The random variable $X$ takes the values $1, 2, 3, \ldots, m$. If $P(X=n) = \frac{1}{m}$ for each $n$,then the variance of $X$ is

If a discrete random variable $X$ has the probability distribution $P(X=x) = k \frac{2^{2x+1}}{(2x+1)!}$ for $x = 0, 1, 2, \ldots, \infty$,then $k =$

$A$ 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 $P(X=0) + P(X=1) + P(X=2) =$

If the probability distribution function of a random variable $X$ is given as follows:

$X=x_i$	$-2$	$-1$	$0$	$1$	$2$
$P(X=x_i)$	$0.2$	$0.3$	$0.15$	$0.25$	$0.1$

Then $F(0)$ is equal to:

In a game of throwing $3$ coins, a player loses $₹ 5$ for each head and gains $₹ 10$ for each tail. If a random variable $X: S \rightarrow R$ is defined as $X(a) = \text{net gain } (a \in S)$, then the mean of the random variable is (in rupees):

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..........