If the mean of the following probability distribution of a random variable $X$ is $\frac{46}{9}$,then the variance of the distribution is:

$X$	$0$	$2$	$4$	$6$	$8$
$P(X)$	$a$	$2a$	$a+b$	$2b$	$3b$

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 the $c.d.f.$ (cumulative distribution function) is given by $F(x) = \frac{x-25}{10}$,then $P(27 \leq x \leq 33) = \_\_\_\_$

$A$ random variable $X$ has the range $\{0, 1, 2, \ldots\}$. If $P(X=r) = k(1+r) 3^{-r}$ for $r=0, 1, 2, \ldots$,where $k > 0$ is a real number,then $P(X=0) + P(X=1) + P(X=2) =$

$A$ random variable $X$ takes values $0, 1, 2, 3$ with probabilities $\frac{2a + 1}{30}, \frac{8a - 1}{30}, \frac{4a + 1}{30}, b$ respectively,where $a, b \in R$. Let $\mu$ and $\sigma$ respectively be the mean and standard deviation of $X$ such that $\sigma^{2} + \mu^{2} = 2$. Then $\frac{a}{b}$ is equal to:

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..........