Let two cards be drawn at random from a pack of $52$ playing cards. Let $X$ be the number of aces obtained. Then the value of $E(X)$ is

If the error involved in making a certain measurement is a continuous random variable $X$ with probability density function $f(x) = k(4 - x^2)$ for $-2 \leq x \leq 2$ and $f(x) = 0$ otherwise,then $P[-1 < X < 1] = $

In a book consisting of $600$ pages, there are $60$ typographical errors. The probability that a randomly chosen page will contain at most two errors is

$A$ random variable $X$ has the probability distribution as given below. Let $E = \{X \mid X \text{ is a prime number}\}$ and $F = \{X \mid X < 4\}$,then $P(E \cup F) = $
$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline X & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline P(X) & K & 2K & K^2 & 2K^2 & 5K^2 & K & K & 2K \\ \hline \end{array}$

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

$X$	$0$	$1$	$2$
$P(X)$	$\frac{25}{36}$	$k$	$\frac{1}{36}$

If the mean of the random variable $X$ is $\frac{1}{3}$,then the variance is:

If the probability distribution of a discrete random variable $X$ is given by $P(X=k) = \frac{2^{-k}(3k+1)}{2^c}$ for $k = 0, 1, 2, \ldots, \infty$,then find the value of $P(X \leq c)$. Note: The expression provided in the prompt is likely $P(X=k) = \frac{(3k+1)}{2^{k+c}}$. Given $\sum_{k=0}^{\infty} P(X=k) = 1$,we have $\frac{1}{2^c} \sum_{k=0}^{\infty} (3k+1) \left(\frac{1}{2}\right)^k = 1$.