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

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

Determine $P(X > 6)$.

$A$ random variable $X$ has the following probability distribution:
| $X$ | $1$ | $2$ | $3$ | $4$ | $5$ |
|---|---|---|---|---|---|
| $P(X)$ | $k^2$ | $2k$ | $k$ | $2k$ | $5k^2$ |
Then $P(X \geq 2)$ is equal to:

$A$ box contains $10$ pens of which $3$ are defective. $A$ sample of $2$ pens is drawn at random and let $X$ denote the number of defective pens. Then the variance of $X$ is

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:

$A$ 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 $\sigma^2 + \mu^2 =$

From a lot of $12$ items containing $3$ defectives,a sample of $5$ items is drawn at random. Let the random variable $X$ denote the number of defective items in the sample. Let items in the sample be drawn one by one without replacement. If variance of $X$ is $\frac{m}{n}$,where $\operatorname{gcd}(m, n)=1$,then $n-m$ is equal to..........