Question

A bag contains 4 red and 6 black balls. Three balls are drawn at random. Find the probability distribution of the number of red balls.

Answer 1

Let X denote the number of red balls in a sample of 3 balls drawn from a bag containing 4 red and 6 black balls. Then, X can take the values 0, 1, 2 and 3.
Now,

\[P\left( X = 0 \right)\]
\[ = P\left( \text{ no red ball } \right)\]
\[ = \frac{{}^6 C_3}{{}^{10} C_3}\]
\[ = \frac{20}{120}\]
\[ = \frac{1}{6}\]
\[P\left( X = 1 \right)\]
\[ = P\left( 1 \text{ red ball }  \right)\]
\[ = \frac{{}^4 C_1 \times^6 C_2}{{}^{10} C_3}\]
\[ = \frac{60}{120}\]
\[ = \frac{1}{2}\]
\[P\left( X = 2 \right)\]
\[ = P\left( 2 \text{ red balls }  \right)\]
\[ = \frac{{}^4 C_2 \times^6 C_1}{{}^{10} C_3}\]
\[ = \frac{36}{120}\]
\[ = \frac{3}{10}\]
\[P\left( X = 3 \right)\]
\[ = P\left( 3 \text { red balls } \right)\]
\[ = \frac{{}^4 C_3}{{}^{10} C_3}\]
\[ = \frac{4}{120}\]
\[ = \frac{1}{30}\]

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