Question

Three cards are drawn at random (without replacement) from a well shuffled pack of 52 cards. Find the probability distribution of number of red cards. Hence, find the mean of the distribution .  

Answer 1

Let X denotes the number of red cards drawn.
Then, X can take the values 0, 1, 2 or 3.
Now,

\[P\left( X = 0 \right) = P\left( BBB \right) = \frac{26}{52} \times \frac{25}{51} \times \frac{24}{50} = \frac{2}{17}, \]
\[P\left( X = 1 \right) = P\left( RBB \text{ or }BRB \text{ or } BBR \right) = 3 \times \frac{26}{52} \times \frac{26}{51} \times \frac{25}{50} = \frac{13}{34}, \]
\[P\left( X = 2 \right) = P\left( RRB \text{ or } RBR \text{ or } BRR \right) = 3 \times \frac{26}{52} \times \frac{25}{51} \times \frac{26}{50} = \frac{13}{34}, \]
\[P\left( X = 3 \right) = P\left( RRR \right) = \frac{26}{52} \times \frac{25}{51} \times \frac{24}{50} = \frac{2}{17}\]

Thus, the probability distribution of X is given by

X	P(X)
0	\[\frac{2}{17}\]
1	\[\frac{13}{34}\]
2	\[\frac{13}{34}\]
3	\[\frac{2}{17}\]

\[\text{ Mean } = \sum p_i x_i = 0 \times \frac{2}{17} + 1 \times \frac{13}{34} + 2 \times \frac{13}{34} + 3 \times \frac{2}{17}\]
\[ = 0 + \frac{13}{34} + \frac{26}{34} + \frac{6}{17}\]
\[ = \frac{51}{34}\]
\[ = \frac{3}{2}\]
\[ = 1 . 5\]