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(D) The probability of getting a head is $P(H) = \frac{1}{2}$ and the probability of not getting a head is $P(T) = \frac{1}{2}$.$A$ wins if he gets a

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(D) The probability of getting a head is $P(H) = \frac{1}{2}$ and the probability of not getting a head is $P(T) = \frac{1}{2}$.$A$ wins if he gets a head on his $1^{st}$ turn,or if $A$ and $B$ both fail and $A$ gets a head on his $2^{nd}$ turn,and so on.This is a geometric series: $P(A 	ext{ wins}) = P(H) + P(T)P(T)P(H) + P(T)P(T)P(T)P(T)P(H) + \dots$$P(A 	ext{ wins}) = \frac{1}{2} + (\frac{1}{2})^2 \cdot \frac{1}{2} + (\frac{1}{2})^4 \cdot \frac{1}{2} + \dots$$P(A 	ext{ wins}) = \frac{1}{2} + (\frac{1}{2})^3 + (\frac{1}{2})^5 + \dots$This is an infinite geometric series with first term $a = \frac{1}{2}$ and common ratio $r = (\frac{1}{2})^2 = \frac{1}{4}$.The sum is $S = \frac{a}{1-r} = \frac{1/2}{1 - 1/4} = \frac{1/2}{3/4} = \frac{1}{2} 	imes \frac{4}{3} = \frac{2}{3}$.

$A$ and $B$ toss a coin alternatively,the first to show a head being the winner. If $A$ starts the game,the chance of his winning is

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