1 problem found
I have a Penny Black stamp which I want to sell to my friend Jim, but we cannot agree a price. So I put the stamp under one of two cups, jumble them up, and let Jim guess which one it is under. If he guesses correctly, I add a third cup, jumble them up, and let Jim guess correctly, adding another cup each time. The price he pays for the stamp is \(\pounds N,\) where \(N\) is the number of cups present when Jim fails to guess correctly. Find \(\mathrm{P}(N=k)\). Show that \(\mathrm{E}(N)=\mathrm{e}\) and calculate \(\mathrm{Var}(N).\)
Solution: \begin{align*} && \mathbb{P}(N = k) &= \mathbb{P}(\text{guesses }k-1\text{ correctly then 1 wrong})\\ &&&= \frac12 \cdot \frac{1}{3} \cdots \frac{1}{k-1} \frac{k-1}{k} \\ &&&= \frac{k-1}{k!} \\ &&\mathbb{E}(N) &= \sum_{k=2}^\infty k \cdot \mathbb{P}(N=k) \\ &&&= \sum_{k=2}^{\infty} \frac{k(k-1)}{k!} \\ &&&= \sum_{k=0}^{\infty} \frac{1}{k!} = e \\ && \textrm{Var}(N) &= \mathbb{E}(N^2) - \mathbb{E}(N)^2 \\ && \mathbb{E}(N^2) &= \sum_{k=2}^{\infty} k^2 \mathbb{P}(N=k) \\ &&&= \sum_{k=2}^{\infty} \frac{k^2(k-1)}{k!} \\ &&&= \sum_{k=0}^{\infty} \frac{k+2}{k!} \\ &&&= \sum_{k=0}^{\infty} \frac{1}{k!} + 2 \sum_{k=0}^{\infty} \frac{1}{k!} = 3e \\ \Rightarrow && \textrm{Var}(N) &= 3e-e^2 \end{align*}