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. $\frac{1}{3}$
• B. $\frac{2}{3}$
• C. $\frac{2}{5}$
• D. $\frac{3}{5}$

Answer 1

The correct option is A

$\frac{1}{3}$

Finding the probability:

Let the “$p$” be the probability of getting a head and “$q$” be the probability of not getting a head.

The head appears the first time on the even number of throws are $2or4or6.$

Thus,

$$\begin{gathered} \frac{2}{5}=qp+q^{3}p+q^{5}p+\dots \\ \frac{2}{5}=\frac{qp}{1-q^2} \end{gathered}$$

We know that, $q=1-p$

$$\begin{gathered} \frac{2}{5}=\frac{(1-p)p}{1-{(1-p)}^2} \\ \Rightarrow \frac{1-p}{2-p}=\frac{2}{5} \\ \Rightarrow 4-2p=5-5p \\ \Rightarrow -2p+5p=5-4 \\ \Rightarrow 3p=1 \\ \Rightarrow p=\frac{1}{3} \end{gathered}$$

Hence, the value of $p$ is $\frac{1}{3}$