TITLE: Chance of meeting probability
QUESTION [6 upvotes]: Two people arrive at a train station some random time between $5$ and $6$ a.m.. Arrival times are uniformly distributed. They stay for exactly five minutes and leave. What's the chance that they meet on a particular day?
My given answer is $23/144$.

Equivalently, given $X, Y \sim U(0, 60)$, we want $P(|X - Y| \leq 5)$. I know that there's a geometric approach, but I want to know what's wrong with my following approach:
I will compute $P(X \leq Y + 5)$ and adjust the answer to account for both cases.
$$P(X \leq Y + 5) = \int P(X \leq Y + 5 \mid Y = k)P(Y = k) = \int P(X \leq k + 5)f_{Y}(k) \mathop{dk} $$
$$= \frac{1}{60} \int P(X \leq k + 5).$$
For $0 \leq k \leq 55$, we have $P(X \leq k + 5) = (k + 5)/60$. For $55 < k \leq 60$, we have $P(X \leq k + 5) = 1$. Thus,
$$P(X \leq Y + 5) = \frac{1}{60} \left[\int_{0}^{55} \frac{k + 5}{60} \mathop{dk} + \int_{55}^{60} 1 \mathop{dk}\right]= \frac{1}{60}\left[\frac{715}{24} + 5\right] = \frac{167}{288}$$
This must be wrong because there's no clear way to go from this to $23/144$. Can someone please point out my mistake?

REPLY [3 votes]: Your region of integration is wrong.
$P(|X-Y|\leq 5)=P((X-Y\leq 5)\cap(X>Y))+P(Y-X\leq 5\cap(X<Y))=\\2\cdot P((X-Y\leq 5)\cap(X>Y))$.

This is the region you have to integrate over to get the probability. I hope now you can modify the limits of the integral on your own . If you directly go by area of triangle then you will get the answer as required. That is $\frac{60\cdot 60-55\cdot 55}{60\cdot 60}$