Thank you for reporting, we will resolve it shortly
Q.
A biased coin with probability p, $0
AIEEEAIEEE 2002
Solution:
Let $q=1-p .$ Since, head appears first time in an even throw 2 or 4 or $6 \ldots .$
$ \therefore \frac{2}{5}=q p+q^{3} p+q^{5} p+\ldots$
$\Rightarrow \frac{2}{5}=\frac{q p}{1-q^{2}}$
$\Rightarrow \frac{2}{5}=\frac{(1-p) p}{1-(1-p)^{2}}$
$\Rightarrow \frac{2}{5}=\frac{1-p}{2-p}$
$\Rightarrow 4-2 p=5-5 p$
$\Rightarrow 3 p=1$
$\Rightarrow p=\frac{1}{3}$