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Q.

A biased coin with probability $P, 0 < P < 1$ of heads is tossed until a head appears for the first time. If the probability that the number of tosses required is even is 2/ 5, then the value of P is.

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a

1/4

b

1/6

c

1/3

d

1/2

answer is C.

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Detailed Solution

$Let q = 1 - p$

Since, head appears first time in an even throw i.e. 2 or 4 or 6 etc.

$\begin{array}{ll} ∴ \frac{2}{5} = q p + q^{3} p + q^{5} p + \dots \\ \Rightarrow \frac{2}{5} = \frac{q p}{1 - q^{2}} \\ \Rightarrow \frac{2}{5} = \frac{(1 - p) p}{1 - (1 - p)^{2}} \\ \Rightarrow \frac{2}{5} = \frac{1 - p}{2 - p} \end{array}$

$\begin{array}{ccc} \Rightarrow & 2 (2 - p) = 5 (1 - p) \\ \Rightarrow & 4 - 2 p = 5 - 5 p \\ \Rightarrow & 3 p = 1 \\ ∴ & p = \frac{1}{3} \end{array}$

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