Question

An urn contains 5 red and 2 black balls. Two balls are randomly drawn, without replacement. Let X represent the number of black balls drawn. What are the possible values of X ? Is X a random variable ? If yes, then find the mean and variance of X.      

Answer 1

As, X represent the number of black balls drawn.
So, it can take values 0, 1 and 2. Yes, X is a random variable.
Now,

\[P\left( X = 0 \right) = P\left( RR \right) = \frac{5}{7} \times \frac{4}{6} = \frac{10}{21}, \]
\[P\left( X = 1 \right) = P\left( RB \text{ or }  BR \right) = 2 \times \frac{5}{7} \times \frac{2}{6} = \frac{10}{21}, \]
\[P\left( X = 2 \right) = P\left( BB \right) = \frac{2}{7} \times \frac{1}{6} = \frac{1}{21}\]
\[\text{ Mean }  = \sum p_i x_i = 0 \times \frac{10}{21} + 1 \times \frac{10}{21} + 2 \times \frac{1}{21}\]
\[ = \frac{10}{21} + \frac{2}{21}\]
\[ = \frac{12}{21}\]
\[ = \frac{4}{7}\]
\[\text{ Also } , \sum p_i {x_i}^2 = 0^2 \times \frac{10}{21} + 1^2 \times \frac{10}{21} + 2^2 \times \frac{1}{21}\]
\[ = \frac{10}{21} + \frac{4}{21}\]
\[ = \frac{14}{21}\]
\[ = \frac{2}{3}\]
\[\text{ So, variance } = \sum p_i {x_i}^2 - \left( \text{ Mean } \right)^2 \]
\[ = \frac{2}{3} - \left( \frac{4}{7} \right)^2 \]
\[ = \frac{2}{3} - \frac{16}{49}\]
\[ = \frac{98 - 48}{147}\]
\[ = \frac{50}{147}\]