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If the mean and variance of a binomial variable $\mathrm{X}$ are 2 and 1 respectively, then $\mathrm{P}(\mathrm{X}>1)=$
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Verified Answer
The correct answer is:
$\frac{11}{16}$
$\because$ Mean $=2 \Rightarrow n p=2$
$\begin{aligned} & \text { and Variance }=1 \Rightarrow n p q=1 \\ & \Rightarrow 2(1-P)=1 \Rightarrow 2-2 p=1 \\ & \Rightarrow p=1 / 2\end{aligned}$
$\therefore \frac{1}{2} n=2 \Rightarrow n=4$
$\begin{aligned} & \text { Now, } P(X>1)=1-P(X \leq 1) \\ & =1-P(X=0)-P(X=1)\end{aligned}$
$\begin{aligned} & =1-{ }^4 \mathrm{C}_0 \mathrm{p}^0 \mathrm{q}^{4-0}-{ }^4 \mathrm{C}_{\mathrm{i}} \mathrm{pq}^3 \\ & =1-1.1\left(\frac{1}{2}\right)^4 4 \cdot \frac{1}{2} \times \frac{1}{2^3} \\ & \Rightarrow \mathrm{P}(\mathrm{X} \geq 1)=\frac{15}{16}-\frac{4}{16}=\frac{11}{16}\end{aligned}$
$\begin{aligned} & \text { and Variance }=1 \Rightarrow n p q=1 \\ & \Rightarrow 2(1-P)=1 \Rightarrow 2-2 p=1 \\ & \Rightarrow p=1 / 2\end{aligned}$
$\therefore \frac{1}{2} n=2 \Rightarrow n=4$
$\begin{aligned} & \text { Now, } P(X>1)=1-P(X \leq 1) \\ & =1-P(X=0)-P(X=1)\end{aligned}$
$\begin{aligned} & =1-{ }^4 \mathrm{C}_0 \mathrm{p}^0 \mathrm{q}^{4-0}-{ }^4 \mathrm{C}_{\mathrm{i}} \mathrm{pq}^3 \\ & =1-1.1\left(\frac{1}{2}\right)^4 4 \cdot \frac{1}{2} \times \frac{1}{2^3} \\ & \Rightarrow \mathrm{P}(\mathrm{X} \geq 1)=\frac{15}{16}-\frac{4}{16}=\frac{11}{16}\end{aligned}$
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