A and B alternately throw a pair of dice. A wins if he throws a sum of 5 before B throws a sum of 8, and B wins if he throws a sum of 8 before A throws a sum of 5. The probability, that A wins if A makes the first throw, is

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$\frac{8}{19}$

$\frac{8}{17}$

$\frac{9}{19}$

$\frac{9}{17}$

answer is B.

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Detailed Solution

$\begin{array}{l} p (S_{5}) = \frac{1}{9} \\ p (S_{5}) = \frac{5}{36} \end{array}$
required prob $= \frac{1}{9} + \frac{8}{9} \cdot \frac{31}{36} \cdot \frac{1}{9} + {(\frac{8}{9} \cdot \frac{31}{36})}^{2} \cdot \frac{1}{9} + \dots \infty$
$= \frac{\frac{1}{9}}{1 - \frac{62}{81}} = \frac{9}{19}$

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