Example 3: A and B agree to meet at a certain place within 10 days. The person who arrives first will wait for the other, but can leave after waiting for 3 days. If they are equally likely to arrive at the destination within the limit, find the probability that the two will meet.

 

If they arrive at the destination within the limit, they are equally likely to do so. Let's denote the time when A and B arrive as \(x\) and \(y\) respectively, where \(0 \leq x, y \leq 10\). They will meet if the difference in their arrival times is less than or equal to 3 days, i.e., \(|x - y| \leq 3\).

 

To find the probability, we can visualize the problem on a coordinate plane where the x-axis represents A's arrival time and the y-axis represents B's arrival time. The region where they meet is the area between the lines \(y = x + 3\) and \(y = x - 3\) within the square defined by \(0 \leq x, y \leq 10\).

 

The total area of the square is \(10 \times 10 = 100\). The area where they do not meet consists of two triangles, each with a base and height of 7. The area of each triangle is \(\frac{1}{2} \times 7 \times 7 = 24.5\). Therefore, the total area where they do not meet is \(2 \times 24.5 = 49\).

 

The area where they meet is \(100 - 49 = 51\). Thus, the probability that A and B will meet is:

\[
\frac{51}{100} = 0.51
\]

So, the probability that A and B will meet is \(0.51\).
Let's assume that person A and person B arrive at the destination on the $x$-th and $y$-th day, respectively. $0 \leqslant x \leqslant 10, 0 \leqslant y \leqslant 10$. The necessary and sufficient condition for them to meet is $|x-y| \leqslant 3$.

The point $(x, y)$ is distributed in the square shown in Figure 18-2. The event of the two people meeting is within the shaded region between the lines $x-y= \pm 3$. Therefore, the required probability is
$$
P=\frac{100-(10-3)^{2}}{100}=\frac{51}{100}
$$