两路公交：平均等车时间

两车构成 30 分钟周期：在每个周期内，先来 #1（距离上一班 #2 已过 10 分钟），再过 20 分钟来 #2。

设到站时刻 $X\sim U(0,30)$（相对某班 #2 的到达）。若 $X\le 10$，等到 #1 需 $10-X$；若 $X>10$，等到 #2 需 $30-X$。

因此

$$\mathbb{E}[W]=\frac{1}{30}\left(\int_0^{10}(10-x)dx+\int_{10}^{30}(30-x)dx\right)=\boxed{\frac{25}{3}}\text{ 分钟}.$$

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英文解析

Consider a time interval between two arrivals of bus #2. This interval lasts for 30 minutes. If a customer arrives in the first 10 minutes of this interval, the customer will take bus #1. Otherwise, the customer will take bus #2.

Let $X$ be the uniform random variable on [0, 30] representing the arrival time of the customer. The waiting time for the bus is

$$W = (10 - X) \cdot \mathbf{1}_{X \leq 10} + (30 - X) \cdot \mathbf{1}_{X > 10},$$

where $\mathbf{1}_{X \leq 10}$ and $\mathbf{1}_{X > 10}$ are the indicator functions of the events $\{X \leq 10\}$ and $\{X > 10\}$ .

10}, respectively, e.g.,

$$\mathbf{1}_{X\leq 10} = \left\{ \begin{array}{ll}1, & \mathrm{if} X\leq 10 \\ 0, & \mathrm{if} X > 10. \end{array} \right.$$

The expression for $W$ can also be written as

$$\begin{array}{c}W = (10 - X) \cdot (\mathbf{1}_{X\leq 10} + \mathbf{1}_{X > 10}) + 20 \cdot \mathbf{1}_{X > 10} \\ = 10 - X + 20 \cdot \mathbf{1}_{X > 10}, \end{array}$$

and therefore

$$\mathbb{E}[W] = 10 - E[X] + 20P(X > 10).$$

Note that $\mathbb{E}[X] = 15$ and $\mathbb{P}(X > 10) = \frac{2}{3}$ , since uniform on $[0,30]$ . From (2.43), it follows that

$$\mathbb{E}[W] = 10 - 15 + 20 \cdot \frac{2}{3} = \frac{25}{3}.$$

The expected waiting time to get on a bus is $\frac{25}{3}$ minutes, that is, 8 minutes and 20 seconds.