9. Two people, 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 3 days. If they are equally likely to arrive at the destination within the limit, then the probability that the two will meet is $\qquad$ .
9. $\frac{51}{100}$.

Let A and B arrive at a place 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 within the square $O A B C$ as shown in Figure 3, and the basic event $S_{1}$ is within the shaded area between the two lines $x-y = \pm 3$.
Thus, the required probability is $P=\frac{100-(10-3)^{2}}{100}=\frac{51}{100}$.