For the following cumulative distribution function $F(x)$ of a random variable $X$,find $P(3 < X \leq 5)$.

$x$	$1$	$2$	$3$	$4$	$5$	$6$
$F(x)$	$0.2$	$0.37$	$0.48$	$0.62$	$0.85$	$1$

For a probability distribution $P(x) = \frac{c}{3} \binom{4}{x}$,where $x = 1, 2, 3, 4$,the value of $c$ is . . . . . . .

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

$X = x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(x)$	$0.01$	$0.10$	$0.26$	$0.33$	$0.18$	$0.06$	$K$	$0.04$

Then $P(X \geq 3) - P(X < 6) =$

If the probability distribution is given by $P(x) = C \binom{4}{x}$ for $x = 0, 1, 2, 3, 4$,then find the value of $C$.

$A$ 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

$A$ random variable $X$ has the following probability distribution:
| $x$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |
|---|---|---|---|---|---|---|---|---|
| $P(x)$ | $0.15$ | $0.23$ | $0.12$ | $0.10$ | $0.20$ | $0.08$ | $0.07$ | $0.05$ |
For the events $E = \{x \text{ is a prime number}\}$ and $F = \{x < 4\}$,the probability $P(E \cup F)$ is: