Suppose $X$ has the following probability mass function $P(X=0)=0.2, P(X=1)=0.5, P(X=2)=0.3$. What is $E[X^2]$?

$A$ random variable $X$ has the following probability distribution:

$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X=x)$	$0$	$k$	$2k$	$2k$	$3k$	$k^2$	$2k^2$	$7k^2+k$

Find the value of $P(0 < X < 6)$.

If the probability mass function (p.m.f.) of a random variable $X$ is $P(X=x) = \frac{1}{10}$ for $x = 1, 2, 3, \ldots, 10$,and $0$ otherwise,then $\operatorname{Var}(X)$ is equal to:

If the probability function of a random variable $X$ is given by $P(X=k) = \frac{3^{ck}}{k!}$ for $k = 1, 2, 3, \ldots$ (where $c$ is a constant),then $c =$

The c.d.f. $F(x)$ associated with the p.d.f. $f(x)$ is given by:
$f(x) = \begin{cases} 12x^2(1-x), & \text{if } 0 < x < 1 \\ 0, & \text{otherwise} \end{cases}$

If $X$ is a Poisson variate such that $\alpha = P(X=1) = P(X=2)$, then $P(X=4)$ is equal to