Two persons A and B take turns in throwing a | vedclass

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(B) Let $p$ be the probability of throwing a sum of $9$ with two dice. The total outcomes are $36$. The favorable outcomes for a sum of $9$ are $(3,6)

Vedclass · Accepted Answer

$\frac{8}{17}$

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(B) Let $p$ be the probability of throwing a sum of $9$ with two dice. The total outcomes are $36$. The favorable outcomes for a sum of $9$ are $(3,6), (4,5), (5,4), (6,3)$,so $p = \frac{4}{36} = \frac{1}{9}$.The probability of not throwing a sum of $9$ is $q = 1 - p = 1 - \frac{1}{9} = \frac{8}{9}$.$A$ throws first. $B$ wins if $A$ fails,then $B$ succeeds,or $A$ fails,$B$ fails,$A$ fails,$B$ succeeds,and so on.The probability that $B$ wins is $P(B) = qp + q^3p + q^5p + \dots$This is a geometric series with first term $a = qp = \frac{8}{9} 	imes \frac{1}{9} = \frac{8}{81}$ and common ratio $r = q^2 = (\frac{8}{9})^2 = \frac{64}{81}$.The sum of the infinite geometric series is $S = \frac{a}{1-r} = \frac{8/81}{1 - 64/81} = \frac{8/81}{17/81} = \frac{8}{17}$.

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

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