$A$ 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 $P(E \cup F)$.

The p.m.f of a random variable $X$ is $P(X=x)=\frac{1}{2^5}\binom{5}{x}$,where $x=0, 1, 2, 3, 4, 5$ and $P(X=x)=0$ otherwise. Then:

If $X$ is a Poisson variate such that $\alpha = P(X=1) = P(X=2)$, then $P(X=4)$ is equal to

$A$ random variable $X$ takes the values $0, 1, 2, 3$ and its mean is $1.3$. If $P(X = 3) = 2P(X = 1)$ and $P(X = 2) = 0.3$, 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:

In a box,there are $20$ cards,out of which $10$ are labelled as $A$ and the remaining $10$ are labelled as $B$. Cards are drawn at random,one after the other and with replacement,until a second $A$-card is obtained. The probability that the second $A$-card appears before the third $B$-card is