If $m$ and $\sigma^2$ are the mean and variance of the random variable $X$,whose distribution is given by:

$X=x$	$0$	$1$	$2$	$3$
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$

Then:

The cumulative distribution function of a continuous random variable $X$ is given by $F(x) = \frac{\sqrt{x}}{2}$ for $0 \leq x \leq 4$. Then $P[X > 1]$ is

The probability distribution of a random variable $X$ is given below:

$X=k$	$0$	$1$	$2$	$3$	$4$
$P(X=k)$	$0.1$	$0.4$	$0.3$	$0.2$	$0$

The variance of $X$ is:

Let $X$ be a continuous random variable with the probability density function (p.d.f.) given by $f(x) = \begin{cases} kx, & 0 \leq x < 1 \\ k, & 1 \leq x < 2 \\ -kx + 3k, & 2 \leq x < 3 \\ 0, & \text{otherwise} \end{cases}$. Find $P(2 < X \leq 3)$.

If a random variable $x$ has the probability distribution as follows:

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

Then $P(3 < x \leq 6)$ is equal to:

$A$ random variable $X$ has the probability distribution as given below. Let $E = \{X \mid X \text{ is a prime number}\}$ and $F = \{X \mid X < 4\}$,then $P(E \cup F) = $
$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline X & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline P(X) & K & 2K & K^2 & 2K^2 & 5K^2 & K & K & 2K \\ \hline \end{array}$