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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Probability

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

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

 

1 Answer

$ 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 by thanvigandhi_1
 

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