Two persons A and B take turns in throwing a pair of dice. The first person to through 9 from both dice will be awarded the prize. If A throws first then the probability that B wins the game is

\frac{9}{17}

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\frac{8}{17}

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\frac{8}{9}

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\frac{1}{9}

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Solution

The correct option is B $\frac{8}{17}$
The probability of throwing 9 with two dice = $\frac{4}{36} = \frac{1}{9}$
$∴$ The probability of not throwing 9 with two dice = $\frac{8}{9}$
If A is to win he should throw 9 in $1^{s t}$ or $3^{r d}$ or $5^{t h}$ attempt
If B is to win, he should throw, 9 in $2^{n d}$ , $4^{t h}$ attempt
B’s chances = $(\frac{8}{9}) . \frac{1}{9} + {(\frac{8}{9})}^{3} . \frac{1}{9} + . . . . . = \frac{\frac{8}{9} \times \frac{1}{9}}{1 - {(\frac{8}{9})}^{2}} = \frac{8}{17}$

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