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(A) Let $p$ be the probability of getting a '$6$' in a single throw,so $p = \frac{1}{6}$.Let $q$ be the probability of not getting a '$6$',so $q = 1 -

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$\frac{5}{11}$

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(A) Let $p$ be the probability of getting a '$6$' in a single throw,so $p = \frac{1}{6}$.Let $q$ be the probability of not getting a '$6$',so $q = 1 - p = 1 - \frac{1}{6} = \frac{5}{6}$.$A$ throws first. $B$ wins if $A$ fails,then $B$ succeeds,or $A$ fails,$B$ fails,$A$ fails,$B$ succeeds,and so on.$P(B 	ext{ winning}) = qp + qqqp + qqqqqp + \dots$This is an infinite geometric series with first term $a = qp = \frac{5}{6} 	imes \frac{1}{6} = \frac{5}{36}$ and common ratio $r = q^2 = (\frac{5}{6})^2 = \frac{25}{36}$.The sum of an infinite geometric series is $S = \frac{a}{1 - r}$.$P(B 	ext{ winning}) = \frac{5/36}{1 - 25/36} = \frac{5/36}{11/36} = \frac{5}{11}$.

$2$ persons $A$ and $B$ throw a die alternately until $1$ of them gets a '$6$' and wins the game. Find the probability of $B$ winning.

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