The probability function of a random variable $X$ is given by $P(X=k)=c k^2$,where $c$ is a constant and $k \in\{0,1,2,3,4\}$. If $\sigma^2$ is the variance of $X$ and $\mu$ is the mean of $X$,then $\sigma^2+\mu^2=$

India plays two matches each against Australia and West Indies. The probabilities of India scoring $0, 1,$ and $2$ points in a match are $0.45, 0.05,$ and $0.50$ respectively. Assuming the results are independent,what is the probability that India scores at least $7$ points?

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:

If $P(X=x)=c\left(\frac{2}{3}\right)^x$ for $x=1, 2, 3, 4, \ldots$ is a probability distribution function of a random variable $X$, then the value of $c$ is

The expected value of the sum of the two numbers obtained on the uppermost faces,when two fair dice are rolled,is

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 $\frac{1}{2^c} \sum_{k=0}^{\infty} (3k+1) \left(\frac{1}{2}\right)^k = 1$.