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

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Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Mean }=2 	ext { and variance }=1 \ & \Rightarrow \mathrm{np}=2 	ext { and } \mathrm{npq}=1 \ 	herefore \quad & \frac{\mathrm{npq}}{\mathrm{np}}=\frac{1}{2} \Rightarrow \mathrm{q}=\frac{1}{2} \ & \Rightarrow \mathrm{p}=1-\frac{1}{2}=\frac{1}{2} \ & n p=2 \ & \Rightarrow \mathrm{n}\left(\frac{1}{2}ight)=2 \Rightarrow \mathrm{n}=4\end{aligned}$ $\begin{array}{ll}	herefore & P(X\gt1) \ & =P(X=2)+P(X=3)+P(X=4)\end{array}$ $={ }^4 \mathrm{C}_2\left(\frac{1}{2}ight)^2\left(\frac{1}{2}ight)^2+{ }^4 \mathrm{C}_3\left(\frac{1}{2}ight)^3\left(\frac{1}{2}ight)+{ }^4 \mathrm{C}_4\left(\frac{1}{2}ight)^4\left(\frac{1}{2}ight)^0$ $\begin{aligned} & =6\left(\frac{1}{2}ight)^4+4\left(\frac{1}{2}ight)^4+1\left(\frac{1}{2}ight)^4 \ & =\frac{1}{2^4}(6+4+1)=\frac{11}{16}\end{aligned}$

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

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