$A$ random variable $X$ has the probability distribution
$\begin{array}{|c|c|c|c|c|c|c|}\hline X=x_i & 1 & 2 & 3 & 4 & 5 & 6 \\\hline P(X=x_i) & 0.2 & 0.3 & 0.12 & 0.1 & 0.2 & 0.08 \\\hline \end{array}$
If $A=\{x_i \mid x_i \text{ is a prime number}\}$ and $B=\{x_i \mid x_i < 4\}$ are two events,then $P(A \cup B) = $

$A$ 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:

Suppose three coins are tossed simultaneously. If $X$ denotes the number of heads,then the probability distribution of $X$ is:

State which of the following is not the probability distribution of a random variable. Give reasons for your answer.

$Y$	$-1$	$0$	$1$
$P(Y)$	$0.6$	$0.1$	$0.2$

In a city,$10$ accidents take place in a span of $50$ days. Assuming that the number of accidents follows the Poisson distribution,the probability that three or more accidents occur in a day is:

The cumulative distribution function of a discrete random variable $X$ is given by the following table:

$X = x$	$-4$	$-2$	$0$	$2$	$4$	$6$	$8$	$10$
$F(X = x)$	$0.1$	$0.3$	$0.5$	$0.65$	$0.75$	$0.85$	$0.90$	$1$

Then,calculate $\frac{P(X \leqslant 0)}{P(X > 0)}$.