Answer: Option B
:
B
The probability of throwing 9 with two dice = $\frac{4}{36}=\frac{1}{9}$
$\therefore$ The probability of not throwing 9 with two dice = $\frac{8}{9}$
If A is to win he should throw 9 in $1^{st}$ or $3^{rd}$ or $5^{th}$ attempt
If B is to win, he should throw, 9 in $2^{nd}$, $4^{th}$ attempt
B’s chances = $\left(\frac{8}{9}\right).\frac{1}{9}+{\left(\frac{8}{9}\right)}^3.\frac{1}{9}+.....=\frac{\frac{8}{9}\times \frac{1}{9}}{1-{\left(\frac{8}{9}\right)}^2}=\frac{8}{17}$