<s>[INST] 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 ?
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\begin{align}
&\sum_{a = 1}^{9}{1 \over 9}\sum_{b = 1}^{9}{1 \over 9}\bracks{a - 2b + 10 > 0} =
{1 \over 81}\sum_{b = 1}^{9}\sum_{a = 1}^{9}\bracks{a > 2b - 10}
\\[5mm] = &\
{1 \over 81}\sum_{b = 1}^{9}\braces{%
\bracks{2b - 10 < 1}\sum_{a = 1}^{9}1 +
\bracks{1 \leq 2b - 10 \leq 9}\sum_{a = 2b - 9}^{9}1}
\\[5mm] = &\
{1 \over 81}\sum_{b = 1}^{9}\braces{%
\bracks{b < {11 \over 2}}9 +
\bracks{{11 \over 2} \leq b \leq {19 \over 2}}\pars{19 - 2b}} =
{1 \over 9}\sum_{b = 1}^{5}1 + {1 \over 81}\sum_{b = 6}^{9}\pars{19 - 2b}
\\[5mm] = &\
{5 \over 9} + {1 \over 81}\pars{4 \times 19 - 2 \times 30} =
\bbx{\ds{61 \over 81}}
\end{align}
</s>