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

$X$	$1, 2, 3, 4, 5$
$P(X)$	$K^2, 2K, K, 2K, 5K^2$

Then $P(X > 2)$ is equal to:

If a random variable $X$ follows a Poisson distribution such that $P(X=1) = 3P(X=2)$, then $P(X=3) =$

If a discrete random variable $X$ has the probability distribution $P(X=x) = k \frac{2^{2x+1}}{(2x+1)!}$ for $x = 0, 1, 2, \ldots, \infty$,then $k =$

Rajesh has just bought a $VCR$ from Maharashtra Electronics and the shop offers an after-sales service contract for Rs. $1000$ for the next five years. Considering the experience of $VCR$ users,the following distribution of maintenance expenses for the next five years is formed:

Expenses	$0$	$500$	$1000$	$1500$	$2000$	$2500$	$3000$
Probability	$0.35$	$0.25$	$0.15$	$0.10$	$0.08$	$0.05$	$0.02$

The expected value of the maintenance cost is:

Let $X$ be a random variable with the following probability distribution:

$x$	$-2$	$-1$	$3$	$4$	$6$
$P(X=x)$	$\frac{1}{5}$	$a$	$\frac{1}{3}$	$\frac{1}{5}$	$b$

If the mean of $X$ is $2.3$ and the variance of $X$ is $\sigma^{2}$,then $100 \sigma^{2}$ is equal to:

If the probability distribution of a discrete random variable $X$ is given by $P(X=k) = \frac{2^{-k}(3k+1)}{2^c}$ for $k = 0, 1, 2, \ldots, \infty$,then find the value of $P(X \leq c)$. Note: The expression provided in the prompt is likely $P(X=k) = \frac{(3k+1)}{2^{k+c}}$. Given $\sum_{k=0}^{\infty} P(X=k) = 1$,we have $\frac{1}{2^c} \sum_{k=0}^{\infty} (3k+1) \left(\frac{1}{2}\right)^k = 1$.