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(B) Let $P(S_5)$ be the probability of getting a sum of $5$ with two dice: $P(S_5) = \frac{4}{36} = \frac{1}{9}$.Let $P(S_8)$ be the probability of ge

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$\frac{9}{19}$

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(B) Let $P(S_5)$ be the probability of getting a sum of $5$ with two dice: $P(S_5) = \frac{4}{36} = \frac{1}{9}$.Let $P(S_8)$ be the probability of getting a sum of $8$ with two dice: $P(S_8) = \frac{5}{36}$.Let $q_5 = 1 - \frac{1}{9} = \frac{8}{9}$ be the probability of not getting a sum of $5$.Let $q_8 = 1 - \frac{5}{36} = \frac{31}{36}$ be the probability of not getting a sum of $8$.$A$ wins if he gets $5$ on the $1^{st}$ throw,or if $A$ fails,$B$ fails,and $A$ gets $5$ on the $3^{rd}$ throw,and so on.The probability $P(A)$ is given by the infinite geometric series:$P(A) = P(S_5) + q_5 \cdot q_8 \cdot P(S_5) + (q_5 \cdot q_8)^2 \cdot P(S_5) + \dots$$P(A) = \frac{P(S_5)}{1 - q_5 \cdot q_8} = \frac{1/9}{1 - (8/9 \cdot 31/36)} = \frac{1/9}{1 - 248/324} = \frac{1/9}{76/324} = \frac{1}{9} \cdot \frac{324}{76} = \frac{36}{76} = \frac{9}{19}$.

$A$ 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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