### Probability Question 13

#### Question 13 - 01 February - Shift 1

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

(1) 51

(2) 52

(3) 53

(4) 50

## Show Answer

#### Answer: (2)

#### Solution:

#### Formula: Mean and variance of probability distribution, Probability of an event, Roots of equations, Important results on probability

$n p+n p q=5, n p \cdot n p q=6$

$np(1+q)=5, n^{2} p^{2} q=6$

$n^{2} p^{2}(1+q)^{2}=25, n^{2} p^{2} q=6$

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

$6 q^{2}+12 q+6=25 q$

$6 q^{2}-13 q+6=0$

$6 q^{2}-9 q-4 q+6=0$

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

$q=\frac{2}{3}, \frac{3}{2}, q=\frac{2}{3}$ is accepted

$p=\frac{1}{3} \Rightarrow$ n. $\frac{1}{3}+n \cdot \frac{1}{3} \cdot \frac{2}{3}=5$

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

$n=9$

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