The probability distribution of a random variable $X$ is

x	0	1	2	3
P(X = x)	$\frac{1}{4}$	$\frac{1}{8}$	$\frac{1}{8}$	$\frac{1}{2}$

The variance of $X$ is

Ten cards are numbered from 1 to 10. A and B are playing a game by drawing a card, alternately, with replacement, till one gets a prime number then the probability of B's winning if A started the game is

A pair of dice is thrown, then the probability of getting a sum of 7, is

If A and B are two events such that $P(A) = \frac{1}{2}$, $P(B) = \frac{1}{3}$, $P(B|A) = \frac{1}{4}$, then $P(A|B)$ is:

If A and B are two events such that $P(A|B) = P(B|A)$, and $A \cap B \neq \phi$ then

The probability distribution of the random variable X is given by

The variance of the random variable X is

A random variable X has the following probability distribution:

The value of $P(X\ge 3)$ is