A and B throw a pair of dice alternately. A wins the game if he gets a total of 7 and B wins the game if he gets a total of 10. If A starts the game, then what is the probability of B wins the game.

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$\frac{4}{17}$

$\frac{7}{17}$

$\frac{6}{17}$

$\frac{5}{17}$

answer is C.

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Detailed Solution

$P_{A} = P (g e t t i n g a t o t a l s u m o f 7 o n a p a i r o f d i c e) = \frac{6}{36} = \frac{1}{6}$

$q_{A} = 1 - P_{A} = \frac{5}{6}$

$P_{B} = P (g e t t i n g a t o t a l s u m o f 10 o n a p a i r o f d i c e) = \frac{3}{36} = \frac{1}{12}$

$q_{B} = 1 - P_{B} = \frac{11}{12}$

Given that, A starts the game

$P (B) = P (B w i n s t h e g a m e)$

$\begin{array}{l} = q_{A} P_{B} + q_{A} q_{B} q_{A} P_{B} + ............ \\ = \frac{5}{6} (\frac{1}{12}) + \frac{5}{6} (\frac{11}{12}) \frac{5}{6} (\frac{1}{12}) + .......... \\ = \frac{5}{72} [1 + \frac{55}{72} + {(\frac{55}{72})}^{2} + ..........] \end{array}$