21. The Meeting Problem. Two comrades agreed to meet at a certain place between 12 o'clock and half past one in the afternoon. The one who arrives first waits for the other for 20 minutes, after which he leaves. Find the probability that the comrades will meet, if each of them randomly chooses the moment of their arrival (within the interval from 12 o'clock to half past one) and the moments of arrival of both are independent.
Solution. Let event $A$ be the meeting of friends.

We will find the probability of event $A$ by applying formula (7).

Let the arrival time of one of them be denoted by $x$, in minutes, and the arrival time of the other by $y$, in minutes. For the meeting to occur, it is necessary and sufficient that the condition $|x-y| \leq 20$ is satisfied.

We will represent $x$ and $y$ as Cartesian coordinates of points on a plane; we will use a minute as the unit of scale (Figure 9).

![](https://cdn.mathpix.com/cropped/2024_05_22_c311c9dd990dc55c716eg-027.jpg?height=458&width=484&top_left_y=332&top_left_x=522)

Figure 9

All possible outcomes of the trial are represented by points within the figure bounded by a square with a side length of 30; the area of this square is $S_{G}=30^{2}$.

The inequality $|x-y| \leq 20$ is equivalent to the system of inequalities:

$$
\left\{\begin{array}{l}
x-y \leq 20 \\
x-y \geq-20
\end{array}\right.
$$

Outcomes of the trial that favor event $A$ satisfy the system of inequalities:

$$
\left\{\begin{array}{l}
x-y \leq 20 \\
x-y \geq-20 \\
0 \leq x \leq 30 \\
0 \leq y \leq 30
\end{array}\right.
$$

The solutions to this system of inequalities are the coordinates of all points on the plane located in the shaded area in Figure 9, that is, between the boundary lines: $x-y=20 ; x-y=-20 ; x=0 ; x=30$; $y=0 ; y=30$ and on the boundary lines themselves. Points on the plane that belong to the shaded area characterize outcomes of the trial that favor event $A$. The area of the shaded figure is $S_{g}=30^{2}-(30-20)^{2}$

The desired probability of event $\boldsymbol{A}$ is the ratio of the area of the shaded figure to the area of the entire square:

$$
P(A)=\frac{30^{2}-10^{2}}{30^{2}}=\frac{8}{9}
$$