Q80 / 160 · AP EAMCET 2024 (23 May Shift 1) · Mathematics

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

  • A
    50
  • B
    53
  • Correct answer
    C
    52
  • D
    51

Solution

Mean $=n p$ and Variance $=n p q$ Sum $=n p+n p q=n p(1+q)=5$ ...(i) Product $=n^2 p^2 q=6$ ...(ii) From (i) and (ii) $\begin{aligned} & \frac{n^2 p^2(1+q)^2}{n^2 p^2 q}=\frac{25}{6} \Rightarrow 6 q^2-13 q+6=0 \\ & \Rightarrow q=\frac{2}{3} \quad \Rightarrow p=-\frac{2}{3}=\frac{1}{3} \quad\left(q \neq \frac{3}{2}\right) \end{aligned}$ From (i) $n=9 \Rightarrow 6(n+p-q)=6\left(9+\frac{1}{3}-\frac{2}{3}\right)=52$;

Hence, the correct answer is option: 'C'.

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