For a probability distribution $P(x) = \frac{c}{3} \binom{4}{x}$,where $x = 1, 2, 3, 4$,the value of $c$ is . . . . . . .

The probability mass function of a random variable $X$ is given by $P[X = r] = \begin{cases} \frac{^n C_r}{32}, & r = 0, 1, 2, \dots, n \\ 0, & \text{otherwise} \end{cases}$. Then,$P[X \leq 2] = $

If the p.m.f. of a random variable $X$ is given by the following table,then the standard deviation of $X$ is (given $p+q=1$):

$x$	$0$	$1$	$2$
$P(X=x)$	$q^2$	$2pq$	$p^2$

In a meeting,$70 \%$ of the members favour and $30 \%$ oppose a certain proposal. $A$ member is selected at random. We take $X=0$ if he opposes the proposal and $X=1$ if the member is in favour. Then the variance of $X$ is:

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

$X$	$1$	$2$	$3$	$4$	$5$	$6$	$7$	$8$
$P(X=x)$	$0.15$	$0.23$	$0.12$	$0.10$	$0.20$	$0.08$	$0.07$	$0.05$

For the events $E = \{X \text{ is a prime number}\}$ and $F = \{X < 4\}$,find $P(E \cup F)$.

Rajesh has just bought a $VCR$ from Maharashtra Electronics and the shop offers an after-sales service contract for Rs. $1000$ for the next five years. Considering the experience of $VCR$ users,the following distribution of maintenance expenses for the next five years is formed:

Expenses	$0$	$500$	$1000$	$1500$	$2000$	$2500$	$3000$
Probability	$0.35$	$0.25$	$0.15$	$0.10$	$0.08$	$0.05$	$0.02$

The expected value of the maintenance cost is: