Q. 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 tossed required is even is $\frac{2}{5},\ then\ p\ equals$
A) $\frac{1}{3}$
B) $\frac{2}{3}$
C) $\frac{2}{5}$
D) $\frac{3}{5}$
Explanation
(a)
Let $q = 1 - p$. Since, head appears first time in an even throw 2 or 4 or 6
$\therefore\ \frac{2}{5} = qp + q^{3}p + q^{5}p + ...$
$\therefore\ \frac{2}{5} = \frac{qp}{1 - q^{2}}$
$\Rightarrow \ \frac{2}{5} = \frac{(1 - p)p}{1 - (1 - p)^{2}}$
$\Rightarrow \ \frac{2}{5} = \frac{1 - p}{2 - p}$
$\Rightarrow \ 4 - 2P = 5 - 5P\ \Rightarrow \ p = \frac{1}{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.