If the probability distribution of a random variable $X$ is as follows,then $k=$

$X=x$	$1$	$2$	$3$	$4$
$P(X=x)$	$2k$	$4k$	$3k$	$k$

(in $/10$)

Let $p(x)$ represent the probability mass function of a Poisson distribution. If its mean $\lambda = 3.725$,then the value of $x$ at which $p(x)$ is maximum is

The cumulative distribution function (c.d.f.) $F(x)$ of a discrete random variable $X$ is given by the following table:

$X$	$-3$	$-1$	$0$	$1$	$3$	$5$	$7$	$9$
$F(X)$	$0.1$	$0.3$	$0.5$	$0.65$	$0.75$	$0.85$	$0.90$	$1$

Then,find $P[X=3]$.

The probability distribution of a random variable $X$ is given by the following table:

$X = x$	$1$	$2$	$3$	$\dots$	$n$
$P(X = x)$	$\frac{1}{n}$	$\frac{1}{n}$	$\frac{1}{n}$	$\dots$	$\frac{1}{n}$

Then $\operatorname{Var}(X) = $

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:

Find the probability distribution of the number of successes in two tosses of a die,where a success is defined as a number greater than $4$.