Question

An urn contains 4 red and 3 blue balls. Find the probability distribution of the number of blue balls in a random draw of 3 balls with replacement.

Answer 1

Let X denote the number of blue balls in a sample of 3 balls drawn from a bag containing 4 red and 3 blue balls. Then, X can take values 0, 1, 2 and 3.
Now,

\[P\left( X = 0 \right) = P\left( \text{ no blue ball } \right) = \frac{4}{7} \times \frac{4}{7} \times \frac{4}{7} = \frac{64}{343}\]

\[P\left( X = 1 \right) = P\left( 1 \text{ blue ball }  \right) = \left( \frac{3}{7} \times \frac{4}{7} \times \frac{4}{7} \right) + \left( \frac{4}{7} \times \frac{3}{7} \times \frac{4}{7} \right) + \left( \frac{4}{7} \times \frac{4}{7} \times \frac{3}{7} \right) = \frac{144}{343}\]

\[P\left( X = 2 \right) = P\left( 2 \text{ blue balls } \right) = \left( \frac{3}{7} \times \frac{3}{7} \times \frac{4}{7} \right) + \left( \frac{4}{7} \times \frac{3}{7} \times \frac{3}{7} \right) + \left( \frac{3}{7} \times \frac{4}{7} \times \frac{3}{7} \right) = \frac{108}{343}\]

\[P\left( X = 3 \right) = P\left( 3 \text{ blue balls } \right) = \frac{3}{7} \times \frac{3}{7} \times \frac{3}{7} = \frac{27}{343}\]

Thus, the probability distribution of X is given by

X	P(X)
0	\[\frac{64}{343}\]
1	\[\frac{144}{343}\]
2	\[\frac{108}{343}\]
3	\[\frac{27}{343}\]