$A$ random variable $X$ takes values $0, 1, 2, 3, \dots$ with probabilities $P(X=x) = k(x+1)\left(\frac{1}{2}\right)^x$. If $k$ is a constant,then $P(X=1) = $

The p.d.f. of a random variable $X$ is given by $f(x) = \frac{k}{\sqrt{x}}$ for $0 \leq x \leq 4$ and $f(x) = 0$ otherwise. Then $P(1 < X < 4) = $

For the probability distribution of a discrete random variable $X$ as given below,the mean of $X$ is:

$X = x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$K + \frac{2}{10}$	$K + \frac{3}{10}$	$K + \frac{3}{10}$	$K + \frac{4}{10}$	$K + \frac{2}{10}$

The random variable $X$ has a probability distribution $P(X)$ of the following form,where $k$ is some number:
$P(X) = \begin{cases} k, & \text{if } x=0 \\ 2k, & \text{if } x=1 \\ 3k, & \text{if } x=2 \\ 0, & \text{otherwise} \end{cases}$
Find $P(X < 2)$,$P(X \leq 2)$,and $P(X \geq 2)$.

From a lot of $10$ items,which include $3$ defective items,a sample of $5$ items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. If the variance of $X$ is $\sigma^2$,then $96 \sigma^2$ is equal to....................

Let $X$ be the random variable taking values $1, 2, \ldots, n$ for a fixed positive integer $n$. If $P(X=k) = \frac{1}{n}$ for $1 \leq k \leq n$,then the variance of $X$ is