The difference between mean and variance of a binomial distribution is 4 and the difference of their squares is 32. Then distribution is given by :

$\left(\frac{1}{3}+\frac{2}{3}\right)^{9}$

$\text{Let mean}=m,\;\text{variance}=v.$

$m-v=4.$

$m^2-v^2=(m-v)(m+v)=32.$

$4(m+v)=32 \Rightarrow m+v=8.$

$m=6,\;v=2.$

$\text{For binomial distribution: }m=np,\;v=npq.$

$np=6,\;npq=2 \Rightarrow q=\frac{1}{3},\;p=\frac{2}{3}.$

$n=\frac{6}{2/3}=9.$

$\text{Distribution is }B(9,\frac{2}{3}).$

$\left(\frac{1}{3}+\frac{2}{3}\right)^{9}$