92. The Meeting Problem. Two people agreed to meet at a certain place between 12 o'clock and 1 o'clock. According to the agreement, the one who arrives first waits for the other for 15 minutes, after which he leaves. What is the probability that these people will meet, if each of them chooses the moment of their arrival at the agreed place at random within the interval between 12 o'clock and 1 o'clock?
92. All possible outcomes of the experiment considered in this problem consist in the first person arriving at some moment of time $x$ between 12 o'clock and 1 o'clock and the second person arriving at another moment of time $y$ within the same interval. Thus, these outcomes are determined by a pair of numbers $x, y$, where $0 \leqslant x \leqslant 1$ and $0 \leqslant y \leqslant 1$. Considering these numbers as coordinates of a point on the plane, we can represent the set of all outcomes as the set of points of the square $O A B C$ with side 1 (Fig. 79). Since $x$ and $y$ are chosen randomly between 12 o'clock and 1 o'clock, the probability that $x$ (respectively $y$) falls within any interval on the x-axis (respectively y-axis) is equal to the length of this interval (recall that in our case $O A=O C=1$). Therefore, the probability that the point $(x, y)$ falls within any rectangle inside $O A B C$ is equal to the area of this rectangle. Since any region can be covered with any precision by a grid of small rectangles, the probability that the point $(x, y)$ falls within any region is equal to the area of this region (this property can be taken as a definition of the term "randomly" mentioned in the problem's condition). The favorable outcomes in our problem will obviously correspond to those points $(x, y)$ for which $|x-y| \leqslant \frac{1}{4}$. All such points are located in the shaded area on Fig. 79, bounded by the lines $M N$ and $P Q$, for which $x-y=\frac{1}{4}$ and respectively $y-x=\frac{1}{4}$. Here $O M=B N=O P=B Q=\frac{1}{4} ; \quad$ therefore, the area of $\quad M C N=$ $=$ the area of $P A Q=\frac{1}{2} \cdot\left(\frac{3}{4}\right)^{2}=\frac{9}{32}$, and the desired probability is equal to the area of $P Q B N M O=1$ - the area of $M C N$ - the area of $P A Q=1-\frac{18}{32}=\frac{7}{16}$.