The time a student waits to ask a question during a tutorial sessions is a random variable with exponential distribution, with expectation of 10 minutes. Assume that waiting times of different students are mutually independent and that only one student can ask the demonstrator a question at a given time. Calculate the expected number of students that get a chance to ask the demonstrator a question until there appears a student that had to wait more than 20 minutes (including this student).
Problem from a past paper. This topic wasn't covered in class but I'd still like an idea of how to tackle it. Right now I have none.
Let $X_n\stackrel{\mathrm{i.i.d.}}\sim\mathrm{Expo}(\lambda)$, i.e. $X_1$ has density $f(t) = \lambda e^{-\lambda t}\mathsf 1_{(0,\infty)}(t)$. Let $T>0$ and define $\tau = \inf\{n>0 : X_n > T\}$. Then for each positive integer $n$, $$ \{\tau = n\} = \{X_n > T\}\cap\bigcap_{i=1}^{n-1}\{X_i\leqslant T\}. $$ By independence, we can compute the probability of this event as a product: \begin{align} \mathbb P(\tau=n) &= \mathbb P\left(\{X_n > T\}\cap\bigcap_{i=1}^{n-1}\{X_i\leqslant T\}\right)\\ &= \mathbb P(X_n>T)\prod_{i=1}^{n-1}\mathbb P(X_i\leqslant T)\\ &= \mathbb P(X_1>T)(\mathbb P(X_1\leqslant T))^{n-1}\\ &= e^{-\lambda T}(1- e^{-\lambda T})^{n-1}. \end{align} It follows that $\tau$ has a geometric distribution with parameter $e^{-\lambda T}$. We can compute its mean directly: \begin{align} \mathbb E[\tau] &= \sum_{n=1}^\infty n\cdot\mathbb P(\tau = n)\\ &= \sum_{n=1}^\infty n(1-e^{-\lambda T})^{n-1}e^{-\lambda T}\\ &= e^{-\lambda T}\sum_{n=0}^\infty (n+1)(1-e^{-\lambda T})^n\\ &= \frac{e^{-\lambda T}}{(1-(1-e^{-\lambda T}))^2}\\ &= \frac1{e^{-\lambda T}}\\ &= e^{\lambda T}. \end{align} In this case, $\lambda = \frac1{10}$ and $T=20$, so the expected value is $$e^{\frac1{10}\cdot 20}=e^2\approx 7.389056.$$