$A$ card is selected from a pack of $52$ cards. Calculate the probability that the card is a black card.

Let $A$ and $B$ be events in a sample space $S$ such that $P(A)=0.5$, $P(B)=0.4$ and $P(A \cup B)=0.6$. Observe the following lists. Match List-$I$ with List-$II$ and select the correct option.

List-$I$	List-$II$
$(i) \ P(A \cap B)$	$(1) \ 0.4$
$(ii) \ P(A \cap \bar{B})$	$(2) \ 0.2$
$(iii) \ P(\bar{A} \cap B)$	$(3) \ 0.3$
$(iv) \ P(\bar{A} \cap \bar{B})$	$(4) \ 0.1$

Let $A$ and $B$ be events for which $P(A) = x$,$P(B) = y$,and $P(A \cap B) = z$. Then $P(\bar{A} \cap B)$ equals:

For two events $A$ and $B$,if $P(A \cup B) = \frac{3}{4}$,$P(A \cap B) = \frac{1}{4}$,and $P(A') = \frac{2}{3}$,then find $P(A' \cap B)$.

The probability that $A$ speaks truth is $4/5$,while the probability that $B$ speaks truth is $3/4$. The probability that $A$ and $B$ contradict each other when asked to reveal the fact is

Let $A$ and $B$ be two events such that $P(\overline{A \cup B}) = \frac{1}{6}$,$P(A \cap B) = \frac{1}{4}$ and $P(\bar{A}) = \frac{1}{4}$,where $\bar{A}$ stands for the complement of the event $A$. Then the events $A$ and $B$ are