Two numbers are selected at random (without replacement) from the first five positive integers. Let X denote the larger of the two numbers obtained. Find the mean and variance of X.

Solution

X can take values as 2, 3, 4,5 such that,

P (X =2) = probability that the larger of two number 2.
= prob. of getting 1 in first selection and 2 in second selection getting 2 in first selection and 1 in second selection.

$∴ P (X = 2) = \frac{1}{5} x \frac{1}{4} + \frac{1}{5} x \frac{1}{4} = \frac{2}{20} similarly, ∴ P (X = 3) = \frac{2}{5} x \frac{1}{4} + \frac{1}{5} x \frac{2}{4} = \frac{4}{20} ∴ P (X = 4) = \frac{3}{5} x \frac{1}{4} + \frac{1}{5} x \frac{3}{4} = \frac{6}{20} ∴ P (X = 5) = \frac{4}{5} x \frac{1}{4} + \frac{1}{5} x \frac{4}{4} = \frac{8}{20}$

x	2	3	4	5
P(X)	$\frac{2}{20}$	$\frac{4}{20}$	$\frac{6}{20}$	$\frac{8}{20}$

$E (X) = 2 x \frac{2}{20} + 3 x \frac{4}{40} + 4 x \frac{6}{20} + 5 x \frac{8}{20} = \frac{80}{20} = 4 E (X^{2}) = 4 x \frac{2}{20} + 9 x \frac{4}{20} + 16 x \frac{6}{20} + 25 x \frac{8}{20} = \frac{340}{20} = 17 V (X) = E (X^{2}) - (E (X))^{2} = 17 - 16 = 1$

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Probability

Two numbers are selected at random (without replacement) from the first five positive integers. Let X denote the larger of the two numbers obtained. Find the mean and variance of X.

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