The probability that $a$ and $b$ satisfy the inequality $a-2b+10>0$? Nine identical balls are numbered $1,2,3,.........,9$ are put in a bag.$A$ draws a ball and gets the number $a$ and puts back in the bag. Next $B$ draws a ball and gets the number $b$.
The probability that $a$ and $b$ satisfy the inequality $a-2b+10>0$ ?

My Try :-
Total pairs of $(a,b)$ possible are $81$ .


*

*If $a=1$, then $b = 1,2,3,4,5$. Similarly for $a=2$.

*If $a=3$, then $b = 1,2,3,4,5,6$. Similarly for $a=4$.

*If $a=5$, then $b = 1,2,3,4,5,6,7$. Similarly for $a=6$.

*If $a=7$, then $b = 1,2,3,4,5,6,7,8$. Similarly for $a=8$.

*If $a=9$, then $b = 1,2,3,4,5,6,7,8,9$. 


Total favourable pairs are $61$. 
Hence, Total Probability = $\frac{61}{81}$

However, I don't have an answer for  this. Am I right or missing something ?

Your solution is correct, but tedious.
A quicker way would be to note that the expression can only be non-positive if $b>5$, which gives only four cases for $b$. In these four cases of $b$, there are $2(b-5)$ values for $a$ that give non-positive results. Therefore the total probability for the expression to be positive is
$$p = 1-\frac{2+4+6+8}{81} = \frac{81 - 20}{81} = \frac{61}{81}$$
