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If the p.m.f. of a discrete random variable $X$ is $P(X=x)=\frac{c}{x^3}$, $x=1,2,3=0$, otherwise then $E(X)$ is equal to
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$\frac{294}{251}$
$\begin{aligned} & \because \sum P_i=1 \\ & \Rightarrow \frac{251}{216} C \\ & \Rightarrow C=\frac{216}{251} \\ & \text { now } E(x)=\sum p_i x_i=\frac{49}{36} C=\frac{49}{36} \times \frac{216}{251}=\frac{294}{251}\end{aligned}$
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