$A$ random variable $X$ has its range $\{-1, 0, 1\}$. If its mean is $0.2$ and $P(X=0)=0.2$,then $P(X=1)=$

$A$ random variable $X$ takes the values $0, 1, 2, 3, \dots$ with probability $P(X=x) = k(x+1)\left(\frac{1}{5}\right)^x$,where $k$ is a constant. Then $P(X=0)$ is

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

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

Two numbers are selected at random (without replacement) from the first six positive integers. Let $X$ denote the larger of the two numbers obtained. Find $E(X)$.

$A$ random variable $X$ has the following probability distribution:
$X = 1, 2, 3, 4, 5$
$P(X) = 0.1, 0.2, 0.3, 0.2, 0.2$
For the events $E = \{ X \text{ is a prime number} \}$ and $F = \{ X < 4 \}$, find $P(E \cap F)$.

The probability distribution of $x$ is given by the following table:

$x$	$0$	$1$	$2$	$3$
$P(x)$	$0.2$	$k$	$k$	$2k$

Find the value of $k$.