Problem 9.5 Easy Free 1 variation

Meeting Problem

Alice and Bob each arrive at a restaurant at an independent, uniformly random time between 8:00pm and 9:00pm, and each stays for exactly $10$ minutes before leaving. Find the probability that their visits overlap.

Work it out, then compare with the answer.

SolutionShow the worked solution

Measure times in minutes after 8:00pm. Let Alice arrive at $X$ and Bob at $Y$ , with $X,Y$ i.i.d. $\mathrm{Unif}(0,60)$ . Then $(X,Y)$ is uniform on the square $[0,60]^2$ , which has area $3600$ .

Alice is present during $[X,X+10]$ and Bob during $[Y,Y+10]$ . Two intervals of equal length overlap exactly when their left endpoints are within that length of each other. So the visits overlap if and only if

|X-Y|\le 10 .

This is a diagonal band through the square.

The complement consists of two corner triangles, $\{Y>X+10\}$ and $\{X>Y+10\}$ . Each has legs of length $50$ , so each has area $\tfrac12\cdot 50^2=1250$ . The band therefore has area

3600-2\cdot 1250=60^2-50^2=1100 .

Since the point is uniform on the square, the probability is

\frac{1100}{3600}=\frac{11}{36}\approx 0.306 .

\boxed{\dfrac{11}{36}}

Generalisation. If the window has length $T$ and each person stays for $t\le T$ , the probability is

1-\left(1-\tfrac{t}{T}\right)^2 .

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 9.5a

Alice arrives at a uniformly random time between 4:00pm and 5:00pm, while Bob arrives independently at a uniformly random time between 4:30pm and 5:00pm. Whoever arrives first will wait up to $10$ minutes for the other before leaving. Find the probability that they meet.

Work it out, then compare with the answer.

Solution

Measure minutes after 4pm, with $X\sim\mathrm{Unif}(0,60)$ for Alice and $Y\sim\mathrm{Unif}(30,60)$ for Bob. The point $(X,Y)$ is uniform on the rectangle $[0,60]\times[30,60]$ , which has area $1800$ . Whoever arrives first waits up to $10$ minutes, so they meet if and only if $|X-Y|\le 10$ .

The complement has two pieces.

$\{X<Y-10\}$ : for each $y\in[30,60]$ the cross-section has length $y-10$ , running from $20$ to $50$ . This is a trapezoid of area $30\cdot\frac{20+50}{2}=1050$ .
$\{X>Y+10\}$ : for each $y\in[30,50]$ the cross-section has length $50-y$ . This is a triangle of area $\tfrac12\cdot 20^2=200$ .

The meeting region therefore has area $1800-1050-200=550$ , and the probability is

\frac{550}{1800}=\frac{11}{36}.

It is a coincidence that this equals the answer to the main problem.

\boxed{\dfrac{11}{36}}

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