Two dice are rolled. If a random variable $X$ is defined as the absolute difference of the two numbers that appear on them, then the mean of $X$ is

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

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

Then $P(2 \leq X < 5) = $

$A$ random variable $X$ takes values $0, 1, 2, 3$ with probabilities $\frac{2a + 1}{30}, \frac{8a - 1}{30}, \frac{4a + 1}{30}, b$ respectively, where $a, b \in R$. Let $\mu$ and $\sigma$ respectively be the mean and standard deviation of $X$ such that $\sigma^{2} + \mu^{2} = 2$. Then $\frac{a}{b}$ is equal to:

Which of the following can not be a valid assignment of probabilities for outcomes of sample space $S = \{\omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}, \omega_{5}, \omega_{6}, \omega_{7}\}$? (Note: The provided table is an example of a valid assignment).

If the function $f$ defined by $f(x) = \begin{cases} K(x-x^2) & \text{if } 0 < x < 1 \\ 0 & \text{otherwise} \end{cases}$ is the probability density function (p.d.f.) of a random variable $X$,then the value of $P(X < \frac{1}{2})$ is

$A$ random variable $X$ has the following probability distribution:
$X = 1, 2, 3, 4, 5$
$P(X) = 0.1, 0.2, 0.3, 0.2, 0.2$
For the events $E = \{X \text{ is a prime number}\}$ and $F = \{X < 4\}$, find $P(E \cup F)$.