The probability mass function of a random variable $X$ is given by $P(X=x) = \frac{{}^{5}C_{x}}{2^{5}}$ for $x = 0, 1, 2, 3, 4, 5$ and $0$ otherwise. Then,$P(X \leq 2)$ is equal to:

$A$ random variable $X$ takes values $-1, 0, 1, 2$ with probabilities $\frac{1+3p}{4}, \frac{1-p}{4}, \frac{1+2p}{4}, \frac{1-4p}{4}$ respectively,where $p$ varies over $\mathbb{R}$. Then the minimum and maximum values of the mean of $X$ are respectively.

If $f(x) = \frac{x+2}{18}$ for $-2 < x < 4$ and $f(x) = 0$ otherwise,is the probability density function (p.d.f.) of a random variable $X$,then the value of $P(|X| < 2)$ is

For a random variable $X$,if $P(X=k) = \frac{(k+1)a}{3^k}$ for $k=0, 1, 2, \ldots$,then $a = $

For the probability distribution given below,find $\operatorname{Var}(X)$.

$X$	$5$	$6$	$7$	$8$	$9$	$10$	$11$
$P(X=x)$	$0.07$	$0.2$	$0.3$	$k$	$0.07$	$0.04$	$0.02$

India plays two matches each with West Indies and Australia. In any match,the probabilities of India getting $0, 1,$ and $2$ points are $0.45, 0.05,$ and $0.50$ respectively. Assuming that the outcomes are independent,the probability of India getting at least $7$ points is: