If the probability distribution of a random variable $X$ is as follows,then $k=$

$X=x$	$1$	$2$	$3$	$4$
$P(X=x)$	$2k$	$4k$	$3k$	$k$

(in $/10$)

The discrete random variables $X$ and $Y$ are independent from one another and are defined as $X \sim B(16, 0.25)$ and $Y \sim P(2)$. Then the sum of the variances of $X$ and $Y$ is

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

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

Then the variance of $X$ is:

If $m$ and ${\sigma ^2}$ are the mean and variance of a random variable $X$,whose distribution is given by:

$X=x$	$0$	$1$	$2$	$3$	$4$
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$	$0$

,then:

The range of a discrete random variable $X$ is $\{1, 2, 3\}$ and the probabilities of its elements are given by $P(X=1) = 3k^3$, $P(X=2) = 2k^2$, and $P(X=3) = 7 - 19k$. Then $P(X=3) = $

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