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# The mean and variance of a binomial variate X are 2 and 1 respectively. The probability that X takes a value greater than 1 is

$\begin {array} {1 1} (A)\;\large\frac{1}{6} & \quad (B)\;\large\frac{5}{16} \\ (C)\;\large\frac{11}{16} & \quad (D)\;\large\frac{15}{16} \end {array}$

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## 1 Answer

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$np=2$
$n\: pq=1$
$Q= \large\frac{1}{2}$
$n = 4$
$P=\large\frac{1}{2}$
$\therefore$ Required probability = $P( X > 1)$
$1-P ( X \leq 1 )$
$= 1-[P(0)+P(1)]$
$= 1- \bigg[ 4C_0 \bigg( \large\frac{1}{2} \bigg)^4+4C_1 \bigg( \large\frac{1}{2} \bigg)^3 \bigg( \large\frac{1}{2} \bigg) \bigg]$
$= \large\frac{11}{16}$
answered Jan 8, 2014

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