The probability that a man will be alive in $20$ years is $\frac{3}{5}$ and the probability that his wife will be alive in $20$ years is $\frac{2}{3}$. Then the probability that at least one will be alive in $20$ years, is

The probabilities of three mutually exclusive events are $\frac{2}{3} ,  \frac{1}{4}$ and $\frac{1}{6}$. The statement is

Let $S=\{1,2,3, \ldots, 2022\}$. Then the probability, that a randomly chosen number $n$ from the set $S$ such that $\operatorname{HCF}( n , 2022)=1$, is.

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

If $A$ and $B$ are two independent events such that $P(A) > 0.5,\,P(B) > 0.5,\,P(A \cap \bar B) = \frac{3}{{25}},\,P(\bar A \cap B) = \frac{8}{{25}}$ , then $P(A \cap B)$ is 

If the odds in favour of an event be $3 : 5$, then the probability of non-occurrence of the event is