3. Nine small balls numbered $1,2, \cdots, 9$ are placed in a bag. These balls are identical except for their numbers. Person A draws a ball from the bag, with the number being $a$, and then returns it. Person B then draws a ball from the bag, with the number being $b$. The probability of the event that the inequality $a-2 b+10>0$ holds is ( ).
(A) $\frac{52}{81}$
(B) $\frac{59}{81}$
(C) $\frac{60}{81}$
(D) $\frac{61}{81}$
3. D.

Person A and Person B each draw a small ball, with 9 different outcomes for each, so the total number of basic events is $9^{2}=81$.
From the inequality $a-2 b+10>0$, we get $2 b<a+10$.
Thus, when $b=1,2,3,4,5$, in each case $a$ can take every value from 1, $2, \cdots, 9$, making the inequality true, which results in a total of $9 \times 5=$ 45 cases; when $b=6$, $a$ can take every value from $3,4, \cdots, 9$, giving 7 cases; when $b=7$, $a$ can take every value from $5,6, \cdots, 9$, giving 5 cases; when $b=8$, $a$ can take every value from $7,8,9$, giving 3 cases; when $b=9$, $a$ can only take 9, giving 1 case. Therefore, the probability of the desired event is
$$
\frac{45+7+5+3+1}{81}=\frac{61}{81} \text {. }
$$