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:

$A$ random variable $X$ has the probability distribution as given below. Let $E = \{X \mid X \text{ is a prime number}\}$ and $F = \{X \mid X < 4\}$,then $P(E \cup F) = $
$\begin{array}{|c|c|c|c|c|c|c|c|c|} \hline X & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline P(X) & K & 2K & K^2 & 2K^2 & 5K^2 & K & K & 2K \\ \hline \end{array}$

An unbiased die is tossed until a number greater than $4$ appears. The probability that an even number of tosses is needed is

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

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$
$P(X)$	$k$	$3k$	$5k$	$7k$	$9k$	$11k$	$13k$

Then find $P(X \ge 2)$.

The variance of the following probability distribution is:

$x$	$0$	$1$	$2$
$P(X)$	$\frac{9}{16}$	$\frac{3}{8}$	$\frac{1}{16}$

If the range of a random variable $X$ is $\{0, 1, 2, 3, 4, \ldots\}$ with $P(X=k) = \frac{(k+1)a}{3^k}$ for $k \geq 0$,then $a$ is equal to