The following is the probability distribution of $X$:

$X$	$0$	$1$	$2$	$3$
$P(X=x)$	$\frac{1+p}{5}$	$\frac{2-2p}{5}$	$\frac{2-p}{5}$	$\frac{2p}{5}$

For a minimum value of $p$,the value of $5 E(X)$ is:

If $X$ is a Poisson variate with $P(X=0)=0.8$,then the variance of $X$ is

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

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

Then $F(4) = $

$A$ 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:

Four defective oranges are accidentally mixed with sixteen good ones. Three oranges are drawn from the mixed lot. The probability distribution of defective oranges is

State whether the following table represents a probability distribution of a random variable $X$. Give reasons for your answer.

$X$	$0$	$1$	$2$
$P(X)$	$0.4$	$0.4$	$0.2$