The p.d.f. of a random variable $X$ is given by $f(x) = \frac{k}{\sqrt{x}}$ for $0 \leq x \leq 4$ and $f(x) = 0$ otherwise. Then $P(1 < X < 4) = $

The probability distribution of a random variable $X$ is given below:

$x$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X=x)$	$a$	$a$	$a$	$b$	$b$	$0.3$

If the mean of $X$ is $4.2$, then $a$ and $b$ are respectively equal to:

In a game,a man wins $₹ 40$ if he gets $5$ or $6$ on a throw of a fair die and loses $₹ 20$ for getting any other number on the die. If he decides to throw the die either until he gets a $5$ or $6$ or to a maximum of $3$ throws,then his expected gain/loss (in rupees) is:

$A$ random variable $X$ has the following distribution.

$X = x_{i}$	$-2$	$-1$	$0$	$1$	$2$	$3$
$P(X = x_{i})$	$0.1$	$k$	$0.2$	$2k$	$3k$	$k$

Then the variance of this distribution 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:

$A$ 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