The p.m.f of a random variable $X$ is given by $P(X) = \frac{2x}{n(n+1)}$ for $x = 1, 2, 3, \ldots, n$,and $0$ otherwise. Then $E(X) = $

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

The probability mass function of a random variable $X$ is given by $P[X = r] = \begin{cases} \frac{^n C_r}{32}, & r = 0, 1, 2, \dots, n \\ 0, & \text{otherwise} \end{cases}$. Then,$P[X \leq 2] = $

$A$ random variable $X$ has the probability distribution given below. Its variance is:

$X$	$1$	$2$	$3$	$4$	$5$
$P(X=x)$	$K$	$2K$	$3K$	$2K$	$K$

If $X$ is a random variable with cumulative distribution function $F(x)$ and its probability distribution is given by the following table:

$X = x$	$-1.5$	$-0.5$	$0.5$	$1.5$	$2.5$
$P(X = x)$	$0.05$	$0.2$	$0.15$	$0.25$	$0.35$

Then,find the value of $F(1.5) - F(-0.5)$.

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: