等间隔公交 II

Bus Wait II

专题: Probability / 概率
难度: L4
来源: QuantQuestion

题目详情

公交车按固定时刻表每 10 分钟一班。

但司机在两次到站之间会独立地以 10% 概率去加油；若加油则该趟行程额外增加 1 小时（即本趟间隔变为 70 分钟）。

你在一天中随机时刻到站（均匀随机）。问：到下一班车到来的期望等待时间是多少（分钟）？

Imagine you are waiting on a bus. The bus schedule is fixed, appearing every $10$ minutes. Although, the driver independently between appearances, may want to refill on gas. The probability the driver refills on gas is $10\%$ probability per trial, independently between trials. Note, if he fills up on gas, $1$ hour is added to his travel time. If you arrive at a uniformly random time throughout the day, what is the expected time until the next bus appears, in minutes?

解析

这是“长度偏置”的等待时间问题。

一个周期内有 9 个 10 分钟间隔与 1 个 70 分钟间隔，总时长

9\cdot 10+70=160\ \text{分钟}.

随机到站落在 70 分钟长间隔的概率为 $70/160=7/16$ ；落在 10 分钟间隔的概率为 $9/16$ 。

条件期望等待：

落在 70 分钟间隔内，期望等待 $70/2=35$ 。
落在 10 分钟间隔内，期望等待 $10/2=5$ 。

因此总体期望为

35\cdot\frac{7}{16}+5\cdot\frac{9}{16}=\frac{145}{8}=18.125.