A baised coin with probability p, 0 < p < 1 of heads is tossed until a heat appears for the first time. If the probability that the number of tosses required is even is 2/5, then p is equal to

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$\frac{2}{3}$

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{1}{4}$

answer is C.

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Detailed Solution

$P (H) = P P (T) = 1 - P$

P (tosses required even )

$\begin{array}{rcl} = & P (T) P (H) + P (T) P (T) P (T) P (H) . . . . . \\ \frac{2}{5} & = & P (1 - P) + {(1 - P)}^{3} P + . . . . . \\ = & \frac{P (1 - P)}{1 - {(1 - P)}^{2}} \Rightarrow \frac{2}{5} = \frac{P (1 - P)}{(2 - P) P} \\ \Rightarrow & 4 - 2 P = 5 - 5 P \Rightarrow 3 P = 1 \Rightarrow P = \frac{1}{3} \end{array}$

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