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 $P(A \cup B)$.

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

In a certain population,$10\%$ of the people are rich,$5\%$ are famous,and $3\%$ are both rich and famous. The probability that a person picked at random from the population is either famous or rich but not both,is equal to

If $A$ and $B$ are two independent events,then the probability of occurrence of at least one of $A$ and $B$ is given by $1 - P(A') P(B')$. Is this statement true or false?

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:

If $P(A) = \frac{1}{4}$,$P(B) = \frac{5}{8}$ and $P(A \cup B) = \frac{3}{4}$,then $P(A \cap B) = $