If the probability mass function (p.m.f.) of | vedclass

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(D) The sum of probabilities must be equal to $1$: $\sum P(X=x) = P(X=1) + P(X=2) + P(X=3) = 1$ $\frac{c}{1^3} + \frac{c}{2^3} + \frac{c}{3^3} = 1$ $c

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$\frac{294}{251}$

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(D) The sum of probabilities must be equal to $1$: $\sum P(X=x) = P(X=1) + P(X=2) + P(X=3) = 1$ $\frac{c}{1^3} + \frac{c}{2^3} + \frac{c}{3^3} = 1$ $c \left( 1 + \frac{1}{8} + \frac{1}{27} ight) = 1$ $c \left( \frac{216 + 27 + 8}{216} ight) = 1$ $c \left( \frac{251}{216} ight) = 1 \implies c = \frac{216}{251}$ Now,the expected value $E(X) = \sum x \cdot P(X=x)$: $E(X) = 1 \cdot \frac{c}{1^3} + 2 \cdot \frac{c}{2^3} + 3 \cdot \frac{c}{3^3}$ $E(X) = c \left( 1 + \frac{2}{8} + \frac{3}{27} ight) = c \left( 1 + \frac{1}{4} + \frac{1}{9} ight)$ $E(X) = c \left( \frac{36 + 9 + 4}{36} ight) = c \left( \frac{49}{36} ight)$ Substituting $c = \frac{216}{251}$: $E(X) = \frac{216}{251} 	imes \frac{49}{36} = 6 	imes \frac{49}{251} = \frac{294}{251}$

$X$	$0$	$1$	$2$
$P(X)$	$0.4$	$0.4$	$0.2$

$X$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X=x)$	$0.15$	$0.23$	$0.12$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

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

If the probability mass function (p.m.f.) of a discrete random variable $X$ is given by $P(X=x) = \frac{c}{x^3}$ for $x = 1, 2, 3$ and $0$ otherwise,then $E(X)$ is equal to:

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