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

Answer: (A)

Let $q=1-p.$ Since, head appears first time in an even throw 2 or 4 or $6\dots .$
$\therefore \frac{2}{5}=qp+q^{3}p+q^{5}p+\dots$
$\Rightarrow \frac{2}{5}=\frac{qp}{1-q^2}$
$\Rightarrow \frac{2}{5}=\frac{(1-p)p}{1-(1-p{)}^2}$
$\Rightarrow \frac{2}{5}=\frac{1-p}{2-p}$
$\Rightarrow 4-2p=5-5p$
$\Rightarrow 3p=1$
$\Rightarrow p=\frac{1}{3}$