If the mean and the variance of Binomial variate $X$ are 2 and 1 respectively, then the probability that X takes a value greater than or equal to one is

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Mean }=\mathrm{np}=2 	ext { and variance }=\mathrm{npq}=1 \ & 	herefore \quad \mathrm{q}=\frac{1}{2} \end{aligned}$ $\begin{aligned} & 	ext { Also, } p=1-q=1-\frac{1}{2}=\frac{1}{2} \ & \mathrm{n}=4 \end{aligned}$ $\begin{aligned} 	herefore \quad P(X \geq 1) & =1-P(X=0) \ & =1-{ }^4 C_0 p^0 q^4 \ & =1-(1)\left(\frac{1}{2}ight)^0\left(\frac{1}{2}ight)^4 \ & =1-\frac{1}{16} \ & =\frac{15}{16} \end{aligned}$

If the mean and the variance of Binomial variate $X$ are 2 and 1 respectively, then the probability that X…

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