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# If the Mean and Variance of a Binomial Variate X Are 2 and 1 Respectively, Then the Probability that X Takes a Value Greater than 1 is (A) 2/3 (B) 4/5 (C) 7/8 (D) 15/16 - Mathematics

ConceptBernoulli Trials and Binomial Distribution

#### Question

If the mean and variance of a binomial variate X are 2 and 1 respectively, then the probability that X takes a value greater than 1 is

• 2/3

• 4/5

• 7/8

• 15/16

#### Solution

15/16

Mean =2 and variance =1

$\Rightarrow np = 2 \text{ and npq } = 1$
$\Rightarrow q = \frac{1}{2}$
$\Rightarrow p = 1 - \frac{1}{2} = \frac{1}{2}$
$n = \frac{\text{ Mean} }{p}$
$\Rightarrow n = 4$
$\text{ Hence, the distribution is given by }$
$P\left( X = r \right) =^{4}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{4 - r} , r = 0, 1, 2, 3, 4$
$\therefore P(X \geq 1) = 1 - P(X = 0)$
$= 1 - \frac{1}{2^4}$
$= \frac{15}{16}$

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Solution If the Mean and Variance of a Binomial Variate X Are 2 and 1 Respectively, Then the Probability that X Takes a Value Greater than 1 is (A) 2/3 (B) 4/5 (C) 7/8 (D) 15/16 Concept: Bernoulli Trials and Binomial Distribution.
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