Question

There are 4 cards numbered 1 to 4, one number on one card. Two cards are drawn at random without replacement. Let X denote the sum of the numbers on the two drawn cards. Find the mean and variance of X.

Answer 1

It is given that X denote the sum of the numbers on the two drawn cards. Clearly, X can take the values 3, 4, 5, 6 and 7.
P(X = 3) = Probability of getting 3 as sum
= P[(Getting 1 in the first draw and 2 in the second draw) Or (Getting 2 in the first draw and 1 in the second draw)] =

\[\frac{1}{4} \times \frac{1}{3} + \frac{1}{4} \times \frac{1}{3} = \frac{2}{12} = \frac{1}{6}\]

Similarly,

\[P\left( X = 4 \right) = \frac{1}{4} \times \frac{1}{3} + \frac{1}{4} \times \frac{1}{3} = \frac{2}{12} = \frac{1}{6}\]

\[P\left( X = 5 \right) = \frac{1}{4} \times \frac{1}{3} + \frac{1}{4} \times \frac{1}{3} + \frac{1}{4} \times \frac{1}{3} + \frac{1}{4} \times \frac{1}{3} = \frac{4}{12} = \frac{2}{6}\]

\[P\left( X = 6 \right) = \frac{1}{4} \times \frac{1}{3} + \frac{1}{4} \times \frac{1}{3} = \frac{2}{12} = \frac{1}{6}\]

\[P\left( X = 7 \right) = \frac{1}{4} \times \frac{1}{3} + \frac{1}{4} \times \frac{1}{3} = \frac{2}{12} = \frac{1}{6}\]

Thus, the probability distribution of X is given below:

x	3	4	5	6	7
P(X	\[\frac{1}{6}\]	\[\frac{1}{6}\]	\[\frac{2}{6}\]	\[\frac{1}{6}\]	\[\frac{1}{6}\]

Now,

\[\sum^{}_{} P_i x_i = \frac{1}{6} \times 3 + \frac{1}{6} \times 4 + \frac{2}{6} \times 5 + \frac{1}{6} \times 6 + \frac{1}{6} \times 7\]

\[ = \frac{1}{6} \times \left( 3 + 4 + 10 + 6 + 7 \right)\]

\[ = \frac{1}{6} \times 30\]

\[ = 5\]

\[\sum^{}_{} P_i {x_i}^2 = \frac{1}{6} \times 9 + \frac{1}{6} \times 16 + \frac{2}{6} \times 25 + \frac{1}{6} \times 36 + \frac{1}{6} \times 49\]

\[ = \frac{1}{6} \times \left( 9 + 16 + 50 + 36 + 49 \right)\]

\[ = \frac{1}{6} \times 160\]

\[ = \frac{80}{3}\]

\[\therefore \text { Mean }\left[ E\left( X \right) \right] = \sum^{}_{} P_i x_i = 5\]

\[\text { Variance } = \sum P_i {x_i}^2 - \left( \sum^{}_{} P_i x_i \right)^2 = \frac{80}{3} - 25 = \frac{5}{3}\]