If $X$ is a random variable with the following probability distribution:

$X=x$	$-3$	$6$	$9$
$P(X=x)$	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{1}{3}$

Then the variance of $X$ is:

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

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X = x)$	$k$	$3k$	$5k$	$7k$	$8k$	$k$

Then $P(2 \leq X < 5) = $

Let a sample space be $S = \{\omega_{1}, \omega_{2}, \ldots, \omega_{6}\}$. Which of the following assignments of probabilities to each outcome is valid?

Outcome	$\omega_1$	$\omega_2$	$\omega_3$	$\omega_4$	$\omega_5$	$\omega_6$
$(a)$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{6}$

$A$ person who tosses an unbiased coin gains two points for turning up a head and loses one point for a tail. If three coins are tossed and the total score $X$ is observed, then the range of $X$ is

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 $F(1.5) - F(-0.5)$.

From a lot of $12$ items containing $3$ defectives,a sample of $5$ items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. Let items in the sample be drawn one by one without replacement. If variance of $X$ is $\frac{m}{n}$,where $\operatorname{gcd}(m, n)=1$,then $n-m$ is equal to..........