Question

Let the p.m.f. (probability mass function) of random variable $x$ be
$P\left(x\right)=\{\begin{array}{c}\left(\frac{4}{x}\right){\left(\frac{5}{9}\right)}^x{\left(\frac{4}{x}\right)}^{4-x},x=0,1,2,3,4\\ 0,\text{otherwise}\end{array}$
Find $E\left(x\right)$ and Var$\left(x\right)$.

Answer 1

$P\left(x\right)=\left(\frac{4}{x}\right){\left(\frac{5}{9}\right)}^x{\left(\frac{4}{x}\right)}^{4-x}$, $x=0,1,2,3,4$.
$=0$ otherwise
The probability distribution table of the function is

$x$	$0$	$1$	$2$	$3$	$4$
$P\left(x\right)$	$\infty$	$\frac{1280}{9}$	$\frac{200}{81}$	$\frac{2000}{6561}$	$\frac{625}{6561}$
$x^2$	$0$	$1$	$4$	$9$	$16$
$P\left(x\right).x^2$	$0$	$\frac{1280}{9}$	$\frac{800}{81}$	$\frac{18000}{6561}$	$\frac{10000}{6561}$

$E\left(x\right)=\sum _{x=1}^{4}x.P\left(x\right)$

$=0\times \infty +1\times \frac{1280}{9}+2\times \frac{200}{81}+3\times \frac{2000}{6561}+4\times \frac{625}{6561}$
$=148.4560$

Var$\left(x\right)=\sum x^{2}P\left(x\right)-{(\sum xP(x\left)\right)}^2=(0+\frac{1280}{9}+\frac{800}{81}+\frac{18000}{6561}+\frac{10000}{6561})-(148.5660{)}^2$
$=156.366-(148.4560{)}^2$