Two persons A and B toss two coins one after another. The person who throws one head and one tail is the winner. If A starts the game the probability that B wins the game is

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$\frac{1}{2}$

$\frac{2}{3}$

$\frac{1}{3}$

$\frac{1}{4}$

answer is C.

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Detailed Solution

two coins thrown

$P (Wins) = P (1 H, 1 T) = \frac{2}{4} = \frac{1}{2} P (Lose) = \frac{1}{2}$

A start the game

$\begin{array}{rcl} P (Bwins) & = & P (L) P (W) + P (L) P (L) PLL) P (W) . . . . \\ = & {(\frac{1}{2})}^{2} + {(\frac{1}{2})}^{4} + . . . . \\ = & \frac{{(\frac{1}{2})}^{2}}{1 - {(\frac{1}{2})}^{2}} = \frac{1}{3} \end{array}$

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