If the function $P[X = x] = \begin{cases} \frac{K \cdot 2^x}{x!}, & x = 0, 1, 2, 3 \\ 0, & \text{otherwise} \end{cases}$ forms a probability mass function (p.m.f.),then the value of $K$ is:

$A$ 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:

$A$ random variable $X$ has the following probability distribution:
$X = 1, P(X) = 0.15$
$X = 2, P(X) = 0.20$
$X = 3, P(X) = 0.25$
$X = 4, P(X) = 0.30$
$X = 5, P(X) = 0.10$
For the event $E = \{ X \text{ is a prime number} \}$ and $F = \{ X < 4 \}$, find $P(E \cup F)$.

Let a pair of dice be thrown and the random variable $X$ be the sum of the numbers that appear on the two dice. Find the mean or expectation of $X$.

Find the probability distribution of the number of successes in two tosses of a die,where a success is defined as 'six appears on at least one die'.

$A$ random variable $X$ has the probability distribution as shown below. For the events $E = \{ X \text{ is a prime number} \}$ and $F = \{ X < 4 \}$,the probability $P(E \cup F)$ is:

$X$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X)$	$0.15$	$0.23$	$0.12$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$