Two persons are equally likely to arrive at a certain place…

Two persons are equally likely to arrive at a certain place at any moment of a time interval of length $T$. Determine the probability that one person will wait for the other no longer than $t$.

Step 1: The model

Arrival times $x$, $y$ are independent and uniform on $[0,T]$: $(x,y)$ is uniform in the square $[0,T]^2$ (area $T^2$).

Step 2: The event

Whoever comes first waits $|x-y|$. We need $|x-y|\le t$ — the band between the lines $y=x-t$ and $y=x+t$.

Step 3: The area

Outside the band there are two triangles with legs $T-t$, total area $(T-t)^2$:

\[ \boxed{P(|x-y|\le t) = \frac{T^2-(T-t)^2}{T^2} = 1-\left(1-\frac tT\right)^2 = \frac{t(2T-t)}{T^2}.} \]

Step 4: Example

Two friends agree to meet between 12:00 and 13:00 ($T=60$ min), each waiting $t=15$ min for the other:

\[ P = 1-\left(\tfrac{45}{60}\right)^2 = 1-\tfrac9{16} = \tfrac7{16}\approx0.44. \]

To meet with probability at least $\frac12$ they must be willing to wait $t\ge T(1-\frac1{\sqrt2})\approx0.29\,T$, i.e. about $18$ minutes.