Bus Wait With Random Delays

Let $L$ be the gap between consecutive arrivals. The gaps are i.i.d. with

L=\begin{cases}10 & \text{with probability } 0.9,\\ 70 & \text{with probability } 0.1,\end{cases} \qquad \mathbb{E}[L]=9+7=16,\qquad \mathbb{E}[L^2]=0.9\cdot 100+0.1\cdot 4900=580 .

We treat the day as long compared with the gaps, so we may use long-run proportions.

Which gap you land in. Consider $n$ consecutive gaps. About $0.9n$ of them have length $10$ and about $0.1n$ have length $70$ . The total time is therefore about $16n$ , of which about $7n$ lies in long gaps. A uniformly random arrival time thus falls in a $70$ -minute gap with probability

\frac{0.1\cdot 70}{\mathbb{E}[L]}=\frac{7}{16}.

It falls in a $10$ -minute gap with probability $\frac{9}{16}$ . In general, a gap of length $\ell$ is selected with probability $\ell\,\mathbb{P}(L=\ell)/\mathbb{E}[L]$ , which is proportional to its length.

Wait within the gap. Given the gap you land in, your position in it is uniform. The expected wait until its end is therefore half its length: $35$ minutes in a long gap and $5$ minutes in a short one.

By the law of total expectation,

\mathbb{E}[W]=\frac{7}{16}\cdot 35+\frac{9}{16}\cdot 5=\frac{245+45}{16}=\frac{145}{8}.

This agrees with the general formula

\mathbb{E}[W]=\sum_\ell \frac{\ell\,\mathbb{P}(L=\ell)}{\mathbb{E}[L]}\cdot\frac{\ell}{2}=\frac{\mathbb{E}[L^2]}{2\,\mathbb{E}[L]}=\frac{580}{32}=\frac{145}{8}.

\boxed{\mathbb{E}[W]=\dfrac{145}{8}\approx 18.1\text{ minutes}}

Key idea. A random arrival is far more likely to fall in a long gap. The answer is therefore not the naive $\mathbb{E}[L]/2=8$ minutes, but

\frac{\mathbb{E}[L^2]}{2\,\mathbb{E}[L]}=\frac{\mathbb{E}[L]}{2}+\frac{\operatorname{Var}(L)}{2\,\mathbb{E}[L]}.