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

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

Two coins are tossed simultaneously. Then,the value of $E(X)$,where $X$ denotes the number of heads is

$A$ random variable $X$ has the following probability distribution:
| $X$ | $1$ | $2$ | $3$ | $4$ | $5$ |
|---|---|---|---|---|---|
| $P(X)$ | $k^2$ | $2k$ | $k$ | $2k$ | $5k^2$ |
Then $P(X \geq 2)$ is equal to:

Let the mean and the standard deviation of the probability distribution be $\mu$ and $\sigma$,respectively. If $\sigma - \mu = 2$,then $\sigma + \mu$ is equal to:

$X$	$\alpha$	$1$	$0$	$-3$
$P(X)$	$\frac{1}{3}$	$K$	$\frac{1}{6}$	$\frac{1}{4}$

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

At a telephone enquiry system,the number of phone calls regarding relevant enquiries follows a Poisson distribution with an average of $5$ phone calls during $10$-minute time intervals. The probability that there is at most one phone call during a $10$-minute time period is: