If $X$ is a Poisson variate such that $P(X=1)=P(X=2)$,then $P(X=4)$ is equal to

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

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

The value of $p$ is:

$A$ fair die is tossed twice in succession. If $X$ denotes the number of fours in $2$ tosses,then the probability distribution of $X$ is given by

If a random variable $X$ has the following probability distribution,then its variance is

$X=x$	$1$	$3$	$5$	$2$
$P(X=x)$	$3 K^2$	$K$	$K^2$	$2 K$

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

$X = x$	$-4$	$-2$	$0$	$2$	$4$	$6$	$8$	$10$
$F(X = x)$	$0.1$	$0.3$	$0.5$	$0.65$	$0.75$	$0.85$	$0.90$	$1$

Then,calculate $\frac{P(X \leqslant 0)}{P(X > 0)}$.

$A$ person throws two fair dice. He wins $Rs.\, 15$ for throwing a doublet (same numbers on the two dice),wins $Rs.\, 12$ when the throw results in the sum of $9$,and loses $Rs.\, 6$ for any other outcome on the throw. Then the expected gain/loss (in $Rs.$) of the person is