In a binomial distribution $B(n,p)$, the sum and the product of the mean and the variance are 5 and 6 respectively, then $6(n+p-q)$ is equal to :
Solution
$$
\begin{aligned}
& \text { Given } \\\\
& \mathrm{np}+\mathrm{npq}=5 \\\\
& \Rightarrow \mathrm{np}(1+\mathrm{q})=5 ........(i) \\\\
& \text { and (np) (npq) }=6 \\\\
& \Rightarrow \mathrm{n}^2 \mathrm{p}^2 \mathrm{q}=6 ........(ii) \\\\
& (\mathrm{i})^2 \div(\mathrm{ii}) \\\\
& \frac{(1+q)^2}{9}=\frac{25}{6} \\\\
& \Rightarrow 6 \mathrm{q}^2-13 \mathrm{q}+6=0 \\\\
& \Rightarrow \mathrm{q}=\frac{2}{3}, \frac{3}{2} \text { (rejected) } \\\\
& \mathrm{p}=1-\frac{2}{3}=\frac{1}{3} \\\\
& \frac{n}{3}\left(1+\frac{2}{3}\right)=5 \\\\
& \Rightarrow \mathrm{n}=9 \\\\
& 6(\mathrm{n}+\mathrm{p}-\mathrm{q})=52
\end{aligned}
$$
About this question
Subject: Mathematics · Chapter: Probability · Topic: Classical and Axiomatic Probability
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