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(B) Let $p$ be the probability of getting a head,so $p = \frac{1}{2}$.Let $q$ be the probability of getting a tail,so $q = 1 - p = \frac{1}{2}$.$A$ st

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$\frac{2}{3}$

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(B) Let $p$ be the probability of getting a head,so $p = \frac{1}{2}$.Let $q$ be the probability of getting a tail,so $q = 1 - p = \frac{1}{2}$.$A$ starts the game. $A$ wins if $A$ gets a head in the $1^{st}$ throw,or $A$ gets a tail,$B$ gets a tail,and $A$ gets a head in the $3^{rd}$ throw,and so on.The probability of $A$ winning is given by the sum of an infinite geometric series:$P(A) = p + qqp + qqqqp + \dots$$P(A) = p(1 + q^2 + q^4 + \dots)$This is an infinite geometric series with first term $a = p$ and common ratio $r = q^2$.The sum is $S = \frac{a}{1 - r} = \frac{p}{1 - q^2}$.Substituting the values $p = \frac{1}{2}$ and $q = \frac{1}{2}$:$P(A) = \frac{1/2}{1 - (1/2)^2} = \frac{1/2}{1 - 1/4} = \frac{1/2}{3/4} = \frac{1}{2} 	imes \frac{4}{3} = \frac{2}{3}$.

$A$ and $B$ throw a coin alternately till $1$ of them gets a head and wins the game. If $A$ starts the game,find the probability of $A$ winning.

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