# If the Mean and Variance of a Binomial Variate X Are 2 and 1 Respectively, Find P (X > 1). - Mathematics

#### Question

If the mean and variance of a binomial variate X are 2 and 1 respectively, find P (X > 1).

#### Solution

$\text{ Mean = 2 , Variance } = 1$
$\therefore q = \frac{\text{ Variance} }{\text{ Mean } } = \frac{1}{2}$
$\text{ and p } = 1 - \frac{1}{2} = \frac{1}{2}$
$n = \frac{\text{ Mean} }{p} = \frac{2}{\frac{1}{2}} = 4$
$\text{ The binomial distribution is given by }$
$P(X = r) = ^ {4}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{4 - r}$
$\therefore P(X = 0) = ^{4}{}{C}_0 \left( \frac{1}{2} \right)^0 \left( \frac{1}{2} \right)^{4 - 0} , r = 0, 1, 2, 3, 4$
$= \left( \frac{1}{2} \right)^4$
$P(X > 1) = 1 - P(X = 0)$
$= 1 - \left( \frac{1}{2} \right)^4$
$= \frac{15}{16}$

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#### APPEARS IN

RD Sharma Solution for Mathematics for Class 12 (Set of 2 Volume) (2018 (Latest))
Chapter 33: Binomial Distribution
Very Short Answers | Q: 7 | Page no. 27

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If the Mean and Variance of a Binomial Variate X Are 2 and 1 Respectively, Find P (X > 1). Concept: Bernoulli Trials and Binomial Distribution.