The probability that an individual suffers a bad reaction from an injection is $0.001$. The probability that out of $2000$ individuals exactly three will suffer a bad reaction is:

$A$ 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) = $

If $X$ is a random variable with the following probability distribution:

$X=x$	$-3$	$6$	$9$
$P(X=x)$	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{1}{3}$

Then the variance of $X$ 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 $\sigma^2 + \mu^2 =$

If random variable $X$ is the waiting time in minutes for a bus and the probability density function of $X$ is given by $f(x) = \begin{cases} \frac{1}{5}, & 0 \leq x \leq 5 \\ 0, & \text{otherwise} \end{cases}$,then the probability of the waiting time being not more than $4$ minutes is = . . . . . . .

In a game,a man wins $Rs. 100$ if he gets $5$ or $6$ on a throw of a fair die and loses $Rs. 50$ for getting any other number on the die. If he decides to throw the die either until he gets a five or a six or to a maximum of three throws,then his expected gain/loss (in rupees) is