The cumulative distribution function of a continuous random variable $X$ is given by $F(x) = \frac{\sqrt{x}}{2}$ for $0 \leq x \leq 4$. Then $P[X > 1]$ is

If the probability function of a random variable $X$ is given by $P(X=j) = \frac{1}{2^j}$ for $j = 1, 2, 3, \ldots, \infty$,then the variance of $X$ is:

The probability distribution of a discrete random variable $X$ is given below. If $E(X^2) = \Sigma x^2 P(X=x)$, then $6 E(X^2) - \operatorname{Var}(X) =$

$X=x$	$-1$	$0$	$1$	$2$
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{3}$

From a lot of $10$ items,which include $3$ defective items,a sample of $5$ items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. If the variance of $X$ is $\sigma^2$,then $96 \sigma^2$ is equal to....................

The p.m.f. of a random variable $X$ is given by $P(X=x) = \begin{cases} \frac{\binom{5}{x}}{2^5}, & x = 0, 1, 2, 3, 4, 5 \\ 0, & \text{otherwise} \end{cases}$. Then which of the following is not correct?

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