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.

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Solution

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,
$P (X = 0) = P (no red ball) = \frac{^{6} C_{3}}{^{10} C_{3}} = \frac{20}{120} = \frac{1}{6} P (X = 1) = P (1 red ball) = \frac{^{4} C_{1} \times^{6} C_{2}}{^{10} C_{3}} = \frac{60}{120} = \frac{1}{2} P (X = 2) = P (2 red balls) = \frac{^{4} C_{2} \times^{6} C_{1}}{^{10} C_{3}} = \frac{36}{120} = \frac{3}{10} P (X = 3) = P (3 red balls) = \frac{^{4} C_{3}}{^{10} C_{3}} = \frac{4}{120} = \frac{1}{30}$

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 $\frac{1}{30}$

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