A box contains 15 green and 10 yellow balls. IF 10 balls are randomly drawn, one-by-one, with replacement, then the variance of the number of green balls drawn, is

$\frac{12}{5}$

Let X-denote the number of green balls in a draw 10  balls one-by-one with replacement. Then, X is a binomial variance with

$n = 10, p =\frac{15}{25}=\frac{3}{5}$ and $q=\frac{2}{5}$

∴ Variance $X=npq= 10×\frac{3}{5}×\frac{2}{5}=\frac{12}{5}$