Two dice are rolled. If a random variable $X$ denotes the sum of the numbers on them and $\mu$ denotes the mean of $X$,then $\mu+P(X < 5)+P(X>9)+P(X=7)=$

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 a random variable $x$ has the probability distribution as follows:

$x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(x)$	$0$	$2k$	$k$	$3k$	$2k^2$	$2k$	$k^2+k$	$7k^2$

Then $P(3 < x \leq 6)$ is equal to:

$A$ 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:

Let $X$ denote the number of hours you study on a Sunday. It is known that $P(X=x) = \begin{cases} 0.1 & \text{if } x=0 \\ kx & \text{if } x=1, 2 \\ k(5-x) & \text{if } x=3, 4 \\ 0 & \text{otherwise} \end{cases}$ where $k$ is a constant. Then the probability that you study at least two hours on a Sunday is

For the given probability distribution,find $E(X^2)$.

$X$	$1$	$2$	$3$	$4$
$P(X)$	$\frac{1}{10}$	$\frac{1}{5}$	$\frac{3}{10}$	$\frac{2}{5}$