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

The probability distribution of a random variable $X$ is given by

$X = x$	$0$	$1$	$2$
$P(X = x)$	$\frac{1}{5}$	$\frac{2}{5}$	$\frac{2}{5}$

Then the variance of $X$ is

Let a random variable $X$ take values $\{0, 1, 2, 3\}$ with $P(X=0) = P(X=1) = p$,$P(X=2) = P(X=3) = q$,and $E(X^2) = 2E(X)$. Then the value of $8p - 1$ is:

The probability distribution of a random variable $X$ is given below:

$X = x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X = x)$	$0$	$k$	$2k$	$2k$	$3k$	$k^2$	$2k^2$	$7k^2 + k$

Then,$P(0 < X < 4)$ is equal to:

The cumulative distribution function $F(x)$ of a discrete random variable $X$ is given by the following table:

$X = x$	$-1$	$0$	$1$	$2$
$F(X = x)$	$0.3$	$0.7$	$0.8$	$1$

Then $E(X^2) = $

$A$ bakerman sells $5$ types of cakes. Profit due to sale of each type of cake is respectively $Rs \ 2$,$Rs \ 2.5$,$Rs \ 3$,$Rs \ 1.5$ and $Rs \ 1$. The demands for these cakes are $20 \%$,$5 \%$,$10 \%$,$50 \%$ and $15 \%$ respectively. Then the expected profit per cake is: