A and B take turns in throwing, a pair of dice, the first to throw a sum of 9 wins the prize. If A throws first, then the ratio of the probabilities of A and B winning is:

Correct option is C

​$$\text{We have} \\\text{Probability that A wins the game} = \frac{4}{36} = \frac{1}{9} \\\text{Probability that A doesn't win the game} = 1 - \frac{1}{9} = \frac{8}{9} \\\text{Probability of A losing \& B winning the game} = \frac{8}{9} \times \frac{1}{9} = \frac{8}{81} \\\text{Then, the ratio that A \& B wins the game} = \frac{\frac{1}{9}}{\frac{8}{81}} = \frac{9}{8}$$​​