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

Options

• 1/3

• 2/3

• 2/5

• 3/5

   

Answer 1

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+ .....)

\[= \left( 1 - p \right)p\left( \frac{1}{1 - \left( 1 - p \right)^2} \right)\]
\[ = \left( 1 - p \right)p\left( \frac{1}{- p^2 + 2p} \right)\]

\[P_e = \frac{\left( 1 - p \right)}{\left( 2 - p \right)}\]
\[ P_e = \frac{2}{5}\]
\[\frac{1 - p}{2 - p} = \frac{2}{5}\]
\[5 - 5p = 4 - 2p\]
\[3p = 1\]
\[p = \frac{1}{3}\]