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. [NCERT EXEMPLAR]

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Solution

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,

$P (X = 0) = P (All white balls) = P (W W W W) = \frac{4}{12} \times \frac{3}{11} \times \frac{2}{10} \times \frac{1}{9} = \frac{1}{495}, P (X = 1) = P (One red balls and three white balls) = P (R W W W) + P (W R W W) + P (W W R W) + P (W W W R) = \frac{8}{12} \times \frac{4}{11} \times \frac{3}{10} \times \frac{2}{9} + \frac{4}{12} \times \frac{8}{11} \times \frac{3}{10} \times \frac{2}{9} + \frac{4}{12} \times \frac{3}{11} \times \frac{8}{10} \times \frac{2}{9} + \frac{4}{12} \times \frac{3}{11} \times \frac{2}{10} \times \frac{8}{9} = 4 \times \frac{8}{495} = \frac{32}{495}, P (X = 2) = P (Two red balls and two white balls) = P (R R W W) + P (R W R W) + P (R W W R) + P (W W R R) + P (W R W R) + P (W R R W) = \frac{8}{12} \times \frac{7}{11} \times \frac{4}{10} \times \frac{3}{9} + \frac{8}{12} \times \frac{4}{11} \times \frac{7}{10} \times \frac{3}{9} + \frac{8}{12} \times \frac{4}{11} \times \frac{3}{10} \times \frac{7}{9} + \frac{4}{12} \times \frac{3}{11} \times \frac{8}{10} \times \frac{7}{9} + \frac{4}{12} \times \frac{8}{11} \times \frac{3}{10} \times \frac{7}{9} + \frac{4}{12} \times \frac{8}{11} \times \frac{7}{10} \times \frac{3}{9} = 6 \times \frac{28}{495} = \frac{168}{495}, P (X = 3) = P (Three red balls and one white ball) = P (R R R W) + P (R R W R) + P (R W R R) + P (W R R R) = \frac{8}{12} \times \frac{7}{11} \times \frac{6}{10} \times \frac{4}{9} + \frac{8}{12} \times \frac{7}{11} \times \frac{4}{10} \times \frac{6}{9} + \frac{8}{12} \times \frac{4}{11} \times \frac{7}{10} \times \frac{6}{9} + \frac{4}{12} \times \frac{8}{11} \times \frac{7}{10} \times \frac{6}{9} = 4 \times \frac{56}{495} = \frac{224}{495}, P (X = 4) = P (All red balls) = P (R R R R) = \frac{8}{12} \times \frac{7}{11} \times \frac{6}{10} \times \frac{5}{9} = \frac{70}{495}$

So, the probability distribution of X is as follows:

X 0 1 2 3 4

P(X) $\frac{1}{495}$ $\frac{32}{495}$ $\frac{168}{495}$ $\frac{224}{495}$ $\frac{70}{495}$

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