$A$ manufacturing company noticed that $1 \%$ of its products are defective. If a dealer orders for $300$ items from this company,then the probability that the number of defective items is at most one is

If $m$ and $\sigma^2$ are the mean and variance of the random variable $X$,whose distribution is given by:

$X=x$	$0$	$1$	$2$	$3$
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$

Then:

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

$X = x$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X = x)$	$0.15$	$0.23$	$k$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

For the events $E = \{x : x \text{ is a prime number}\}$ and $F = \{x : x < 4\}$,then $P(E \cup F) = $

The probability distribution of a random variable $X$ is given by the following table:

$X = x$	$1$	$2$	$3$	$\dots$	$n$
$P(X = x)$	$\frac{1}{n}$	$\frac{1}{n}$	$\frac{1}{n}$	$\dots$	$\frac{1}{n}$

Then $\operatorname{Var}(X) = $

If $X$ is a Poisson variate such that $P(X=1) = 2P(X=2)$,then $P(X=3)$ is equal to:

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..........