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If the mean and variance of a binomial variate $X$ are 2 and 1 respectively, then the probability that $X$ takes a value greater than 1 , is
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Verified Answer
The correct answer is:
${ }^{\frac{15}{16}}$
We have mean $(X)=n p=2$
and variance $(X)=n p q=1 \Rightarrow q=\frac{1}{2}$ or $p=\frac{1}{2}$ and $n=4$
Thus
$p(X \geq 1)=1-p(X=0)$ $=1-{ }^4 C_0\left(\frac{1}{2}\right)^4=\frac{15}{16}$
and variance $(X)=n p q=1 \Rightarrow q=\frac{1}{2}$ or $p=\frac{1}{2}$ and $n=4$
Thus
$p(X \geq 1)=1-p(X=0)$ $=1-{ }^4 C_0\left(\frac{1}{2}\right)^4=\frac{15}{16}$
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