Question

Four balls are to be drawn without replacement from a box containing 8 red and 4 white balls. If X denotes the number of red balls drawn, then find the probability distribution of X.                         

Answer 1

As, four balls are to be drawn without replacement and X denotes the number of red balls drawn
So, X is a random variable that can take values 0, 1, 2, 3 or 4
Now,

\[P\left( X = 0 \right) = P\left( \text{ All white balls }  \right) = P\left( WWWW \right) = \frac{4}{12} \times \frac{3}{11} \times \frac{2}{10} \times \frac{1}{9} = \frac{1}{495}, \]
\[P\left( X = 1 \right) = P\left( \text{ One red balls and three white balls }  \right) = P\left( RWWW \right) + P\left( WRWW \right) + P\left( WWRW \right) + P\left( WWWR \right)\]
\[ = \frac{8}{12} \times \frac{4}{11} \times \frac{3}{10} \times \frac{2}{9} + \frac{4}{12} \times \frac{8}{11} \times \frac{3}{10} \times \frac{2}{9} + \frac{4}{12} \times \frac{3}{11} \times \frac{8}{10} \times \frac{2}{9} + \frac{4}{12} \times \frac{3}{11} \times \frac{2}{10} \times \frac{8}{9}\]
\[ = 4 \times \frac{8}{495} = \frac{32}{495}, \]
\[P\left( X = 2 \right) = P\left( \text{ Two red balls and two white balls }  \right) = P\left( RRWW \right) + P\left( RWRW \right) + P\left( RWWR \right) + P\left( WWRR \right) + P\left( WRWR \right) + P\left( WRRW \right)\]
\[ = \frac{8}{12} \times \frac{7}{11} \times \frac{4}{10} \times \frac{3}{9} + \frac{8}{12} \times \frac{4}{11} \times \frac{7}{10} \times \frac{3}{9} + \frac{8}{12} \times \frac{4}{11} \times \frac{3}{10} \times \frac{7}{9} + \frac{4}{12} \times \frac{3}{11} \times \frac{8}{10} \times \frac{7}{9} + \frac{4}{12} \times \frac{8}{11} \times \frac{3}{10} \times \frac{7}{9} + \frac{4}{12} \times \frac{8}{11} \times \frac{7}{10} \times \frac{3}{9}\]
\[ = 6 \times \frac{28}{495} = \frac{168}{495}, \]
\[P\left( X = 3 \right) = P\left( \text{ Three red balls and one white ball } \right) = P\left( RRRW \right) + P\left( RRWR \right) + P\left( RWRR \right) + P\left( WRRR \right)\]
\[ = \frac{8}{12} \times \frac{7}{11} \times \frac{6}{10} \times \frac{4}{9} + \frac{8}{12} \times \frac{7}{11} \times \frac{4}{10} \times \frac{6}{9} + \frac{8}{12} \times \frac{4}{11} \times \frac{7}{10} \times \frac{6}{9} + \frac{4}{12} \times \frac{8}{11} \times \frac{7}{10} \times \frac{6}{9}\]
\[ = 4 \times \frac{56}{495} = \frac{224}{495}, \]
\[P\left( X = 4 \right) = P\left( \text{ All red balls } \right) = P\left( RRRR \right) = \frac{8}{12} \times \frac{7}{11} \times \frac{6}{10} \times \frac{5}{9}\]
\[ = \frac{70}{495}\]

So, the probability distribution of X is as follows:

x	0	1	2	3	4
P(X)	\[\frac{1}{495}\]	\[\frac{32}{495}\]	\[\frac{168}{495}\]	\[\frac{224}{495}\]	\[\frac{70}{495}\]