If $P(X = x) = 5r^x$,$x = 1, 2, 3, \dots$ is the probability mass function of a discrete random variable $X$,then $r = $

Given below is the distribution of a random variable $X$:

$X=x$	$1$	$2$	$3$	$4$
$P(X=x)$	$\lambda$	$2\lambda$	$3\lambda$	$4\lambda$

If $\alpha=P(X < 3)$ and $\beta=P(X>2)$,then $\alpha: \beta=$

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

$X=k$	$0$	$1$	$2$	$3$	$4$
$P(X=k)$	$0.1$	$0.4$	$0.3$	$0.2$	$0$

The variance of $X$ is:

The c.d.f. $F(x)$ associated with the p.d.f. $f(x)$ is given by:
$f(x) = \begin{cases} 12x^2(1-x), & \text{if } 0 < x < 1 \\ 0, & \text{otherwise} \end{cases}$

$A$ random variable $X$ takes the values $0, 1, 2, 3, \dots$ with probability $P(X=x) = k(x+1)\left(\frac{1}{5}\right)^x$,where $k$ is a constant. Then $P(X=0)$ is

Which of the following cannot be a valid assignment of probabilities for outcomes of sample space $S = \{\omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}, \omega_{5}, \omega_{6}, \omega_{7}\}$?

Outcome	$\omega_{1}$	$\omega_{2}$	$\omega_{3}$	$\omega_{4}$	$\omega_{5}$	$\omega_{6}$	$\omega_{7}$
Probability	$-0.1$	$0.2$	$0.3$	$0.4$	$-0.2$	$0.1$	$0.3$