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A random variable $X \sim B(n, p)$, if values of mean and variance of $\mathrm{X}$ are 18,12 respectively, then $\mathrm{n}=$
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Verified Answer
The correct answer is:
54
We have $n p=18$ and $n p q=12$
$$
\begin{aligned}
& \therefore \mathrm{q}=\frac{12}{18}=\frac{2}{3} \quad \Rightarrow \mathrm{p}=1-\frac{2}{3}=\frac{1}{3} \\
& \therefore \mathrm{n}\left(\frac{1}{3}\right)=18 \quad \Rightarrow \mathrm{n}=54
\end{aligned}
$$
$$
\begin{aligned}
& \therefore \mathrm{q}=\frac{12}{18}=\frac{2}{3} \quad \Rightarrow \mathrm{p}=1-\frac{2}{3}=\frac{1}{3} \\
& \therefore \mathrm{n}\left(\frac{1}{3}\right)=18 \quad \Rightarrow \mathrm{n}=54
\end{aligned}
$$
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