If $X$ is a Poisson variable such that $3 P(X=4)=\frac{1}{2} P(X=2)+P(X=0)$,then the mean of $X$ is

The following table shows the probability distribution of smart phones sold in a shop per day:

Number of smart phones $(x)$	$0$	$1$	$2$	$3$	$4$	$5$
Probability $(P(x))$	$k$	$0.3$	$0.15$	$0.15$	$0.1$	$2k$

Then $E(x) = ?$

$A$ 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.

From a lot of $12$ items containing $3$ defectives,a sample of $5$ items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. Let items in the sample be drawn one by one without replacement. If variance of $X$ is $\frac{m}{n}$,where $\operatorname{gcd}(m, n)=1$,then $n-m$ is equal to..........

Three rotten apples are accidentally mixed with fifteen good apples. Assuming the random variable $X$ to be the number of rotten apples in a draw of two apples,the variance of $X$ is

If the probability density function of a continuous random variable $X$ is $f(x) = \frac{x^3}{3}$ for $-1 < x < 2$,and $f(x) = 0$ otherwise,then the cumulative distribution function $F(x)$ for $-1 < x < 2$ is: