and light, the strings are light and inextensible, the particles move vertically and the pulleys labelled with \(P\) are fixed. The masses of the particles are as indicated on the diagrams.
A game in a casino is played with a fair coin and an unbiased cubical die whose faces are labelled \(1, 1, 1, 2, 2\) and \(3.\) In each round of the game, the die is rolled once and the coin is tossed once. The outcome of the round is a random variable \(X\). The value, \(x\), of \(X\) is determined as follows. If the result of the toss is heads then \(x= \vert ks -1\vert\), and if the result of the toss is tails then \(x=\vert k-s\vert\), where \(s\) is the number on the die and \(k\) is a given number. Show that \(\mathbb{E}(X^2) = k +13(k-1)^2 /6\). Given that both \(\mathbb{E}(X^2)\) and \(\mathbb{E}(X)\) are positive integers, and that \(k\) is a single-digit positive integer, determine the value of \(k\), and write down the probability distribution of \(X\). A gambler pays \(\pounds 1\) to play the game, which consists of two rounds. The gambler is paid:
Solution: \begin{align*} && \mathbb{E}(X^2) &= \frac12 \left (\frac16 \left ( 3(k -1)^2+2(2k-1)^2+(3k-1)^2 \right) +\frac16 \left ( 3(k -1)^2+2(k-2)^2+(k-3)^2 \right) \right) \\ &&&= \frac12 \left (\frac16 \left (20k^2-20k+6 \right) + \frac16 \left ( 6k^2-20k+20\right) \right) \\ &&&= \frac1{12} \left (26k^2-40k+ 26\right) \\ &&&= \frac{13}{6} (k^2+1) - \frac{10}{3}k \\ &&&= \frac{13}{6}(k-1)^2+k \end{align*} Since \(k\) a single digit positive number and \(\mathbb{E}(X^2)\) is an integer, \(6 \mid k-1 \Rightarrow k = 1, 7\). \begin{align*} \mathbb{E}(X | k=1) &= \frac12 \left (\frac16 \left ( 2+2 \right) +\frac16 \left ( 2+2 \right) \right) = \frac23 \not \in \mathbb{Z}\\ \mathbb{E}(X | k=7) &= \frac12 \left (\frac16 \left ( 3\cdot6+2\cdot13+20 \right) +\frac16 \left ( 3\cdot6+2\cdot5+4 \right) \right) = 8 \end{align*} Therefore \(k = 7\) The probability distribution is \begin{align*} && \mathbb{P}(X=4) = \frac1{12} \\ && \mathbb{P}(X=5) = \frac1{6} \\ && \mathbb{P}(X=6) = \frac12 \\ && \mathbb{P}(X=13) = \frac1{6} \\ && \mathbb{P}(X=20)= \frac1{12} \\ \end{align*} The only ways to score more than \(25\) are: \(20+6, 20+13, 20+20, 13+13\) The only ways to score exactly \(25\) are \(20+5\) \begin{align*} \mathbb{P}(>25) &= \frac1{12} \cdot\left(2\cdot \frac12+2\cdot\frac16+\frac1{12}\right) + \frac{1}{6^2} \\ &= \frac{7}{48} \\ \mathbb{P}(=25) &= \frac{2}{12 \cdot 6} = \frac{1}{36} \\ \\ \mathbb{E}(\text{payout}) &= \frac{7}{48}w + \frac{1}{36} = \frac{21w+4}{144} \end{align*} The casino needs \(\frac{21w+4}{144} < 1 \Rightarrow 21w< 140 \Rightarrow w < \frac{20}{3}\)
A continuous random variable \(X\) has a triangular distribution, which means that it has a probability density function of the form \[ \f(x) = \begin{cases} \g(x) & \text{for \(a< x \le c\)} \\ \h(x) & \text{for \(c\le x < b\)} \\ 0 & \text{otherwise,} \end{cases} \] where \(\g(x)\) is an increasing linear function with \(\g(a)=0\), \(\h(x)\) is a decreasing linear function with \(\h(b) =0\), and \(\g(c)=\h(c)\). Show that \(\g(x) = \dfrac{2(x-a)}{(b-a)(c-a)}\) and find a similar expression for \(\h(x)\).
Solution: Since \(\int f(x) \, dx = 1\), and \(f(x)\) is a triangle with base \(b-a\), it must have height \(\frac{2}{b-a}\) in order to have the desired area. Since \(g(a) = 0, g(c) = \frac{2}{b-a}\), \(g(x) = A(x-a)\) and \(\frac{2}{b-a} = A (c-a) \Rightarrow g(x) = \frac{2(x-a)}{(b-a)(c-a)}\) as required. Similarly, \(h(x) = B(x-b)\) and \(\frac{2}{b-a} = B(c-b) \Rightarrow h(x) = \frac{2(b-x)}{(b-a)(b-c)}\) The mean of the distribution will be: \begin{align*} \int_a^b xf(x) \, dx &= \int_a^c xg(x) \, dx + \int_c^b xh(x) \, dx \\ &= \frac{2}{(b-a)(c-a)} \int_a^c x(x-a) dx + \frac{2}{(b-a)(b-c)} \int_c^b x(b-x) \, dx \\ &= \frac{2}{(b-a)} \l \frac{1}{c-a} \left [ \frac{x^3}{3} - a\frac{x^2}{2} \right ]_a^c + \frac{1}{b-c} \left [ b\frac{x^2}{2} - \frac{x^3}{3} \right ]_c^b\r \\ &= \frac{2}{(b-a)} \l \frac{1}{c-a} \l \frac{c^3}{3} - a\frac{c^2}{2} - \frac{a^3}{3} + \frac{a^3}{2} \r + \frac{1}{b-c} \l \frac{b^3}{2} - \frac{b^3}{3} - \frac{bc^2}{2} + \frac{c^3}{3} \r \r \\ &= \frac{2}{(b-a)} \l \l \frac{c^2+ac+a^2}{3} - \frac{a(a+c)}{2} \r +\l \frac{b(b+c)}{2} - \frac{b^2+bc+c^2}{3} \r\r \\ &= \frac{2}{(b-a)} \l \frac{2c^2+2ac+2a^2}{6} - \frac{3a^2+3ac}{6} + \frac{3b^2+3bc}{6} - \frac{2b^2+2bc+2c^2}{6} \r \\ &= \frac{2}{(b-a)} \l \frac{-a^2+b^2-ac+bc}{6} \r \\ &= \frac{a+b+c}{3} \\ \end{align*} The median \(M\) satisfies: \begin{align*} && \int_a^M f(x) \, dx &= \frac12 \\ \end{align*} The left hand triangle will have area: \(\frac{c-a}{b-a}\) which will be \(\geq \frac12\) if \(c \geq \frac{a+b}{2}\). In this case we need \begin{align*} && \frac{(M-a)^2}{(b-a)(c-a)} &= \frac12 \\ \Rightarrow && M &= a + \sqrt{\frac12 (b-a)(c-a)} \end{align*} Otherwise, we need: \begin{align*} && \frac{(b-M)^2}{(b-a)(b-c)} &= \frac12 \\ \Rightarrow && M &= b - \sqrt{\frac12 (b-a)(b-c)} \end{align*} These are consistent, if \(c = \frac{b+a}{2}\)