$A$ fair coin is tossed $2$ times. $A$ person receives $₹ X^{3}$ if he gets $X$ number of heads. His expected gain is $=$

If the function $f$ defined by $f(x) = \begin{cases} K(x-x^2) & \text{if } 0 < x < 1 \\ 0 & \text{otherwise} \end{cases}$ is the probability density function (p.d.f.) of a random variable $X$,then the value of $P(X < \frac{1}{2})$ is

Two dice are rolled. If a random variable $X$ denotes the sum of the numbers on them and $\mu$ denotes the mean of $X$, then $\mu+P(X < 5)+P(X>9)+P(X=7)=$

The p.d.f. of a continuous random variable $X$ is given by $f(x) = \frac{x}{8}$ for $0 < x < 4$ and $f(x) = 0$ otherwise. Then $P(X \leq 2)$ is:

$A$ player tosses two fair coins. He wins $Rs. 5$ if two heads appear, $Rs. 3$ if one head appears, and $Rs. 2$ if no head appears. The variance of the winning amount is:

The variance of the random variable $X$ having the following distribution is:

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