If the probability distribution of a random variable $X$ is as follows,then $P(X \leq 2) = $

$x_i$	$0$	$1$	$2$	$3$	$4$
$P(X = x_i)$	$3K$	$5K$	$3k^2$	$4k^2 + k$	$3k^2$

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

If a discrete random variable $X$ takes the values $1, 2, 3, 4$ such that $2P(X = 1) = 3P(X = 2) = P(X = 3) = 5P(X = 4)$, then $P(X = 4) = ...$ (in $61$)

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:

Let $X$ be a random variable,and let $P(X=x)$ denote the probability that $X$ takes the value $x$. Suppose that the points $(x, P(X=x))$ for $x=0,1,2,3,4$ lie on a fixed straight line in the $xy$-plane,and $P(X=x)=0$ for all $x \in \mathbb{R} \setminus \{0,1,2,3,4\}$. If the mean of $X$ is $\frac{5}{2}$,and the variance of $X$ is $\alpha$,then the value of $24\alpha$ is:

The following table represents the probability distribution of a random variable $X$ for some $k \in Q$. Find the mean of $X$.
$\begin{array}{|c|c|c|c|c|c|c|} \hline X=x & -2 & -1 & 0 & 1 & 2 & 3 \\ \hline P(X=x) & 0.1 & k & 0.2 & 2k & 0.3 & k \\ \hline \end{array}$