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(D) The two positive integers can be selected from the first six positive integers without replacement in $6 	imes 5 = 30$ ways.$X$ represents the la

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$14/3$

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(D) The two positive integers can be selected from the first six positive integers without replacement in $6 	imes 5 = 30$ ways.$X$ represents the larger of the two numbers obtained. Therefore,$X$ can take the values $2, 3, 4, 5,$ or $6$.For $X=2$,the possible pairs are $(1, 2)$ and $(2, 1)$. Thus,$P(X=2) = \frac{2}{30} = \frac{1}{15}$.For $X=3$,the possible pairs are $(1, 3), (2, 3), (3, 1),$ and $(3, 2)$. Thus,$P(X=3) = \frac{4}{30} = \frac{2}{15}$.For $X=4$,the possible pairs are $(1, 4), (2, 4), (3, 4), (4, 3), (4, 2),$ and $(4, 1)$. Thus,$P(X=4) = \frac{6}{30} = \frac{1}{5} = \frac{3}{15}$.For $X=5$,the possible pairs are $(1, 5), (2, 5), (3, 5), (4, 5), (5, 4), (5, 3), (5, 2),$ and $(5, 1)$. Thus,$P(X=5) = \frac{8}{30} = \frac{4}{15}$.For $X=6$,the possible pairs are $(1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 5), (6, 4), (6, 3), (6, 2),$ and $(6, 1)$. Thus,$P(X=6) = \frac{10}{30} = \frac{1}{3} = \frac{5}{15}$.The probability distribution is:$X$$2$$3$$4$$5$$6$$P(X)$$1/15$$2/15$$3/15$$4/15$$5/15$$E(X) = \sum X_i P(X_i) = 2(\frac{1}{15}) + 3(\frac{2}{15}) + 4(\frac{3}{15}) + 5(\frac{4}{15}) + 6(\frac{5}{15})$$E(X) = \frac{2 + 6 + 12 + 20 + 30}{15} = \frac{70}{15} = \frac{14}{3}$.

$X = x$	$0$	$1$	$2$	$3$	$4$
$P(X = x)$	$k$	$2k$	$4k$	$2k$	$k$

Two numbers are selected at random (without replacement) from the first six positive integers. Let $X$ denote the larger of the two numbers obtained. Find $E(X)$.

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