If $A$ and $B$ are two events of a sample space $S$ such that $P(A)=0.2$,$P(B)=0.6$ and $P(A \mid B)=0.5$,then $P(A^{\prime} \mid B) = $

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

$A$ family has $3$ children. The probability that all the three children are girls,given that at least one of them is a girl is:

Given $P(A)=0.5, P(B)=0.4, P(A \cap B)=0.3$,then $P(A^{\prime} / B^{\prime})$ is equal to

$A$ bag contains $3$ black and $4$ white balls. Two balls are drawn one by one at random without replacement. The probability that the second drawn ball is white,is

Let $A, B$ and $C$ be three events,which are pairwise independent and $\bar{E}$ denotes the complement of an event $E$. If $P(A \cap B \cap C) = 0$ and $P(C) > 0$,then $P[(\bar{A} \cap \bar{B})|C]$ is equal to