In a binomial distribution $\mathrm{B}(n,p)$, the sum and product of the mean & variance are $5$ and $6$ respectively, then find $6(n+p-q)$ is equal to :-

For the binomial distribution $B(n, p)$,

Mean $=np$ and Variance $=npq$

Sum of mean and variance $=np+npq=5$

$\Rightarrow np(1+q)=5$

$\Rightarrow n^2{p}^2(1+q{)}^2=25$     ....(1)

Product of mean and variance $=np.npq=6$

$\Rightarrow n^2{p}^{2}q=6$      ....(2)

Put the value of $n^2{p}^2$ from equation (2) into equation (1):

$\frac{6}{q}(1+q{)}^2=25$

$\Rightarrow 6q^2+12q+6=25q$

$\Rightarrow 6q^2-13q+6=0$

$\Rightarrow (3q-2)(2q-3)=0$

$\Rightarrow q=\frac{2}{3}, \frac{3}{2}$ where $q=\frac{2}{3}$ accepted

$\therefore p=\frac{1}{3}\Rightarrow n·\frac{1}{3}+n·\frac{1}{3}·\frac{2}{3}=5$

$\Rightarrow \frac{3n+2n}{9}=5$

$\Rightarrow n=9$

Therefore,

$6(n+p-q)=6\left(9+\frac{1}{3}-\frac{2}{3}\right)=52$