If the mean of the following probability distribution of a random variable $X$ is $\frac{46}{9}$,then the variance of the distribution is:

$X$	$0$	$2$	$4$	$6$	$8$
$P(X)$	$a$	$2a$	$a+b$	$2b$	$3b$

Let $F(x)$ be the cumulative distribution function (c.d.f.) of a continuous random variable $X$. If $F(b) = 0.7$ and $P(X > a) = 0.4$, then the value of $P(a < X < b)$ is ...

$A$ student studies for $X$ number of hours during a randomly selected school day. The probability distribution of $X$ is given by the following form,where $k$ is a constant:
$P(X=x) = \begin{cases} 0.2, & \text{if } x=0 \\ kx, & \text{if } x=1 \text{ or } 2 \\ k(6-x), & \text{if } x=3 \text{ or } 4 \\ 0, & \text{otherwise} \end{cases}$
The probability that the student studies for at most two hours is:

If $X$ is a random variable with probability distribution $P(X=k) = \frac{(2k+3)c}{3^k}$, $k=0, 1, 2, \ldots, \infty$, then $P(X=3) =$

Three fair coins with faces numbered $1$ and $0$ are tossed simultaneously. The variance $(X)$ of the probability distribution of random variable $X$,where $X$ is the sum of numbers on the uppermost faces,is

$A$ biased coin with probability $p$ $(0 < p < 1)$ of getting a head is tossed until a head appears for the first time. If the probability that the number of tosses required is even is $\frac{2}{5}$, then $p=$