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(C) Given mean $= np = 6$ ...$(i)$ Variance $= npq = 2$ ...(ii) Dividing (ii) by $(i)$,we get $q = \frac{2}{6} = \frac{1}{3}$. Since $p + q = 1$,we ha

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$1-\frac{19}{3^9}$

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(C) Given mean $= np = 6$ ...$(i)$ Variance $= npq = 2$ ...(ii) Dividing (ii) by $(i)$,we get $q = \frac{2}{6} = \frac{1}{3}$. Since $p + q = 1$,we have $p = 1 - \frac{1}{3} = \frac{2}{3}$. Substituting $p$ in $(i)$,$n 	imes \frac{2}{3} = 6 \Rightarrow n = 9$. We need to find $P(X \geq 2) = 1 - [P(X=0) + P(X=1)]$. $P(X=0) = {}^9C_0 	imes (\frac{2}{3})^0 	imes (\frac{1}{3})^9 = \frac{1}{3^9}$. $P(X=1) = {}^9C_1 	imes (\frac{2}{3})^1 	imes (\frac{1}{3})^8 = 9 	imes \frac{2}{3} 	imes \frac{1}{3^8} = \frac{18}{3^9}$. $P(X \geq 2) = 1 - [\frac{1}{3^9} + \frac{18}{3^9}] = 1 - \frac{19}{3^9}$.

For a binomial distribution with mean $6$ and variance $2$,find $P(X \geq 2)$.

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