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,
$$\begin{gathered} P\left(X=0\right)=P\left(\mathrm{no}\mathrm{blue}\mathrm{ball}\right)=\frac{4}{7}\times \frac{4}{7}\times \frac{4}{7}=\frac{64}{343} \\ P\left(X=1\right)=P\left(1\mathrm{blue}\mathrm{ball}\right)=\left(\frac{3}{7}\times \frac{4}{7}\times \frac{4}{7}\right)+\left(\frac{4}{7}\times \frac{3}{7}\times \frac{4}{7}\right)+\left(\frac{4}{7}\times \frac{4}{7}\times \frac{3}{7}\right)=\frac{144}{343} \\ P\left(X=2\right)=P\left(2\mathrm{blue}\mathrm{balls}\right)=\left(\frac{3}{7}\times \frac{3}{7}\times \frac{4}{7}\right)+\left(\frac{4}{7}\times \frac{3}{7}\times \frac{3}{7}\right)+\left(\frac{3}{7}\times \frac{4}{7}\times \frac{3}{7}\right)=\frac{108}{343} \\ P\left(X=3\right)=P\left(3\mathrm{blue}\mathrm{balls}\right)=\frac{3}{7}\times \frac{3}{7}\times \frac{3}{7}=\frac{27}{343} \end{gathered}$$

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	$\frac{27}{343}$