The probability distribution of a discrete random variable $X$ is given by the table below:

$X$	$0$	$1$	$2$	$3$	$4$
$P(X=x)$	$2k$	$k$	$2k$	$4k$	$k$

If $a = P(X < 3)$ and $b = P(2 < X < 4)$,then:

Consider the probability distribution
$\begin{array}{|r|c|c|c|c|c|} \hline X=x & 1 & 2 & 3 & 4 & 5 \\ \hline P(X=x) & K & 2K & K^2 & 2K & 5K^2 \\ \hline \end{array}$
Then the value of $P(X > 2)$ is

For the random variable $X$ with the probability distribution given by the table:

$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$K$	$K + \frac{1}{7}$	$2K$	$\frac{2}{5}$

The mean of $X$ is:

$A$ jar contains $7$ white marbles and $3$ blue marbles. Given that $4$ marbles are chosen from the jar at the same time,the standard deviation of the number of blue marbles chosen is $\frac{\sqrt{a}}{b}$,where $a$ and $b$ are co-prime numbers and $a$ is square-free. Then $a + b$ is:

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

Let a sample space be $S = \{\omega_{1}, \omega_{2}, \dots, \omega_{6}\}$. Which of the following assignments of probabilities to each outcome is valid?

Outcome	$\omega_1$	$\omega_2$	$\omega_3$	$\omega_4$	$\omega_5$	$\omega_6$
$(e)$	$0.1$	$0.2$	$0.3$	$0.4$	$0.5$	$0.6$