$p:$ probability of getting head

$q:$ probability of getting tail

Such that, $p+q=1$

According to question we can get head in $2^{\mathrm{n}\mathrm{d}}, 4^{\mathrm{t}\mathrm{h}}, 6^{\mathrm{t}\mathrm{h}}$ … tail.

Hence required probability $=qp\left(1+q^2+q^4+\dots \right)=\frac{2}{5}$

$\Rightarrow$ $\frac{qp}{1-q^2}=\frac{2}{5}$

$\Rightarrow 5pq=2-2q^2$

$\Rightarrow 5p\left(1-p\right)=2-2{\left(1-p\right)}^2$

$\Rightarrow 5p-5p^2=2-2\left(1+p^2-2p\right)$

$\Rightarrow 5p-5p^2=2-2-2p^2+4p$

$\Rightarrow 3p^2=p$

$\Rightarrow p\left(3p-1\right)=0$

$\Rightarrow p=0,p=\frac{1}{3}$

$\because p\ne 0$

$\therefore p=\frac{1}{3}$

$\Rightarrow 9p=3$