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 :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 9 : 8
Total outcomes = 36
P(getting 9 as sum) = $\frac{4}{36}=\frac{1}{9}$
P(A wins) = P(A gets 9 in first throw) + P(A gets 9 in third throw)
$=\frac{1}{9}+(1-\frac{1}{9})(1-\frac{1}{9})(\frac{1}{9})+......$
$=\frac{1}{9}(1+\frac{64}{81}+\frac{64^2}{81^2}+\frac{64^3}{81^3})$
$=\frac{1}{9}(\frac{1}{1-\frac{64}{81}})=\frac{9}{17}$
P(B wins) = P(B gets 9 in 2nd throw) + P(B gets 9 in 4th throw)
$=(1-\frac{1}{9})\frac{1}{9}+(1-\frac{1}{9})(1-\frac{1}{9})(1-\frac{1}{9})(\frac{1}{9})....$
$=\frac{8}{81}(1+\frac{64}{81}+\frac{64^2}{81^3}.....)$
$=\frac{8}{17}$
so P(A win) : P(B win) = 9 : 8