The random variable $X$ takes the values $1, 2, 3, \ldots, m$. If $P(X=n) = \frac{1}{m}$ for each $n$,then the variance of $X$ is

For the probability distribution of a discrete random variable $X$ as given below,the mean of $X$ is:

$X = x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$K + \frac{2}{10}$	$K + \frac{3}{10}$	$K + \frac{3}{10}$	$K + \frac{4}{10}$	$K + \frac{2}{10}$

Three rotten apples are mixed accidentally with seven good apples and four apples are drawn one by one without replacement. Let the random variable $X$ denote the number of rotten apples. If $\mu$ and $\sigma^2$ represent the mean and variance of $X$,respectively,then $10(\mu^2 + \sigma^2)$ is equal to

$A$ coin is biased so that the head is $3$ times as likely to occur as tail. If the coin is tossed twice,find the probability distribution of number of tails.

The following is the p.d.f. of a continuous random variable $X$: $f(x) = \begin{cases} \frac{x}{8} & , \text{if } 0 < x < 4 \\ 0 & , \text{otherwise} \end{cases}$
Then $F(0.5)$,$F(1.7)$,and $F(5)$ are respectively:

$A$ random variable $X$ has the following probability distribution:
| $x$ | $1$ | $2$ | $3$ | $4$ | $5$ | $6$ | $7$ | $8$ |
|---|---|---|---|---|---|---|---|---|
| $P(x)$ | $0.15$ | $0.23$ | $0.12$ | $0.10$ | $0.20$ | $0.08$ | $0.07$ | $0.05$ |
For the events $E = \{x \text{ is a prime number}\}$ and $F = \{x < 4\}$,the probability $P(E \cup F)$ is: