From a bag containing $4$ white and $5$ red balls,if $3$ balls are drawn at random,then the mean of the number of red balls among the balls drawn is:

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:

In a game,a man wins $₹ 40$ if he gets $5$ or $6$ on a throw of a fair die and loses $₹ 20$ for getting any other number on the die. If he decides to throw the die either until he gets a $5$ or $6$ or to a maximum of $3$ throws,then his expected gain/loss (in rupees) is:

$A$ fair $n$ $(n > 1)$ faced die is rolled repeatedly until a number less than $n$ appears. If the mean of the number of tosses required is $\frac{n}{9}$,then $n$ is equal to

If the probability function of a random variable $X$ is defined by $P(X=k) = a \left( \frac{k+1}{2^k} \right)$ for $k = 0, 1, 2, 3, 4, 5$,then the probability that $X$ takes a prime value is

If $X$ is a random variable with the following probability distribution:

$X=x$	$-3$	$6$	$9$
$P(X=x)$	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{1}{3}$

Then the variance of $X$ is: