Find the mean number of heads in three tosses of a fair coin. (in $.5$)

$A$ random variable $X$ takes the values $1, 2, 3$ and $4$ such that $2 P(X=1) = 3 P(X=2) = P(X=3) = 5 P(X=4)$. If $\sigma^2$ is the variance and $\mu$ is the mean of $X$, then $\sigma^2 + \mu^2 =$

The probability distribution of a random variable $X$ is given by

$X=x$	$0$	$1$	$2$	$3$	$4$
$P(X=x)$	$0.4$	$0.3$	$0.1$	$0.1$	$0.1$

The variance of $X$ is

$A$ die is loaded in such a way that each odd number is twice as likely to occur as each even number. If $E$ is the event that a number greater than or equal to $4$ occurs on a single toss of the die,then $P(E)$ is equal to:

The cumulative distribution function of a discrete random variable $X$ is given by the following table:

$X = x$	$-4$	$-2$	$0$	$2$	$4$	$6$	$8$	$10$
$F(X = x)$	$0.1$	$0.3$	$0.5$	$0.65$	$0.75$	$0.85$	$0.90$	$1$

Then,calculate $\frac{P(X \leqslant 0)}{P(X > 0)}$.

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: