Find the mean number of heads in three tosses of a fair coin. (in $.5$)

$A$ random variable $X$ has the following probability distribution:

$X=x_i$	$-2$	$-1$	$0$	$1$	$2$
$P(X=x_i)$	$1/6$	$k$	$1/4$	$k$	$1/6$

The variance of this random variable is

$A$ random variable $X$ takes the values $1, 2, 3$ and $4$ such that $2 P(X=1) = 3 P(X=2) = P(X=3) = 5 P(X=4)$. If $\sigma^2$ is the variance and $\mu$ is the mean of $X$,then $\sigma^2 + \mu^2 =$

$A$ player tosses $2$ fair coins. He wins $Rs. 5$ if $2$ heads appear,$Rs. 2$ if $1$ head appears,and $Rs. 1$ if no head appears. Then,the variance of his winning amount is

Given below is the probability distribution of a discrete random variable $X$:

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

Then $P(X \geq 4) = $

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: