Problem 9 (Meeting Problem). Two students agreed to meet at a certain place between 12:00 and 13:00. The one who arrives first waits for the other for $\alpha$ minutes ( $\alpha<60$ ), after which he leaves. What is the probability of their meeting (event $A$), if each of them can arrive at any time during the specified hour, and the times of their arrivals are independent? Consider the special case $\alpha=10$ minutes.
Solution. Let the arrival time (counting from 12 o'clock) of one of the students be denoted by $x$, and the other student's arrival time by $y$. For the meeting to occur, it is necessary and sufficient that the inequality $|x-y| \leqslant \alpha$ holds, which is equivalent to the following: $-\alpha \leqslant x-y \leqslant \alpha$, or the system:

$$
\left\{\begin{array}{l}
x-y-\alpha \leqslant 0 \\
x-y+\alpha \geqslant 0
\end{array}\right.
$$

Let's represent $x$ and $y$ as Cartesian coordinates on a plane, choosing 1 minute as the unit of scale (Fig. 6.2). All possible outcomes will be represented by points within a square with a side length of 60, and the outcomes favorable for the meeting will be represented by points in the shaded area, satisfying the conditions.

![](https://cdn.mathpix.com/cropped/2024_05_22_446938f1e4808e7d6e07g-133.jpg?height=389&width=421&top_left_y=524&top_left_x=2056)

Fig. 6.2 According to the formula in point $5^{\circ}$, we get:

$$
p(A)=\frac{60^{2}-(60-\alpha)^{2}}{60^{2}}=1-\left(1-\frac{\alpha}{60}\right)^{2}
$$

In particular, if $\alpha=10$ minutes, then $p(A)=\frac{11}{36}$.