A bag contains 4 white and 6 black balls. Three balls are drawn at random from the bag. Let X be the number of white balls, among the drawn balls. If $$\sigma^2$$ is the variance of X, then $$100\sigma^2$$ is equal to

We have a bag with 4 white and 6 black balls. Three balls are drawn at random, and $$X$$ is the number of white balls drawn. We need $$100\sigma^2$$ where $$\sigma^2$$ is the variance of $$X$$.

$$X$$ follows a hypergeometric distribution. The total number of ways to draw 3 balls from 10 is $$\binom{10}{3} = 120$$.

$$P(X = 0) = \frac{\binom{4}{0}\binom{6}{3}}{\binom{10}{3}} = \frac{1 \cdot 20}{120} = \frac{20}{120} = \frac{1}{6}$$

$$P(X = 1) = \frac{\binom{4}{1}\binom{6}{2}}{\binom{10}{3}} = \frac{4 \cdot 15}{120} = \frac{60}{120} = \frac{1}{2}$$

$$P(X = 2) = \frac{\binom{4}{2}\binom{6}{1}}{\binom{10}{3}} = \frac{6 \cdot 6}{120} = \frac{36}{120} = \frac{3}{10}$$

$$P(X = 3) = \frac{\binom{4}{3}\binom{6}{0}}{\binom{10}{3}} = \frac{4 \cdot 1}{120} = \frac{4}{120} = \frac{1}{30}$$

Now we compute $$E(X)$$:

$$E(X) = 0 \cdot \frac{1}{6} + 1 \cdot \frac{1}{2} + 2 \cdot \frac{3}{10} + 3 \cdot \frac{1}{30}$$

$$= 0 + \frac{1}{2} + \frac{3}{5} + \frac{1}{10} = \frac{5 + 6 + 1}{10} = \frac{12}{10} = \frac{6}{5}$$

Now $$E(X^2)$$:

$$E(X^2) = 0 + 1 \cdot \frac{1}{2} + 4 \cdot \frac{3}{10} + 9 \cdot \frac{1}{30}$$

$$= \frac{1}{2} + \frac{12}{10} + \frac{9}{30} = \frac{1}{2} + \frac{6}{5} + \frac{3}{10} = \frac{5 + 12 + 3}{10} = \frac{20}{10} = 2$$

Therefore: $$\sigma^2 = E(X^2) - [E(X)]^2 = 2 - \left(\frac{6}{5}\right)^2 = 2 - \frac{36}{25} = \frac{50 - 36}{25} = \frac{14}{25}$$

So $$100\sigma^2 = 100 \cdot \frac{14}{25} = \frac{1400}{25} = 56$$.

Hence, the correct answer is 56.