Let $A, B,$ and $C$ be events such that $P(A) = P(B) = P(C) = \frac{1}{4}, P(AB) = P(CB) = 0,$ and $P(AC) = \frac{1}{8}.$ Then $P(A \cup B) = \dots$

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

When two dice are thrown once,find the odds against getting a sum of $7$ on the two dice.

If $P(A)=\frac{6}{11}, P(B)=\frac{5}{11}$ and $P(A \cup B)=\frac{7}{11},$ find $P(A \cap B).$

The probability of occurrence of an event $A$ is $1/2$ and that of $B$ is $3/10$. If $A$ and $B$ are mutually exclusive, then the probability of occurrence of neither $A$ nor $B$ is (in $/5$)

If the odds against $A$ solving a problem are $4:3$ and the odds in favor of $B$ solving the problem are $7:5$,what is the probability that only one of them solves the problem?