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 $P(A \cup B)$.

$A$ person $P$ speaks the truth in $75\%$ of cases and another person $R$ in $80\%$ of cases. What is the probability that they are likely to contradict each other in narrating the same event?

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

For two given events $A$ and $B$,$P(A \cap B)$ is:

For two events $A$ and $B$,the probability that at least one of them occurs is $0.6$ and the probability that both occur is $0.2$. Then $P(A) + P(B) = \dots$

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