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# If the Mean and Variance of a Random Variable X with a Binomial Distribution Are 4 and 2 Respectively, Find P (X = 1). - Mathematics

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Short Note

If the mean and variance of a random variable X with a binomial distribution are 4 and 2 respectively, find P (X = 1).

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#### Solution

$\text{ Given, np = 4 and npq } = 2$
$\therefore p = 1 - \frac{\text{ Variance } }{\text{ Mean}}$
$= 1 - \frac{2}{4}$
$= \frac{1}{2}$
$\text{ and } q = \frac{1}{2}$
$\text{ and } n = \frac{np}{p}$
$= \frac{4}{\frac{1}{2}}$
$= 8$
$\text{ Hence, the binomial distribution is given by }$
$P\left( X = r \right) = ^ {8}{}{C}_r \left( \frac{1}{2} \right)^r \left( \frac{1}{2} \right)^{8 - r} , r = 0, 1, 2, 3 . . . . . . . 8$
$\therefore P(X = 1) = 8 \left( \frac{1}{2} \right)^8$
$= \frac{1}{32}$

Concept: Bernoulli Trials and Binomial Distribution
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 33 Binomial Distribution
Very Short Answers | Q 6 | Page 27

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