Two fair dice, each with faces numbered 1 , 2 , 3 , 4 , 5 and 6 , are rolled together and the sum of the…

Perfect Square can be $4$ or $9$, which can be obtained by $\left(1,3\right),\left(3,1\right),\left(2,2\right),\left(6,3\right)\left(3,6\right),\left(5,4\right),\left(4,5\right)$.

Prime Number can be $2$ or $3$ or $5$ or $7$ or $11$, which can be obtained by 

$\left(1,1\right),\left(1,2\right),\left(2,1\right),\left(1,4\right),\left(4,1\right),\left(2,3\right),\left(3,2\right),\left(1,6\right),\left(6,1\right),\left(5,2\right),\left(2,5\right),\left(3,4\right),\left(4,3\right),\left(6,5\right),\left(5,6\right)$

$P$(square or prime)$=\frac{22}{36}$

$\mathrm{P}\left(\frac{\text{ Perfect square is odd }}{\text{ perfect square before prime }}\right)=\frac{{\displaystyle \frac{4}{36}}+{\displaystyle \frac{14}{36}}.{\displaystyle \frac{4}{36}}+{\displaystyle \frac{14}{36}}.{\displaystyle \frac{14}{36}}.{\displaystyle \frac{4}{36}}+....}{{\displaystyle \frac{7}{36}}+{\displaystyle \frac{14}{36}}.{\displaystyle \frac{7}{36}}+{\displaystyle \frac{14}{36}}.{\displaystyle \frac{14}{36}}.{\displaystyle \frac{7}{36}}+......}$

$=\frac{\left({\displaystyle \frac{{\displaystyle \frac{4}{36}}}{1-{\displaystyle \frac{14}{36}}}}\right)}{\left(\frac{{\displaystyle \frac{7}{36}}}{1-{\displaystyle \frac{14}{36}}}\right)}=\frac{4}{7}=P$

$\Rightarrow 14 P=8$