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.

Three fair coins,each numbered $1$ and $0$,are tossed simultaneously. The variance $\operatorname{Var}(X)$ of the probability distribution of the random variable $X$,where $X$ is the sum of the numbers on the uppermost faces,is:

If $X$ is a Poisson variable such that $3 P(X=4)=\frac{1}{2} P(X=2)+P(X=0)$,then the mean of $X$ is

If the probability that the random variable $X$ takes values $x$ is given by $P(X = x) = k(x + 1)3^{-x}$ for $x = 0, 1, 2, 3, \ldots$,where $k$ is a constant,then $P(X \geq 2)$ is equal to

$A$ random variable $X$ 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

$A$ random variable $X$ has the following probability distribution:
$X$: $1, 2, 3, 4$
$P(X)$: $0.2, 0.4, 0.3, 0.1$
The mean and variance of $X$ are respectively: