If $X$ is a Poisson variate such that $2 P(X=1)=5 P(X=5)+2 P(X=3)$,then the standard deviation of $X$ is

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

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X = x)$	$0.15$	$0.23$	$0.10$	$0.12$	$0.20$	$0.08$	$0.07$	$0.05$

For the event $E = \{ X \text{ is a prime number} \}$,$F = \{ X < 4 \}$,then $P(E \cup F)$ is

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?

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

The value of the constant $c$,so that $P(x)=c\left(\frac{2}{3}\right)^{x}$,$x=1,2,3, \ldots$ is the probability distribution function of a discrete random variable $X$ is

$A$ player tosses two fair coins. He wins $Rs. 5$ if two heads appear, $Rs. 3$ if one head appears, and $Rs. 2$ if no head appears. The variance of the winning amount is: