Question

Five bad oranges are accidently mixed with 20 good ones. If four oranges are drawn one by one successively with replacement, then find the probability distribution of number of bad oranges drawn. Hence find the mean and variance of the distribution.

Answer 1

Let X be the random variable denoting the number of bad oranges drawn.
P (getting a good orange) = \[\frac{20}{25} = \frac{4}{5}\]

P (getting a bad orange) = \[\frac{5}{25} = \frac{1}{5}\]

The probability distribution of X is given by

X	0	1	2	3	4
P(X)	\[\left( \frac{4}{5} \right)^4\] =\[\frac{256}{625}\]	\[^{4}{}{C}_1 \left( \frac{4}{5} \right)^3 \left( \frac{1}{5} \right)\] =\[\frac{256}{625}\]	\[^{4}{}{C}_2 \left( \frac{4}{5} \right)^2 \left( \frac{1}{5} \right)^2\] =\[\frac{96}{625}\]	\[^{4}{}{C}_3 \left( \frac{4}{5} \right) \left( \frac{1}{5} \right)^3\] =\[\frac{16}{625}\]	\[\left( \frac{1}{5} \right)^4\] =\[\frac{1}{625}\]

Mean of X is given by

\[\overline{X} = \sum P_i X_i\]

\[= 0 \times \frac{256}{625} + 1 \times \frac{256}{625} + 2 \times \frac{96}{625} + 3 \times \frac{16}{625} + 4 \times \frac{1}{625}\]

\[ = \frac{1}{625}\left( 256 + 192 + 48 + 4 \right)\]

\[ = \frac{4}{5}\]

Variance of X is given by \[\text{ Var } (X) = \sum P_i {X_i}^2 - \left( \sum P_i X_i \right)^2\]

\[= 0 \times \frac{256}{625} + 1 \times \frac{256}{625} + 4 \times \frac{96}{625} + 9 \times \frac{16}{625} + 16 \times \frac{1}{625} - \left( \frac{4}{5} \right)^2 \]

\[ = \frac{1}{625}\left( 256 + 384 + 144 + 16 \right) - \frac{16}{25}\]

\[ = \frac{800}{625} - \frac{16}{25}\]

\[ = \frac{400}{625}\]

\[ = \frac{16}{25}\]

Thus, the mean and vairance of the distribution are \[\frac{4}{5}\] and  \[\frac{16}{25}\] , respectively.