If a Poisson variate $X$ satisfies the relation $P(X=3)=P(X=5)$, then $P(X=4)=$

The following table shows the probability distribution of smart phones sold in a shop per day:

Number of smart phones $(x)$	$0$	$1$	$2$	$3$	$4$	$5$
Probability $(P(x))$	$k$	$0.3$	$0.15$	$0.15$	$0.1$	$2k$

Then $E(x) = ?$

If a random variable $X$ has the following probability distribution, then the mean of $X$ is

$X = x_i$	$1$	$2$	$3$	$5$
$P(X = x_i)$	$2k^2$	$k$	$k$	$k^2$

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

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

For the event $E = \{ X \text{ is a prime number} \}$,$F = \{ X < 4 \}$,then $P(E \cup F)$ is

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

$A$ jar contains $7$ white marbles and $3$ blue marbles. Given that $4$ marbles are chosen from the jar at the same time,the standard deviation of the number of blue marbles chosen is $\frac{\sqrt{a}}{b}$,where $a$ and $b$ are co-prime numbers and $a$ is square-free. Then $a + b$ is: