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प्रश्न
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.
उत्तर १
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
उत्तर २
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,
\[\text { Var }\left( X \right) = E\left( X^2 \right) - \left[ E\left( X \right) \right]^2\]
\[= \frac{101}{3} - \left( \frac{17}{3} \right)^2 \]
\[ = \frac{303 - 289}{9} = \frac{14}{9}\]
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