If $A$ and $B$ are two events such that $P(A) \neq 0$ and $P(B) \neq 1$,then $P\left( \frac{\overline{A}}{\overline{B}} \right) = $

Let $E_{1}$ and $E_{2}$ be two events such that the conditional probabilities $P(E_{1} \mid E_{2}) = \frac{1}{2}$,$P(E_{2} \mid E_{1}) = \frac{3}{4}$ and $P(E_{1} \cap E_{2}) = \frac{1}{8}$. Then:

Find $P(E | F)$ for a die thrown three times,where $E: 4$ appears on the third toss,and $F: 6$ and $5$ appear respectively on the first two tosses.

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

$X$	$0$	$1$	$2$	$3$	$4$
$P(X)$	$k$	$2k$	$4k$	$6k$	$8k$

The value of $P(1 < X < 4 \mid X \leq 2)$ is equal to:

$A$ candidate takes three tests in succession and the probability of passing the first test is $p$. The probability of passing each succeeding test is $p$ if he passes the preceding one, and $\frac{p}{2}$ if he fails the preceding one. The candidate is selected if he passes at least two tests. The probability that the candidate is selected is:

Suppose that $E_1$ and $E_2$ are two events of a random experiment such that $P(E_1) = \frac{1}{4}$,$P(E_2 / E_1) = \frac{1}{2}$ and $P(E_1 / E_2) = \frac{1}{4}$. Observe the lists given below. The correct matching of List-$I$ with List-$II$ is:

List-$I$	List-$II$
$(A)$ $P(E_2)$	$(i)$ $1/4$
$(B)$ $P(E_1 \cup E_2)$	$(ii)$ $5/8$
$(C)$ $P(\bar{E}_1 / \bar{E}_2)$	$(iii)$ $1/8$
$(D)$ $P(E_1 / \bar{E}_2)$	$(iv)$ $1/2$
	$(v)$ $3/8$
	$(vi)$ $3/4$