#### 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

1/3

2/3

2/5

3/5

#### Solution

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*)

^{3}

*p*+(1 −

*p*)

^{5}

*p*+ .....

= (1 −

*p*)

*p*(1 + (1 −

*p*)

^{2}+ (1 −

*p*)

^{4}+ .....)

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

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

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

\[ P_e = \frac{2}{5}\]

\[\frac{1 - p}{2 - p} = \frac{2}{5}\]

\[5 - 5p = 4 - 2p\]

\[3p = 1\]

\[p = \frac{1}{3}\]

Is there an error in this question or solution?

Solution 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 Concept: Bernoulli Trials and Binomial Distribution.