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:

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

If the range of a random variable $X$ is $\{0, 1, 2, 3, 4, \ldots\}$ with $P(X=k) = \frac{(k+1)a}{3^k}$ for $k \geq 0$,then $a$ is equal to

The probability of India winning a test match against West Indies is $\frac{1}{2}$. Assuming independence from match to match,the probability that in a $5$ match series India's second win occurs at the third test,is

$A$ fair $n$ $(n > 1)$ faced die is rolled repeatedly until a number less than $n$ appears. If the mean of the number of tosses required is $\frac{n}{9}$,then $n$ is equal to

$A$ jar contains $7$ white marbles and $3$ blue marbles. Given that $4$ marbles are chosen from the jar at the same time,the standard deviation of the number of blue marbles chosen is $\frac{\sqrt{a}}{b}$,where $a$ and $b$ are co-prime numbers and $a$ is square-free. Then $a + b$ is: