Let X denote the number of tosses required.
Then P(x = r) = (1 − P)^{r − 1} P; r = 1, 2, 3....

Let E denote the event that the number of tosses required is even. Then,

P(E) = P{(x = 2) U (x = 4) U (x = 6) U...}

= P(x = 2) + P(x = 4) + P(x = 6) +....

= p(1 − p) + p(1 − p)³ + p(1 − p)⁵ +..

$=\frac{\mathrm{p}(1-\mathrm{p})}{1-{(1-\mathrm{p})}^2}=\frac{\mathrm{P}(1–\mathrm{p})}{2\mathrm{P}–{\mathrm{p}}^2}=\frac{1–\mathrm{p}}{2–\mathrm{p}}$

As we are given that $\mathrm{P}\left(\mathrm{E}\right)=\frac{2}{5}$, we get

5(1 – p) = 2(2 – p) 

⇒ 1 – 3p = 0

⇒ $\mathrm{P}=\frac{1}{3}$