Two numbers are selected at random,without replacement from the first $6$ positive integers. Let $X$ denote the larger of the two numbers. Then $E(X) = $

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

$X$	$4k$	$\frac{30}{7}k$	$\frac{32}{7}k$	$\frac{34}{7}k$	$\frac{36}{7}k$	$\frac{38}{7}k$	$\frac{40}{7}k$	$6k$
$P(X)$	$\frac{2}{15}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$	$\frac{2}{15}$	$\frac{1}{5}$	$\frac{1}{15}$

If $E(X) = \frac{263}{15}$, then $P(X < 20)$ is 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:

In a communication network, $98 \%$ of messages are transmitted with no error. If a random variable $X$ denotes the number of incorrectly transmitted messages, then the probability that at most one message is transmitted incorrectly out of $500$ messages sent, is

The probability distribution of a discrete random variable $X$ is given below. If $E(X^2) = \Sigma x^2 P(X=x)$, then $6 E(X^2) - \operatorname{Var}(X) =$

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

$A$ random variable $X$ takes values $-1, 0, 1, 2$ with probabilities $\frac{1+3p}{4}, \frac{1-p}{4}, \frac{1+2p}{4}, \frac{1-4p}{4}$ respectively,where $p$ varies over $\mathbb{R}$. Then the minimum and maximum values of the mean of $X$ are respectively.