Q.If the p.m.f. of a r.v. X is $P(x) = \dfrac{c}{x^3}$, for $x = 1, 2, 3$; $= 0$, otherwise, then $E(X) =$ ________.

This is a Binomial$(4,\frac59)$ p.m.f.; $E(X)=np$.

The given p.m.f. $P(X=x)={}^4C_x\left(\frac59\right)^x\left(\frac49\right)^{4-x}$ is a Binomial$(n=4,p=\frac59)$ distribution …

Mean of a discrete uniform distribution on $\{1,\dots,15\}$ is $\dfrac{1+15}{2}$.

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First find $c$ from $\sum P(x)=1$, then compute $E(X)=\sum xP(x)$.

$\sum_{x=1}^3 P(x)=1$: $\ c\left(1+\dfrac18+\dfrac1{27}\right)=1$

$1+\dfrac18+\dfrac1{27} = \dfrac{216+27+8}{216} = \dfrac{251}{216} \implies c=\dfrac{216}{251}$

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$E(X) = \sum x_i P(x_i)$.

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Compute each probability from the p.m.f., then $E(X)=\sum x\,P(x)$.

$P(-1)=\frac{3-(-1)}{10}=\frac{4}{10}=0.4,\quad P(0)=\frac{3}{10}=0.3,\quad P(1)=\frac{2}{10}=0.2,\quad P(2)=\frac{1}{10}=0.1$

(Check: $0.4+0.3+0.2+0.1=1$ ✓, a valid p.m.f.)

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$X\sim B(3,\tfrac12)$; expected value $E(X)=np$.

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