In a binomial distribution B ( n, p ) , the sum and product of the...

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

Mean $= np$ and Variance $= npq$

Sum of mean and variance $= np + npq = 5$

$⇒ np ( 1 + q ) = 5$

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

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

$⇒ {n}^{2}{p}^{2}q = 6$ ....(2)

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

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

$⇒ 6{q}^{2} + 12q + 6 = 25q$

$⇒ 6{q}^{2} - 13q + 6 = 0$

$⇒ ( 3q - 2 ) ( 2q - 3 ) = 0$

$⇒ q = \dfrac{2}{3}, \dfrac{3}{2}$ where $q = \dfrac{2}{3}$ accepted

$∴ p = \dfrac{1}{3} ⇒ n \cdot \dfrac{1}{3} + n \cdot \dfrac{1}{3} \cdot \dfrac{2}{3} = 5$

$⇒ \dfrac{3n + 2n}{9} = 5$

$⇒ n = 9$

Therefore,

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