If a random variable $X$ has the p.d.f. $f(x) = \begin{cases} \frac{k}{x^2+1} & , \text{if } 0 < x < \infty \\ 0 & , \text{otherwise} \end{cases}$,then the c.d.f. of $X$ is:

If $X$ is a Poisson variate such that $P(X=1)=P(X=2)$,then $P(X=4)$ is equal to

The probability distribution of a random variable $X$ is given below.

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

Then the variance of $X$ is

It is known that a box of $8$ batteries contains $3$ defective pieces and a person randomly selects two batteries from the box. If $X$ is the number of defective batteries selected,then $P(X \leq 1) = $

In a game,a man wins $Rs. 100$ if he gets $5$ or $6$ on a throw of a fair die and loses $Rs. 50$ for getting any other number on the die. If he decides to throw the die either until he gets a five or a six or to a maximum of three throws,then his expected gain/loss (in rupees) is

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

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X)$	$0$	$2p$	$2p$	$3p$	$p^2$	$2p^2$	$7p^2$	$2p$

The value of $p$ is: