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

If $P(A) = 2/3$,$P(B) = 1/2$ and $P(A \cup B) = 5/6$,then events $A$ and $B$ are

If $A$ and $B$ are two events with $P(A \cap B) = \frac{1}{3}$,$P(A \cup B) = \frac{5}{6}$,and $P(A^C) = \frac{1}{2}$,then the value of $P(B^C)$ is

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

The two events $A$ and $B$ have probabilities $0.25$ and $0.50$ respectively. The probability that both $A$ and $B$ occur simultaneously is $0.14$. Then the probability that neither $A$ nor $B$ occurs is

$A$,$B$,and $C$ are three events,one of which must and only one can happen. The odds in favor of $A$ are $4 : 6$,and the odds against $B$ are $7 : 3$. Thus,the odds against $C$ are: