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(A) Let $S$ denote the success (getting a '$6$') and $F$ denote the failure (not getting a '$6$').$P(S) = \frac{1}{6}$,$P(F) = \frac{5}{6}$.$A$ wins i

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$P(A) = \frac{6}{11}, P(B) = \frac{5}{11}$

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(A) Let $S$ denote the success (getting a '$6$') and $F$ denote the failure (not getting a '$6$').$P(S) = \frac{1}{6}$,$P(F) = \frac{5}{6}$.$A$ wins if $A$ gets a '$6$' in the $1st, 3rd, 5th, \dots$ throw.$P(A 	ext{ wins}) = P(S) + P(FFS) + P(FFFFS) + \dots$$= \frac{1}{6} + (\frac{5}{6})^2 \cdot \frac{1}{6} + (\frac{5}{6})^4 \cdot \frac{1}{6} + \dots$This is an infinite geometric series with first term $a = \frac{1}{6}$ and common ratio $r = (\frac{5}{6})^2 = \frac{25}{36}$.$P(A 	ext{ wins}) = \frac{a}{1-r} = \frac{1/6}{1 - 25/36} = \frac{1/6}{11/36} = \frac{1}{6} \cdot \frac{36}{11} = \frac{6}{11}$.Since $A$ and $B$ are the only players,$P(B 	ext{ wins}) = 1 - P(A 	ext{ wins}) = 1 - \frac{6}{11} = \frac{5}{11}$.

$A$ and $B$ throw a die alternatively till one of them gets a '$6$' and wins the game. Find their respective probabilities of winning,if $A$ starts first.

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