2. Place nine small balls numbered $1, 2, \cdots, 9$ into 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 that the inequality $a - 2b + 10 > 0$ holds is $\qquad$.
2. $\frac{61}{81}$ Detailed Explanation: Each of person A and B can draw a ball with 9 different outcomes, 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 $5,6,7,8,9$, 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 $\frac{45+7+5+3+1}{81}=\frac{61}{81}$.