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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Probability
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The mean and standard deviation of a binomial variate X are 4 and $ \sqrt 3$ respectively. The $P( X \geq 1 )$ is equal to

$\begin {array} {1 1} (A)\;1- \bigg(\large\frac{1}{4} \bigg)^{16} & \quad (B)\;1- \bigg(\large\frac{3}{4} \bigg)^{16} \\ (C)\;1- \bigg(\large\frac{2}{3} \bigg)^{16} & \quad (D)\;1- \bigg(\large\frac{1}{3} \bigg)^{16} \end {array}$

 

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1 Answer

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Given :
Mean = 4
Variance = $ npq = 3$
On solvng we get,
$ Q=\large\frac{3}{4}$
$n = 16$
$p = \large\frac{1}{4}$
$ P ( X \geq 1 ) = 1- P(x=0)$
$ = 1-nC_0P^0Q^{n-0}$
$ = 1- \bigg( \large\frac{3}{4} \bigg)^{16}$
answered Jan 16, 2014 by thanvigandhi_1
 

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