If the probability mass function (p.m.f.) of a discrete random variable $X$ is given by $P(X=x) = \frac{c}{x^3}$ for $x = 1, 2, 3$ and $0$ otherwise,then $E(X)$ is equal to:

Let $X$ be a random variable,and let $P(X=x)$ denote the probability that $X$ takes the value $x$. Suppose that the points $(x, P(X=x))$ for $x=0,1,2,3,4$ lie on a fixed straight line in the $xy$-plane,and $P(X=x)=0$ for all $x \in \mathbb{R} \setminus \{0,1,2,3,4\}$. If the mean of $X$ is $\frac{5}{2}$,and the variance of $X$ is $\alpha$,then the value of $24\alpha$ is:

$A$ random variable $X$ has the probability distribution as shown below. For the events $E = \{ X \text{ is a prime number} \}$ and $F = \{ X < 4 \}$,the probability $P(E \cup F)$ is:

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

The probability mass function of a random variable $X$ is given by $P(X=x) = \frac{{}^{5}C_{x}}{2^{5}}$ for $x = 0, 1, 2, 3, 4, 5$ and $0$ otherwise. Then,$P(X \leq 2)$ is equal to:

$A$ bag contains $2$ white and $1$ red balls. One ball is drawn at random and then put back in the box after noting its colour. The process is repeated again. If $X$ denotes the number of red balls recorded in the two draws,describe $X$.

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