If $X$ is a Poisson variate such that $P(X=2)=P(X=3)$,then $e^3 P(X=4)$ is

Let $X$ be a discrete random variable. The probability distribution of $X$ is given below:

$X$	$30$	$10$	$-10$
$P(X)$	$\frac{1}{5}$	$A$	$B$

If $E(X) = 4$,then the value of $AB$ is equal to:

$A$ random variable $X$ has the following probability distribution. For events $E = \{X \text{ is a prime number}\}$ and $F = \{X < 4\}$,what is the probability $P(E \cup F)$?

$X$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X)$	$0.15$	$0.23$	$0.12$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

$A$ random variable $X$ takes values $-1, 0, 1, 2$ with probabilities $\frac{1+3p}{4}, \frac{1-p}{4}, \frac{1+2p}{4}, \frac{1-4p}{4}$ respectively,where $p$ varies over $\mathbb{R}$. Then the minimum and maximum values of the mean of $X$ are respectively.

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

If $X$ is a Poisson variate with $P(X=0)=0.8$,then the variance of $X$ is