$A$ random variable $X$ takes values $0, 1, 2, 3, \ldots$ with probability $P(X=x) = K(x+1)\left(\frac{1}{5}\right)^x$,where $K$ is a constant. Then $P(X=0)$ is:

$A$ fair die is tossed twice in succession. If $X$ denotes the number of sixes in $2$ tosses,then the probability distribution of $X$ is given by

The random variable $X$ has a probability distribution $P(X)$ of the following form,where $k$ is some number:
$P(X) = \begin{cases} k, & \text{if } x=0 \\ 2k, & \text{if } x=1 \\ 3k, & \text{if } x=2 \\ 0, & \text{otherwise} \end{cases}$
Determine the value of $k$.

If the probability function of a random variable $X$ is defined by $P(X=k) = a \left( \frac{k+1}{2^k} \right)$ for $k = 0, 1, 2, 3, 4, 5$,then the probability that $X$ takes a prime value is

For a Poisson variate $X$,if $P(X=2)=3 P(X=3)$,then the mean of $X$ is:

$A$ man draws a card from a pack of $52$ playing cards,replaces it,and shuffles the pack. He continues this process until he gets a spade card. The probability that he will fail the first two times is: