For a Binomial distribution, $B(n,p)$, where $p+q=1$, the sum and product of mean and variance are 8 and 12 respectively, when the value of $n$ is:

9

The correct answer is Option (2) → 9

$X\sim B(n,p),\quad p+q=1$

$\text{Mean}=np$

$\text{Variance}=npq$

Given

$np+npq=8$

$np(1+q)=8$

$np(2-p)=8 \quad ...(1)$

$np\cdot npq=12$

$n^2p^2q=12 \quad ...(2)$

From (1):

$n=\frac{8}{p(2-p)}$

Substitute in (2):

$\left(\frac{8}{p(2-p)}\right)^2 p^2(1-p)=12$

$\frac{64(1-p)}{(2-p)^2}=12$

$16(1-p)=3(2-p)^2$

$16-16p=12-12p+3p^2$

$3p^2+4p-4=0$

$p=\frac{2}{3}$

Using (1):

$n\cdot\frac{2}{3}\left(2-\frac{2}{3}\right)=8$

$n\cdot\frac{2}{3}\cdot\frac{4}{3}=8$

$\frac{8n}{9}=8$

$n=9$

The value of $n$ is $9$.