$A$ random variable $X$ has a Poisson distribution with mean $\lambda = 2$. Then $P(X > 1.5)$ is equal to:

$A$ random variable $X$ takes the values $0, 1$ and $2$. If $P(X=1)=P(X=2)$ and $P(X=0)=0.4$, then the mean of the random variable $X$ is

The p.d.f. of a continuous random variable $X$ is given by $f(x) = \frac{x+2}{18}$ for $-2 < x < 4$ and $f(x) = 0$ otherwise. Then $P[|x| < 1] = $

The cumulative distribution function $F(x)$ of a discrete random variable $X$ is given by the following table:

$X = x$	$-1$	$0$	$1$	$2$
$F(X = x)$	$0.3$	$0.7$	$0.8$	$1$

Then $E(X^2) = $

Let $X$ be a random variable such that the probability function of a distribution is given by $P(X=0) = \frac{1}{2}$ and $P(X=j) = \frac{1}{3^j}$ for $j = 1, 2, 3, \ldots, \infty$. Then the mean of the distribution and $P(X \text{ is positive and even})$ respectively are:

If a discrete random variable $X$ is defined as follows:
$P(X=x) = \begin{cases} \frac{k(x+1)}{5^x}, & x=0, 1, 2, \ldots \\ 0, & \text{otherwise} \end{cases}$
then $k=$