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The mean and the variance in a binomial distribution are found to be 2 and 1 respectively. The probability $P(X=0)$ is
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Verified Answer
The correct answer is:
$\frac{1}{16}$
$\mathrm{nP}=2$ $\mathrm{n} \mathrm{Pq}=1$
$\mathrm{q}=\frac{1}{2}$
$\mathrm{P}+\mathrm{q}=1$
$\mathrm{P}=1-\frac{1}{2}=\frac{1}{2}$
$\mathrm{n}\left(\frac{1}{2}\right)=2$
$\therefore \quad \mathrm{n}=4$
${ }^{4} \mathrm{C}_{0}=\left(\frac{1}{2}\right)^{4-0}=\frac{1}{16} .$
$\mathrm{q}=\frac{1}{2}$
$\mathrm{P}+\mathrm{q}=1$
$\mathrm{P}=1-\frac{1}{2}=\frac{1}{2}$
$\mathrm{n}\left(\frac{1}{2}\right)=2$
$\therefore \quad \mathrm{n}=4$
${ }^{4} \mathrm{C}_{0}=\left(\frac{1}{2}\right)^{4-0}=\frac{1}{16} .$
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