Nine balls of the same size and colour, numbered 1, 2,…..9, were put  into a packet. Now A draws a ball from packet, noted that it is of number a, and puts it back. Then B also draws a ball from the pocket and noted that it is of number b. Then probability for the inequality $a-2b+10>0$ to hold is

Since each has equally 9 different possible results for A and B to draw a ball from the packet independently, the total number of possible events is $9^2=81.$  From $a-2b+10>0$  we get  $2b<a+10.$ We find that when $b=1,2,3,4,5$  a can take any value in $1,2,3,\mathrm{.....},9$  to make the inequality hold. Then we have  $9\times 5=45$ admissible events
When b = 6, a can 3, 4, ….., 9 and there are 7 admissible events
When b = 7, a can 5, 6, 7, 8, 9 and there are 5 admissible events
When b = 8, a can 7, 8, 9 and there are 3 admissible events
When b = 9, a can 9 and there are 1 admissible events
So, the required probability is  $\frac{45+7+5+3+1}{81}=\frac{61}{81}$