Question

Two numbers are selected at random (without replacement) from positive integers 2, 3, 4, 5, 6 and 7. Let X denote the larger of the two numbers obtained. Find the mean and variance of the probability distribution of X.

Answer 1

The random variable X, i.e., the larger of the two selected numbers, takes the values 3, 4, 5, 6 or 7.

\[P\left( X = 3 \right) = P\left\{ \left( 2, 3 \right), \left( 3, 2 \right) \right\} = 2 \times \frac{1}{6} \times \frac{1}{5} = \frac{2}{30}\]

\[P\left( X = 4 \right) = P\left\{ \left( 2, 4 \right), \left( 3, 4 \right), \left( 4, 2 \right), \left( 4, 3 \right) \right\} = 4 \times \frac{1}{6} \times \frac{1}{5} = \frac{4}{30}\]

\[P\left( X = 5 \right) = P\left\{ \left( 2, 5 \right), \left( 3, 5 \right), \left( 4, 5 \right), \left( 5, 2 \right), \left( 5, 3 \right), \left( 5, 4 \right) \right\} = 6 \times \frac{1}{6} \times \frac{1}{5} = \frac{6}{30}\]

\[P\left( X = 6 \right) = P\left\{ \left( 2, 6 \right), \left( 3, 6 \right), \left( 4, 6 \right), \left( 5, 6 \right), \left( 6, 2 \right), \left( 6, 3 \right), \left( 6, 4 \right), \left( 6, 5 \right) \right\} = 8 \times \frac{1}{6} \times \frac{1}{5} = \frac{8}{30}\]

\[P\left( X = 7 \right) = P\left\{ \left( 2, 7 \right), \left( 3, 7 \right), \left( 4, 7 \right), \left( 5, 7 \right), \left( 6, 7 \right), \left( 7, 2 \right), \left( 7, 3 \right), \left( 7, 4 \right), \left( 7, 5 \right), \left( 7, 6 \right) \right\} = 10 \times \frac{1}{6} \times \frac{1}{5} = \frac{10}{30}\]

Thus, the probability distribution of X is

Answer 2

The random variable X, i.e., the larger of the two selected numbers, takes the values 3, 4, 5, 6 or 7.

\[P\left( X = 3 \right) = P\left\{ \left( 2, 3 \right), \left( 3, 2 \right) \right\} = 2 \times \frac{1}{6} \times \frac{1}{5} = \frac{2}{30}\]

\[P\left( X = 4 \right) = P\left\{ \left( 2, 4 \right), \left( 3, 4 \right), \left( 4, 2 \right), \left( 4, 3 \right) \right\} = 4 \times \frac{1}{6} \times \frac{1}{5} = \frac{4}{30}\]

\[P\left( X = 5 \right) = P\left\{ \left( 2, 5 \right), \left( 3, 5 \right), \left( 4, 5 \right), \left( 5, 2 \right), \left( 5, 3 \right), \left( 5, 4 \right) \right\} = 6 \times \frac{1}{6} \times \frac{1}{5} = \frac{6}{30}\]

\[P\left( X = 6 \right) = P\left\{ \left( 2, 6 \right), \left( 3, 6 \right), \left( 4, 6 \right), \left( 5, 6 \right), \left( 6, 2 \right), \left( 6, 3 \right), \left( 6, 4 \right), \left( 6, 5 \right) \right\} = 8 \times \frac{1}{6} \times \frac{1}{5} = \frac{8}{30}\]

\[P\left( X = 7 \right) = P\left\{ \left( 2, 7 \right), \left( 3, 7 \right), \left( 4, 7 \right), \left( 5, 7 \right), \left( 6, 7 \right), \left( 7, 2 \right), \left( 7, 3 \right), \left( 7, 4 \right), \left( 7, 5 \right), \left( 7, 6 \right) \right\} = 10 \times \frac{1}{6} \times \frac{1}{5} = \frac{10}{30}\]

Thus, the probability distribution of X is

X	3	4	5	6
P(X)	`2/30`	`4/30`	`6/30`	`8/30`

\[\therefore \text { Mean }, E(X) = \Sigma x_i p_i = 3 \times \frac{2}{30} + 4 \times \frac{4}{30} + 5 \times \frac{6}{30} + 6 \times \frac{8}{30} + 7 \times \frac{10}{30}\]

\[ = \frac{6}{30} + \frac{16}{30} + \frac{30}{30} + \frac{48}{30} + \frac{70}{30}\]

\[ = \frac{170}{30} = \frac{17}{3}\]

\[E( X^2 ) = \Sigma x_i^2 p_i = 3^2 \times \frac{2}{30} + 4^2 \times \frac{4}{30} + 5^2 \times \frac{6}{30} + 6^2 \times \frac{8}{30} + 7^2 \times \frac{10}{30}\]

\[ = \frac{18}{30} + \frac{64}{30} + \frac{150}{30} + \frac{288}{30} + \frac{490}{30}\]

\[ = \frac{1010}{30} = \frac{101}{3}\]

Thus,

\[\text { Var }\left( X \right) = E\left( X^2 \right) - \left[ E\left( X \right) \right]^2\]

\[= \frac{101}{3} - \left( \frac{17}{3} \right)^2 \]

\[ = \frac{303 - 289}{9} = \frac{14}{9}\]