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(D) The probability of getting a head is $P(H) = 1/2$ and the probability of not getting a head is $P(T) = 1/2$.Since $A$ starts the game,$A$ wins if

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$2/3$

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(D) The probability of getting a head is $P(H) = 1/2$ and the probability of not getting a head is $P(T) = 1/2$.Since $A$ starts the game,$A$ wins if he gets a head on the $1^{st}, 3^{rd}, 5^{th}, \dots$ toss.$A$ wins on the $1^{st}$ toss with probability $1/2$.$A$ wins on the $3^{rd}$ toss if the sequence is $T, T, H$ with probability $(1/2) 	imes (1/2) 	imes (1/2) = (1/2)^3$.$A$ wins on the $5^{th}$ toss if the sequence is $T, T, T, T, H$ with probability $(1/2)^5$.Thus,the total probability of $A$ winning is an infinite geometric series:$P(A) = 1/2 + (1/2)^3 + (1/2)^5 + \dots$This is a geometric progression with first term $a = 1/2$ and common ratio $r = (1/2)^2 = 1/4$.The sum of an infinite geometric series is $S = a / (1 - r)$.$P(A) = (1/2) / (1 - 1/4) = (1/2) / (3/4) = (1/2) 	imes (4/3) = 2/3$.

$A$ 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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