两人到站等 5 分钟相遇概率

Meeting Probability

专题: Probability / 概率
难度: L4
来源: QuantQuestion

题目详情

两位银行家分别在 5:00–6:00 之间独立均匀随机到达车站（将 5:00 记为 0 分钟）。每人只等 5 分钟就离开。

问：他们在某天相遇的概率是多少？

Two bankers each arrive at the station at some random time between 5:00–6:00 am (arrival times are independent and uniformly distributed from 0 to 60 minutes). Each stays for 5 minutes and then leaves. What is the probability that they meet on a given day?

解析

设到达时间 $X,Y\sim\mathrm{Unif}(0,60)$ 独立。

相遇当且仅当 $|X-Y|\le 5$ 。

在 $60\times 60$ 的正方形内，剔除两条对角线外的两块直角三角形（边长 55），得到概率：

\frac{60^2-2\cdot\frac12\cdot 55^2}{60^2}=\frac{23}{144}.

Original Explanation

Let $X$ be the arrival time (in minutes after 5:00) for banker A, and $Y$ for banker B. Then $X,Y\sim\text{Uniform}(0,60)$ , i.i.d. The two meet if and only if $|X - Y|\le 5.$

By a geometric argument (square of side 60 in the $X\!Y$ -plane, subtracting two diagonal strips of width 5), the probability is $\frac{60\times60 - 2 \times \Bigl(\tfrac12\times55\times55\Bigr)}{60\times60} = \frac{23}{144}.$