Question

Find the probability distribution of the number of white balls drawn in a random draw of 3 balls without replacement, from a bag containing 4 white and 6 red balls.

Answer 1

Let X denote the number of white balls in a sample of 3 balls drawn from a bag containing 4 white and 6 red balls. Then, X can take the values 0, 1, 2 and 3.
Now,
$$\begin{gathered} P\left(X=0\right) \\ =P\left(\mathrm{no}\mathrm{white}\mathrm{ball}\right) \\ =\frac{{}^6{C}_3}{{}^{10}{C}_3} \\ =\frac{20}{120} \\ =\frac{1}{6} \\ P\left(X=1\right) \\ =P\left(1\mathrm{white}\mathrm{ball}\right) \\ =\frac{{}^4{C}_1{\times }^6{C}_2}{{}^{10}{C}_3} \\ =\frac{60}{120} \\ =\frac{1}{2} \\ P\left(X=2\right) \\ =P\left(2\mathrm{white}\mathrm{balls}\right) \\ =\frac{{}^4{C}_2{\times }^6{C}_1}{{}^{10}{C}_3} \\ =\frac{36}{120} \\ =\frac{3}{10} \\ P\left(X=3\right) \\ =P\left(3\mathrm{white}\mathrm{balls}\right) \\ =\frac{{}^4{C}_3}{{}^{10}{C}_3} \\ =\frac{4}{120} \\ =\frac{1}{30} \end{gathered}$$

Thus, the probability distribution of X is given by

X	P(X)
0	$\frac{1}{6}$
1	$\frac{1}{2}$
2	$\frac{3}{10}$
3	$\frac{1}{30}$