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,

\[P\left( X = 0 \right)\]

\[ = P\left( \text{ no white ball } \right)\]

\[ = \frac{{}^6 C_3}{{}^{10} C_3}\]

\[ = \frac{20}{120}\]

\[ = \frac{1}{6}\]

\[P\left( X = 1 \right)\]

\[ = P\left( 1 \text{ white ball } \right)\]

\[ = \frac{{}^4 C_1 \times^6 C_2}{{}^{10} C_3}\]

\[ = \frac{60}{120}\]

\[ = \frac{1}{2}\]

\[P\left( X = 2 \right)\]

\[ = P\left( 2 \text{ white balls } \right)\]

\[ = \frac{{}^4 C_2 \times^6 C_1}{{}^{10} C_3}\]

\[ = \frac{36}{120}\]

\[ = \frac{3}{10}\]

\[P\left( X = 3 \right)\]

\[ = P\left( 3 \text{ white balls } \right)\]

\[ = \frac{{}^4 C_3}{{}^{10} C_3}\]

\[ = \frac{4}{120}\]

\[ = \frac{1}{30}\]

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	\[\frac{1}{30}\]