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Two persons $A$ and $B$ take turns in throwing a pair of dice. The first person to through $9$ from both dice will be avoided the prize. If $A$ throws first then the probability that $B$ wins the game is
$\frac{9}{{17}}$
$\frac{8}{{17}}$
$\frac{8}{9}$
$\frac{1}{9}$
Solution
(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
$\therefore $ $B$'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}}$.
