For a Binomial distribution B (n, p), $\frac{E(x)}{V(x)}$ is equal to : (symbols have their usual meaning)

$\frac{1}{p^2}$ $1-p$ $\frac{1}{1+p}$ $\frac{1}{1-p}$

$\frac{1}{1-p}$

For B (n, p) $E(x)=np$ $V(x)=np(1-p)$ so $\frac{E(x)}{V(x)}=\frac{np}{np(1-p)}=\frac{1}{1-p}$