If $A$ and $B$ are events such that $P(A \cup B) = \frac{5}{6}$,$P(\bar{A}) = \frac{1}{4}$,and $P(B) = \frac{1}{3}$,then $A$ and $B$ are

One of the two events must occur. If the chance of one is $\frac{2}{3}$ of the other,then the odds in favour of the other are

Statement $- I :$ If the probabilities of solving a problem by $A$ and $B$ are $1/3$ and $1/4$ respectively,then the probability that the problem is solved is $7/12$.
Statement $- II :$ The events described above are independent events.

Given that the events $A$ and $B$ are such that $P(A) = \frac{1}{2}$,$P(A \cup B) = \frac{3}{5}$,and $P(B) = p$. Find $p$ if they are mutually exclusive.

If $A$ and $B$ are two events of a random experiment such that $P(A \cup B) = 0.65$ and $P(A \cap B) = 0.15$,then $P(\overline{A}) + P(\overline{B}) = $

The probability of an event happening is the square of the probability of a second event,but the odds against the first are the cube of the odds against the second. The probabilities of the events are: