Let $A$ and $B$ be independent events with $P(A)=0.3$ and $P(B)=0.4$. Find  $P(A \cap B)$

For the three events $A, B$ and $C, P$ (exactly one of the events $A$ or $B$ occurs) = $P$ (exactly one of the events $B$ or $C$ occurs)= $P$ (exactly one of the events $C$ or $A$ occurs)= $p$ and $P$ (all the three events occur simultaneously) $ = {p^2},$ where $0 < p < 1/2$. Then the probability of at least one of the three events $A, B$ and $C$ occurring is

For an event, odds against is $6 : 5$. The probability that event does not occur, is

Let $A$ and $B$ be two events such that $P\,(A) = 0.3$ and $P\,(A \cup B) = 0.8$. If $A$ and $B$ are independent events, then $P(B) = $

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

Given $P(A)=\frac{3}{5}$ and $P(B)=\frac{1}{5}$. Find $P(A $  or  $B),$ if $A$ and $B$ are mutually exclusive events.