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 is seven is $\frac{2}{5}$, then p equals

$\frac{1}{3}$

Let q= 1-p. Since head appears first time in an even throw i.e., 2 or 4 or 6 etc.

$∴ \frac{2}{5}=qp +q^3 p +q^5p+...$

$⇒\frac{2}{5}=\frac{qp}{1-q^2}$

$⇒\frac{2}{5}=\frac{(1-p)p}{1-(1-p)^2}=\frac{1-p}{2-p}$

$⇒ 2(2-p) = 5 (1- p) ⇒ p = \frac{1}{3}$