The value of $C$ for which $P(X = k) = Ck^2$ can serve as the probability function of a random variable $X$ that takes values $0, 1, 2, 3, 4$ is

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

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

For the events $E = \{x : x \text{ is a prime number}\}$ and $F = \{x : x < 4\}$, then $P(E \cup F) = $

In a game,$3$ coins are tossed. $A$ person is paid $₹150$ if he gets all heads or all tails,and he is supposed to pay $₹50$ if he gets one head or two heads. The amount he can expect to win or lose on an average per game in $₹$ is:

If a random variable $X$ has the probability distribution given by $P(X=0)=3C^3$,$P(X=2)=5C-10C^2$ and $P(X=4)=4C-1$,then the variance of that distribution is

For the following probability distribution,find the expected value of $X$:

$x$	$0$	$1$	$2$	$3$	$4$	$5$
$P(x)$	$k$	$0.3$	$0.15$	$0.15$	$0.1$	$2k$

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