返回题库

等间隔公交

Bus Wait II

专题
Probability / 概率
难度
L4

题目详情

公交车按固定时刻表每 10 分钟一班。

但每一班车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 1010 minutes. Although, the driver independently between appearances, may want to refill on gas. The probability the driver refills on gas is 10%10\% probability per trial, independently between trials. Note, if he fills up on gas, 11 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”的等待问题。

10 趟里约 9 趟是 10 分钟间隔，1 趟是 70 分钟间隔。长间隔更容易被随机时刻落入。

在一个由 9 个 10 分钟与 1 个 70 分钟组成的循环里，总时长为 9⋅10+70=1609\cdot 10+70=160 分钟，其中 70 分钟处于“加油延误间隔”。因此随机到达落在延误间隔的概率为 70/160=7/1670/160=7/16，落在正常间隔的概率为 9/169/16。

  • 若落在 70 分钟间隔中，期望等待为 70/2=3570/2=35。
  • 若落在 10 分钟间隔中，期望等待为 10/2=510/2=5。

因此总体期望等待为

35⋅716+5⋅916=1458≈18.125.35\cdot\frac{7}{16}+5\cdot\frac{9}{16}=\frac{145}{8} \approx 18.125.

Original Explanation

Let's break this problem down. We need to consider the condition on showing up on a turn when the bus refills for gas versus when it does not. On 99 in every 1010 trips there won't be a gas refill and the trip length will be 1010 minutes. Hence 11 of every 1010 trips the bus does a gas refill and the trip length is 7070 minutes.

Therefore, on average, in every 160160 minutes, 7070 of those will be when the bus is on a refill trip. Hence, the probability of you appearing during a refill trip is 716\dfrac{7}{16}. The expected wait of that trip would be 12⋅70=35\dfrac{1}{2} \cdot 70 = 35. Then, with probability 916\dfrac{9}{16} the trip is regular and the expected wait is 55 minutes as per the previous part of this question. By the Law of Total Expectation, the total wait time is

35⋅716+5⋅916=145835 \cdot \dfrac{7}{16} + 5 \cdot \dfrac{9}{16} = \dfrac{145}{8}