Meeting probability of two bankers: uniform distribution puzzle

Question

Two bankers each arrive at the station at some random time between 5PM and 6PM (arrival time for each of them is uniformly distributed). They stay exactly five minutes and then leave. What is the probability that they will meet on a given day?

I am not sure how to go about modelling this problem as uniform distribution and solving it. Appreciate any help.

Here is how I start with it: Assume banker A arrives X minutes after 5PM and B arrives Y minutes after 5PM. Both X and Y are uniformly distributed between 5PM and 6PM. So pdf of X, Y is $\frac{1}{60}$. Now A and B will meet if $|X - Y| < 5$.

So required probability is $P(|X - Y| < 5)$ = Integral of joint distribution function of $|X - Y|$ from $0$ to $5$?

Now not sure how to write the equation from this point onwards and solve it.

Answer: $\frac {23}{144}$

Answer 1

Hint: If $f_{X,Y}$ is the multivariate pdf then you want to solve the following integral $$ \int_{0}^{60}\int_{\max\{0,x-5\}}^{\min\{60,x+5\}} f_{X,Y}(x,y)dydx. $$

Answer 2

Hint:

We will consider the situation in terms of minutes. We notice that that they will meet only if $|X-Y|\leq 5/60$, where $X,Y$ is the time of arrival as a fraction of the hour. We notice that $X,Y\overset{iid}{\sim} \text{unif(0,1)}$. Then they problem becomes $$P(|X-Y| \leq 5/60).$$ It will help to draw a picture. No integration required.

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You can do it considering the $X,Y\overset{iid}\sim\text{unif}(1,60)$ too.

Notice that we want the blue part. Also, notice that we can instead consider the complement \begin{align*} P(|X-Y|<5) &= P(-5<X-Y<5)\\ &=P(\text{blue})\\ &= 1-P(\text{Not blue})\\ &= 1-2\bigg[\underbrace{(1/2)\cdot55\cdot55\cdot 60}_{(1)}/\underbrace{60^3}_{(2)}\bigg]\\ &= 1-\left(\frac{55}{60}\right)^2 \\ &= \frac{23}{144} \end{align*}

where in

1. That is the volume of one wedge
2. That is the entire volume.