The probability of happening at least one of the events $A$ and $B$ is $0.6$. If the events $A$ and $B$ happen simultaneously with a probability of $0.2$,then $P(\bar{A}) + P(\bar{B}) = $

Three coins are tossed once. Let $A$ denote the event 'three heads show',$B$ denote the event 'two heads and one tail show',$C$ denote the event 'three tails show',and $D$ denote the event 'a head shows on the first coin'. Which events are simple?

$A$ and $B$ toss a coin alternatively,the first to show a head being the winner. If $A$ starts the game,the chance of his winning is

$A$ man and his wife appear in an interview for two vacancies. The probability of the husband's selection is $1/7$ and the probability of the wife's selection is $1/5$. What is the probability that only one of them is selected?

If a leap year is selected at random,what is the chance that it will contain $53$ Tuesdays?

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