$A$ random variable $X$ takes the values $0, 1, 2, 3$ and its mean is $1.3$. If $P(X=3)=2 P(X=1)$ and $P(X=2)=0.3$,then $P(X=0)$ is

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

$A$ box contains $8$ red and $N$ green balls. Two balls are drawn at random from it. If $X$ is the random variable representing the number of green balls drawn and $E(X) = 1.2$, then $N = ...$

$A$ random variable $X$ has the following probability distribution:

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

Then $F(4) = $

The p.m.f of a random variable $X$ is given by $P(X) = \frac{2x}{n(n+1)}$ for $x = 1, 2, 3, \ldots, n$,and $0$ otherwise. Then $E(X) = $

$A$ random variable $X$ has the probability distribution
$\begin{array}{|c|c|c|c|c|c|c|}\hline X=x_i & 1 & 2 & 3 & 4 & 5 & 6 \\\hline P(X=x_i) & 0.2 & 0.3 & 0.12 & 0.1 & 0.2 & 0.08 \\\hline \end{array}$
If $A=\{x_i \mid x_i \text{ is a prime number}\}$ and $B=\{x_i \mid x_i < 4\}$ are two events,then $P(A \cup B) = $