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

We need to find $$6(n + p - q)$$ for a binomial distribution $$B(n, p)$$ where the sum of mean and variance is 5 and their product is 6.

Set up the equations for a binomial distribution where the mean is $$np$$ and the variance is $$npq$$ with $$q = 1 - p$$. The sum of mean and variance is 5, giving $$np + npq = 5$$ or $$np(1 + q) = 5$$, and their product is 6, so $$np \cdot npq = 6$$ or $$n^2p^2q = 6$$.

From $$np(1 + q) = 5$$ and $$q = 1 - p$$, it follows that $$np = \frac{5}{1 + q} = \frac{5}{2 - p}$$.

Substituting $$np = \frac{5}{2 - p}$$ into $$n^2p^2q = 6$$ gives $$\left(\frac{5}{2 - p}\right)^2(1 - p) = 6$$, which simplifies to $$\frac{25(1-p)}{(2-p)^2} = 6$$. Multiplying both sides by $$(2-p)^2$$ yields $$25(1-p) = 6(2-p)^2 = 6(4 - 4p + p^2)$$, so $$25 - 25p = 24 - 24p + 6p^2$$. Rearranging gives $$6p^2 + p - 1 = 0$$, factoring to $$(3p - 1)(2p + 1) = 0$$ and thus $$p = \frac{1}{3}$$ (taking the positive root since $$0 < p < 1$$).

With $$p = \frac{1}{3}$$, we have $$q = \frac{2}{3}$$. Substituting into $$np(1 + q) = 5$$ yields $$\frac{n}{3} \times \frac{5}{3} = \frac{5n}{9} = 5$$, so $$n = 9$$.

Finally, $$6(n + p - q) = 6\left(9 + \frac{1}{3} - \frac{2}{3}\right) = 6\left(9 - \frac{1}{3}\right) = 6 \times \frac{26}{3} = 52$$.

Therefore, the value of $$6(n + p - q)$$ is 52, which matches Option B.