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

The probabilities that $A$ and $B$ speak the truth are $\frac{4}{5}$ and $\frac{3}{4}$ respectively. The probability that they contradict each other when asked to speak on a fact is

If $A$ and $B$ are two mutually exclusive events,then $P(A \cup B) = $

$A$ party of $23$ persons take their seats at a round table. The odds against two specific persons sitting together are

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

Two candidates $A$ and $B$ have attended an interview conducted by a recruitment board for two jobs. If the probability that candidate $A$ will get the job is $0.8$ and the probability that candidate $B$ will get the job is $0.7$,then the probability that at least one of them will get the job is