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If mean and variance of a Binomial variate $\mathrm{X}$ are 2 and 1 respectively, then the probability that $\mathrm{X}$ takes a value greater than 1 is
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Verified Answer
The correct answer is:
$\frac{11}{16}$
We have; $\mathrm{np}=2=$ mean $\mathrm{npq}=1=$ variance
$\Rightarrow \mathrm{p}=\frac{1}{2} ; \mathrm{q}=\frac{1}{2} \& \mathrm{n}=4$
$\begin{aligned} \text { Required probability } &=\mathrm{P}(\mathrm{x}>1) \\ &=1-\mathrm{P}(\mathrm{x} \leq 1) \\ &=1-[\mathrm{P}(\mathrm{x}=0)+\mathrm{P}(\mathrm{x}=1)] \\ &=1-\left[{ }^{4} \mathrm{C}_{0} \mathrm{q}^{4}+{ }^{4} \mathrm{C}_{1} \mathrm{q}^{3} \mathrm{p}^{1}\right] \\ &=1-\frac{5}{16}=\frac{11}{16} \end{aligned}$
$\Rightarrow \mathrm{p}=\frac{1}{2} ; \mathrm{q}=\frac{1}{2} \& \mathrm{n}=4$
$\begin{aligned} \text { Required probability } &=\mathrm{P}(\mathrm{x}>1) \\ &=1-\mathrm{P}(\mathrm{x} \leq 1) \\ &=1-[\mathrm{P}(\mathrm{x}=0)+\mathrm{P}(\mathrm{x}=1)] \\ &=1-\left[{ }^{4} \mathrm{C}_{0} \mathrm{q}^{4}+{ }^{4} \mathrm{C}_{1} \mathrm{q}^{3} \mathrm{p}^{1}\right] \\ &=1-\frac{5}{16}=\frac{11}{16} \end{aligned}$
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