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,

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

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

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

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

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	$$\frac{27}{343}$$