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If the mean and variance of a binomial distribution are 4 and 2 respectively, then the probability of 2 successes of that binomial variate $X$, is
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$\frac{7}{64}$
For a binomial distribution, Mean, $\mu=n \rho=4$... (i)
Variance, $\sigma^2=n \mathrm{p} q=2$... (ii)
On solving eqs. (i) and (ii), we get $n=8, \mathrm{p}=\frac{1}{2}, q=\frac{1}{2}$
Probability of getting 2 successes,
$\mathrm{p}(\mathrm{X}=2)={ }^8 \mathrm{C}_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^6=\frac{{ }^8 \mathrm{C}_2}{2^8}=\frac{28}{256}=\frac{7}{64}$
Variance, $\sigma^2=n \mathrm{p} q=2$... (ii)
On solving eqs. (i) and (ii), we get $n=8, \mathrm{p}=\frac{1}{2}, q=\frac{1}{2}$
Probability of getting 2 successes,
$\mathrm{p}(\mathrm{X}=2)={ }^8 \mathrm{C}_2\left(\frac{1}{2}\right)^2\left(\frac{1}{2}\right)^6=\frac{{ }^8 \mathrm{C}_2}{2^8}=\frac{28}{256}=\frac{7}{64}$
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