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(C) The total number of ways to select $2$ numbers from the first $6$ positive integers is $^6C_2 = \frac{6 	imes 5}{2} = 15$.Let $X$ be the random v

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$\frac{14}{9}$

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(C) The total number of ways to select $2$ numbers from the first $6$ positive integers is $^6C_2 = \frac{6 	imes 5}{2} = 15$.Let $X$ be the random variable representing the larger of the two numbers. The possible values for $X$ are $2, 3, 4, 5, 6$.The probability distribution is as follows:For $X=2$: Pairs are $(1,2)$,so $P(X=2) = \frac{1}{15}$.For $X=3$: Pairs are $(1,3), (2,3)$,so $P(X=3) = \frac{2}{15}$.For $X=4$: Pairs are $(1,4), (2,4), (3,4)$,so $P(X=4) = \frac{3}{15}$.For $X=5$: Pairs are $(1,5), (2,5), (3,5), (4,5)$,so $P(X=5) = \frac{4}{15}$.For $X=6$: Pairs are $(1,6), (2,6), (3,6), (4,6), (5,6)$,so $P(X=6) = \frac{5}{15}$.Calculating $E(X) = \sum x_i P_i = 2(\frac{1}{15}) + 3(\frac{2}{15}) + 4(\frac{3}{15}) + 5(\frac{4}{15}) + 6(\frac{5}{15}) = \frac{2+6+12+20+30}{15} = \frac{70}{15} = \frac{14}{3}$.Calculating $E(X^2) = \sum x_i^2 P_i = 4(\frac{1}{15}) + 9(\frac{2}{15}) + 16(\frac{3}{15}) + 25(\frac{4}{15}) + 36(\frac{5}{15}) = \frac{4+18+48+100+180}{15} = \frac{350}{15} = \frac{70}{3}$.$\operatorname{Var}(X) = E(X^2) - [E(X)]^2 = \frac{70}{3} - (\frac{14}{3})^2 = \frac{70}{3} - \frac{196}{9} = \frac{210-196}{9} = \frac{14}{9}$.

Outcome	$\omega_{1}$	$\omega_{2}$	$\omega_{3}$	$\omega_{4}$	$\omega_{5}$	$\omega_{6}$	$\omega_{7}$
Probability	$-0.1$	$0.2$	$0.3$	$0.4$	$-0.2$	$0.1$	$0.3$

Two numbers are selected at random from the first six positive integers. If $X$ denotes the larger of the two numbers,then $\operatorname{Var}(X) = $

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