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

As, X denote the larger of the two numbers obtained.

So, X can take the values 3, 4, 5, 6 and 7.

Now,

\[P\left( X = 3 \right) = P\left\{ \left( 2, 3 \right) \text{or} \left( 3, 2 \right) \right\} = 2 \times \frac{1}{6} \times \frac{1}{5} = \frac{1}{15}, \]
\[P\left( X = 4 \right) = P\left\{ \left( 2, 4 \right) \text{or} \left( 4, 2 \right) \text{or} \left( 3, 4 \right) \text{or} \left( 4, 3 \right) \right\} = 4 \times \frac{1}{6} \times \frac{1}{5} = \frac{2}{15}, \]
\[P\left( X = 5 \right) = P\left\{ \left( 2, 5 \right) \text{or} \left( 5, 2 \right) \text{or} \left( 3, 5 \right) \text{or} \left( 5, 3 \right) \text{ or } \left( 4, 5 \right) \text{ or }  \left( 5, 4 \right) \right\} = 6 \times \frac{1}{6} \times \frac{1}{5} = \frac{1}{5}, \]
\[P\left( X = 6 \right) = P\left\{ \left( 2, 6 \right) \text{ or} \left( 6, 2 \right) \text{ or } \left( 3, 6 \right) \text{ or }  \left( 6, 3 \right) \text{ or }  \left( 4, 6 \right) \text{ or }  \left( 6, 4 \right) \text{ or }  \left( 5, 6 \right) \text{ or }  \left( 6, 5 \right) \right\} = 8 \times \frac{1}{6} \times \frac{1}{5} = \frac{4}{15}, \]
\[P\left( X = 7 \right) = P\left\{ \left( 2, 7 \right) \text{ or}  \left( 7, 2 \right) \text{ or }  \left( 3, 7 \right) \text{ or }  \left( 7, 3 \right) \text{ or }  \left( 4, 7 \right) \text{ or }  \left( 7, 4 \right) \text{ or }  \left( 5, 7 \right) \text{ or }  \left( 7, 5 \right) or \left( 6, 7 \right) \text{ or }  \left( 7, 6 \right) \right\} = 10 \times \frac{1}{6} \times \frac{1}{5} = \frac{1}{3}\]

The probability distribution of X is as follows:

X:	3	4	5	6	7
P(X):	\[\frac{1}{15}\]	\[\frac{2}{15}\]	\[\frac{1}{5}\]	\[\frac{4}{15}\]	\[\frac{1}{3}\]

\[\text{ So, Mean } , E\left( X \right) = 3 \times \frac{1}{15} + 4 \times \frac{2}{15} + 5 \times \frac{1}{5} + 6 \times \frac{4}{15} + 7 \times \frac{1}{3}\]
\[ = \frac{3}{15} + \frac{8}{15} + \frac{15}{15} + \frac{24}{15} + \frac{35}{15}\]
\[ = \frac{85}{15}\]
\[ = \frac{17}{3}\]
\[\text{ Also, } E\left( X^2 \right) = 3^2 \times \frac{1}{15} + 4^2 \times \frac{2}{15} + 5^2 \times \frac{1}{5} + 6^2 \times \frac{4}{15} + 7^2 \times \frac{1}{3}\]
\[ = \frac{9}{15} + \frac{32}{15} + \frac{75}{15} + \frac{144}{15} + \frac{245}{15}\]
\[ = \frac{505}{15}\]
\[ = \frac{101}{3}\]
\[\text{ Var } \left( X \right) = E\left( X^2 \right) - \left[ E\left( X \right) \right]^2 \]
\[ = \frac{101}{3} - \left( \frac{17}{3} \right)^2 \]
\[ = \frac{101}{3} - \frac{289}{9}\]
\[ = \frac{303 - 289}{9}\]
\[ = \frac{14}{9}\]