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(A) The total number of outcomes when throwing two dice is $36$. The outcomes for a sum of $4$ are $(1,3), (3,1), (2,2)$.Therefore,the probability of

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$\frac{11}{23}$

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(A) The total number of outcomes when throwing two dice is $36$. The outcomes for a sum of $4$ are $(1,3), (3,1), (2,2)$.Therefore,the probability of getting a sum of $4$ in a single throw is $p = \frac{3}{36} = \frac{1}{12}$.The probability of not getting a sum of $4$ is $q = 1 - p = 1 - \frac{1}{12} = \frac{11}{12}$.Since $A$ starts the game,$B$ wins if $A$ fails on the $1^{st}$ turn,$B$ succeeds on the $2^{nd}$ turn,or $A$ fails on the $1^{st}, 3^{rd}$ turns and $B$ fails on the $2^{nd}$ turn and succeeds on the $4^{th}$ turn,and so on.The probability that $B$ wins is given by the infinite geometric series:$P(B 	ext{ wins}) = qp + q^3p + q^5p + \dots$This is a geometric series with first term $a = qp = \frac{11}{12} 	imes \frac{1}{12} = \frac{11}{144}$ and common ratio $r = q^2 = (\frac{11}{12})^2 = \frac{121}{144}$.The sum of an infinite geometric series is $S = \frac{a}{1-r}$.$P(B 	ext{ wins}) = \frac{\frac{11}{144}}{1 - \frac{121}{144}} = \frac{\frac{11}{144}}{\frac{144-121}{144}} = \frac{11}{23}$.

Two persons $A$ and $B$ throw a pair of dice alternately until one of them gets the sum of the numbers appeared on the dice as $4$,and the person who gets this result first is declared as the winner. If $A$ starts the game,then the probability that $B$ wins the game is:

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