Question

A biased coin with probability $P(0<p<1)$ of heads is tossed until a head appears for the first time. lf the probability that the number of tosses required is even is $\frac{2}{5}$ then $P=$

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

Answer 1

Answer: (D)

$\frac{1}{3}$

One can get heads after 2 throws or 4 or 6 ... till infinity.
Let probability of a head $=p$ and that of a tail$=(1-p)$.
Hence, probability $=(1-p)\ast p+{(1-p)}^3\ast p+{(1-p)}^5\ast p+...\infty =\frac{(1-p)\ast p}{1-{(1-p)}^2}=\frac{1-p}{2-p}$
$\Rightarrow$ $\frac{1-p}{2-p}=\frac{2}{5}=>p=\frac{1}{3}$