$A$ random variate $X$ takes the values $0, 1, 2, 3$ and its mean is $1.3$. If $P(X=3) = 2 P(X=1)$ and $P(X=2) = 0.3$,then $P(X=0)$ is equal to:

$A$ random variable $X$ has the following probability distribution:

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X)$	$0$	$k$	$2k$	$3k$	$3k^2$	$k^2$	$2k^2$	$7k^2+k$

Determine $P(0 < X < 3)$. (in $/10$)

In a Poisson distribution with unit mean,calculate the value of $\sum_{x=0}^{\infty} |x-\bar{x}| P(X=x)$,where $\bar{x}$ is the mean of the distribution.

The distribution of a random variable $X$ is given below. The value of $k$ is:

$X = x$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$k$	$\frac{1}{5}$	$2k$	$\frac{3}{10}$	$k$

If a random variable $X$ takes the values $x_1, x_2, x_3, \ldots, x_{100}$ with probability $P(X=x_i) = K i(i+1)$,then $200 K=$

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: