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 ?

You seek the probability that $~2b < 10+a$ when selecting the values with replacement and no bias.
Let us see: You have listed values of $b$ that satisfy this for every $a$, counted them, and compared as a ratio of the size of the sample space.
Yes, that is okay; you have arrived at the correct answer by a valid process and made no errors.
