Question

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.

Answer 1

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  \[\frac{8}{9}\]   .