Explanation

(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$\ = 1 - \frac{1}{9} = \frac{8}{9}$
If $A$ is to win, he should throw 9, in $Ist\ or\ 3rd$ or $5th$ and so on.
If $B$ is to win, he should throw, 9 in $2nd,\ 4th$ and so on.
$B$ can be get the chance only when $A$ should not get it.
$\therefore B$’s chances $= \left( \frac{8}{9} \right) \bullet \frac{1}{9} + \left( \frac{8}{9} \right)^{3} \bullet \frac{1}{9} + ...$
$= \frac{\frac{8}{9} \times \frac{1}{9}}{1 - \left( \frac{8}{9} \right)^{2}} = \frac{8}{17}$