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(C) For a Binomial distribution $B(n, p)$,Mean $\mu = np$ and Variance $\sigma^2 = npq$,where $q = 1-p$. Given: Sum: $np + npq = 5 \Rightarrow np(1+q)

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$52$

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(C) For a Binomial distribution $B(n, p)$,Mean $\mu = np$ and Variance $\sigma^2 = npq$,where $q = 1-p$. Given: Sum: $np + npq = 5 \Rightarrow np(1+q) = 5$ ...$(i)$ Product: $(np)(npq) = n^2p^2q = 6$ ...(ii) From $(i)$,$np = \frac{5}{1+q}$. Substituting this into (ii): $\left(\frac{5}{1+q}\right)^2 q = 6 \Rightarrow 25q = 6(1+q)^2 \Rightarrow 25q = 6(1+2q+q^2) \Rightarrow 6q^2 - 13q + 6 = 0$. Solving the quadratic equation: $6q^2 - 9q - 4q + 6 = 0 \Rightarrow 3q(2q-3) - 2(2q-3) = 0 \Rightarrow (3q-2)(2q-3) = 0$. Since $q Substituting $q = \frac{2}{3}$ into $(i)$: $np(1 + \frac{2}{3}) = 5 \Rightarrow np(\frac{5}{3}) = 5 \Rightarrow np = 3$. Since $p = \frac{1}{3}$,$n(\frac{1}{3}) = 3 \Rightarrow n = 9$. Finally,$6(n+p-q) = 6(9 + \frac{1}{3} - \frac{2}{3}) = 6(9 - \frac{1}{3}) = 54 - 2 = 52$.

In a Binomial distribution $B(n, p)$,the sum and product of the mean and the variance are $5$ and $6$ respectively,then $6(n+p-q)=$

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