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

$X = x$	$0$	$1$	$2$
$P(X = x)$	$4k - 10k^2$	$5k - 1$	$3k^3$

Then $P(X < 2)$ is:

$A$ random variate $X$ 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 equal to:

$A$ random variable $X$ has the following probability distribution. Then,$P(2 \leq X < 5) = $ . . . . . .

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X = x)$	$K$	$3K$	$5K$	$7K$	$8K$	$K$

The probability distribution of a random variable $X$ is given below:

$x$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X=x)$	$a$	$a$	$a$	$b$	$b$	$0.3$

If the mean of $X$ is $4.2$, then $a$ and $b$ are respectively equal to:

The probability mass function of a random variable $X$ is given by $P[X = r] = \begin{cases} \frac{^n C_r}{32}, & r = 0, 1, 2, \dots, n \\ 0, & \text{otherwise} \end{cases}$. Then,$P[X \leq 2] = $

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