For the following probability distribution of a random variable $X$, the Expected value and Variance of $X$ are respectively

$X = x$	$1$	$2$	$3$
$P(X = x)$	$1/5$	$2/5$	$2/5$

A fair die is rolled indefinitely. Player A wins if two consecutive rolls show 3 or 5, and player B wins if two consecutive rolls show 1 or 2 or 4 or 6. The probability that player A wins in the long run is

A random variable $X \sim B(n, p)$ follows a binomial distribution with $n = 6$. If $9P(X = 4) = P(X = 2)$, then the probability of success $p$ is..

Let X be a continuous random variable with the probability density function(p.d.f.) given by
$f(x) = \begin{cases} kx, & 0 \leq x < 1 \\ k, & 1 \leq x < 2 \\ -kx + 3k, & 2 \leq x < 3 \\ 0, & \text{otherwise} \end{cases}$
$P(2 < X \leq 3) = \cdots$