The discrete random variable $X$ has probability mass function given by $\mathrm{P}(X=x)=\left\{ \be

Question

The discrete random variable $X$ has probability mass function given by $\mathrm{P}(X=x)=\left\{ \begin{matrix} & \frac{ax}{n(n+1)}\,\,\,\,\mathrm{for}\,\,x=1,2,...,n, \ & 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathrm{otherwise,} \ \end{matrix} ight.$ where $a$ is a constant. Show that $a=2$. [2] Find $\mathrm{E}(X)$ in terms of $n$. You may use the result that $\sum\limits_{r=1}^{n}{{{r}^{2}}=\frac{n}{6}(n+1)(2n+1)}$. [3] Two players ${{P}_{1}}$ and ${{P}_{2}}$ play a game with $n+1$ tokens labelled $1$ to $n+1$. Each player randomly picks one token without replacement and the player who picks the token with the smaller number loses. The amount of money lost by the losing player, in dollars, will be the number on the winning token. For example, if ${{P}_{1}}$ and ${{P}_{2}}$ pick the tokens labelled $5$ and $3$ respectively, ${{P}_{2}}$ loses $\$\,5$ to ${{P}_{1}}$. Explain why the probability of ${{P}_{2}}$ losing a game is $0.5$. [1] Given that ${{P}_{2}}$ loses, find the probability ${{P}_{2}}$ loses $\$\,(m+1)$ in terms of $m$ and $n$, where $m$ is such that $1\le m\le n$. [3] Using the result in part (i), when ${{P}_{2}}$ loses a game, find the amount that he is expected to lose in terms of $n$. [3]

Accepted Answer

(i) $a = 2$ (ii) $\frac{1}{3}(2n+1)$ (v) $\frac{2}{3}n + \frac{4}{3}$

2018 DHS P2 Q10

2018 DHS P2 Q10

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