$A$ 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:

In a meeting,$70 \%$ of the members favour and $30 \%$ oppose a certain proposal. $A$ member is selected at random. We take $X=0$ if he opposes the proposal and $X=1$ if the member is in favour. Then the variance of $X$ is:

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

$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{10}$	$\frac{2}{10}$	$\frac{3}{10}$	$\frac{4}{10}$

Then the variance of $X$ is

Two numbers are selected at random (without replacement) from the first six positive integers. Let $X$ denote the larger of the two numbers obtained. Find $E(X)$.

In a game,$3$ coins are tossed. $A$ person is paid $₹150$ if he gets all heads or all tails,and he is supposed to pay $₹50$ if he gets one head or two heads. The amount he can expect to win or lose on an average per game in $₹$ is:

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) = $