Question

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 p equals
(a) 1/3
(b) 2/3
(c) 2/5
(d) 3/5

Answer 1

(a) 1/3

Probability of heads = p
Probability of Tails = (1 − p)
Probability that first head appears at even turn
P_e = (1 − p)p + (1 − p)³p+(1 − p)⁵p + .....
= (1 − p)p (1 + (1 − p)² + (1 − p)⁴+ .....)

$$\begin{gathered} =\left(1-p\right)p\left(\frac{1}{1-{\left(1-p\right)}^2}\right) \\ =\left(1-p\right)p\left(\frac{1}{-p^2+2p}\right) \end{gathered}$$

$$\begin{gathered} {\mathrm{P}}_{\mathrm{e}}=\frac{\left(1-p\right)}{\left(2-p\right)} \\ {\mathrm{P}}_{\mathrm{e}}=\frac{2}{5} \\ \frac{1-p}{2-p}=\frac{2}{5} \\ 5-5p=4-2p \\ 3p=1 \\ p=\frac{1}{3} \end{gathered}$$