# An Urn Contains 3 White and 6 Red Balls. Four Balls Are Drawn One by One with Replacement from the Urn. Find the Probability Distribution of the Number of Red Balls Drawn. Also - Mathematics

Sum

An urn contains 3 white and 6 red balls. Four balls are drawn one by one with replacement from the urn. Find the probability distribution of the number of red balls drawn. Also find mean and variance of the distribution.

#### Solution

Let X denote the total number of red balls when four balls are drawn one by one with replacement.
P (getting a red ball in one draw) = $\frac{2}{3}$

P (getting a white ball in one draw) =  $\frac{1}{3}$

 X 0 1 2 3 4 P(X) $\left( \frac{1}{3} \right)^4$ $\frac{2}{3} \left( \frac{1}{3} \right)^3 . ^ {4}{}{C}_1$ $\left( \frac{2}{3} \right)^2 \left( \frac{1}{3} \right)^2 . ^{4}{}{C}_2$ $\left( \frac{2}{3} \right)^3 \left( \frac{1}{3} \right) . ^{4}{}{C}_3$ $\left( \frac{2}{3} \right)^4$ $\frac{1}{81}$ $\frac{8}{81}$ $\frac{24}{81}$ $\frac{32}{81}$ $\frac{16}{81}$

Using the formula for mean, we have

$\overline{X} = \sum P_i X_i$

$\text{ Mean } ( \bar{X}) = \left( 0 \times \frac{1}{81} \right) + 1 \left( \frac{8}{81} \right) + 2\left( \frac{24}{81} \right) + 3 \left( \frac{32}{81} \right) + 4 \left( \frac{16}{81} \right)$

$= \frac{1}{81}\left( 8 + 48 + 96 + 64 \right)$

$= \frac{216}{81}$

$= \frac{8}{3}$

Using the formula for variance, we have

$\text{ Var } (X) = \sum P_i {X_i}^2 - \left( \sum P_i X_i \right)^2$

$\text{ Var } (X) = \left\{ \left( 0 \times \frac{1}{81} \right) + 1 \left( \frac{8}{81} \right) + 4\left( \frac{24}{81} \right) + 9 \left( \frac{32}{81} \right) + 16 \left( \frac{16}{81} \right) \right\} - \left( \frac{8}{3} \right)^2$

$= \frac{648}{81} - \frac{64}{9}$

$= \frac{8}{9}$

Hence, the mean of the distribution is  $\frac{8}{3}$  and the variance of the distribution is  $\frac{8}{9}$   .

Concept: Random Variables and Its Probability Distributions
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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 33 Binomial Distribution
Exercise 33.2 | Q 26 | Page 26