If $X$ is a random variable such that $P(X=-2)=P(X=-1)=P(X=2)=P(X=1)=\frac{1}{6}$ and $P(X=0)=\frac{1}{3}$,then the mean of $X$ is

If a random variable $X$ has p.d.f. $f(x) = \begin{cases} \frac{ax^2}{2} + bx & , \text{if } 1 \leqslant x \leqslant 3 \\ 0 & , \text{otherwise} \end{cases}$ and $f(2) = 2$,then the values of $a$ and $b$ are,respectively

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) = $

$A$ person who tosses an unbiased coin gains two points for turning up a head and loses one point for a tail. If three coins are tossed and the total score $X$ is observed, then the range of $X$ is

The cumulative distribution function (c.d.f.) $F(x)$ of a discrete random variable $X$ is given by the following table:

$X$	$-3$	$-1$	$0$	$1$	$3$	$5$	$7$	$9$
$F(X)$	$0.1$	$0.3$	$0.5$	$0.65$	$0.75$	$0.85$	$0.90$	$1$

Then,find $P[X=3]$.

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

$X$	$1, 2, 3, 4, 5$
$P(X)$	$K^2, 2K, K, 2K, 5K^2$

Then $P(X > 2)$ is equal to: