In a game two players A and B take turns in throwing a pair of fair dice starting with player A and total of scores on the two dice, in each throw is noted. A wins the game if he throws a total of 6 before B throws a total of 7 and $\mathrm{B}$ wins the game if he throws a total of 7 before A throws a total of six The game stops as soon as either of the players wins. The probability of A winning the game is:
Correct Option: , 4
$P(6)=\frac{1}{6}, P(7)=\frac{5}{36}$
$P(A)=W+F F W+F F F F W+\ldots . .$
$=\frac{1}{6}+\frac{5}{6} \times \frac{31}{36} \times \frac{1}{6}+\left(\frac{5}{6}\right)^{2}\left(\frac{31}{36}\right)^{2} \frac{1}{6}+\ldots$
$=\frac{\frac{1}{6}}{1-\frac{155}{216}}=\frac{36}{61}$
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