If the probability function of a random variable $X$ is defined by $P(X=k) = a \left( \frac{k+1}{2^k} \right)$ for $k = 0, 1, 2, 3, 4, 5$,then the probability that $X$ takes a prime value is

Let $X$ denote the number of hours you study during a randomly selected school day. The probability that $X$ can take the values $x$ has the following form,where $k$ is some unknown constant.
$P(X=x) = \begin{cases} 0.1, & \text{if } x=0 \\ kx, & \text{if } x=1 \text{ or } 2 \\ k(5-x), & \text{if } x=3 \text{ or } 4 \\ 0, & \text{otherwise} \end{cases}$
What is the probability that you study at least two hours? Exactly two hours? At most two hours?

If the probability distribution of a random variable $X$ is as follows,then $P(X \leq 2) = $

$x_i$	$0$	$1$	$2$	$3$	$4$
$P(X = x_i)$	$3K$	$5K$	$3k^2$	$4k^2 + k$	$3k^2$

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

$X$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X=x_i)$	$\alpha$	$\alpha$	$\alpha$	$\beta$	$\beta$	$0.3$

If $\mu$ and $\sigma^2$ represent the mean and variance of $X$ and $\mu=4.2$,then $\sigma^2+\mu^2=$

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 $X$ assumes values $1, 2, 3, \ldots, n$ with equal probabilities. If $\operatorname{var}(X) : E(X) = 4 : 1$,then $n$ is equal to