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

Option: 1

52


Option: 2

50


Option: 3

51


Option: 4

53


Answers (1)

best_answer

Given

\begin{aligned} & n p+n p q=5 \\ & \Rightarrow n p(1+q)=5 \, \, \, \, \, \, \, \, \, ....(i)\\ & \text { and }(n p)(n p q)=6 \\ & \Rightarrow n^2 p^2 q=6 \, \, \, \, \, \, ...(ii)\\ & \left(i^2 \div(i i)\right. \\ & \frac{(1+9)^2}{9}=\frac{25}{6} \\ & \Rightarrow 6 q^2-13 q+6=0 \\ & \Rightarrow q=\frac{2}{3}, \frac{3}{2} \text { (reject) } \\ & \mathrm{P}=1-\frac{2}{3}=\frac{1}{3} \\ & \frac{n}{3}\left(1+\frac{2}{3}\right)=5 \\ & \Rightarrow n=9 \\ & 6(n+p-q)=52 \end{aligned}

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