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

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

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 $P(A \cup B)$.

There are three events $A, B, C$,one of which must and only one can happen. The odds are $8:3$ against $A$,$5:2$ against $B$,and the odds against $C$ are $43:17k$. Then the value of $k$ is:

The odds in favour of getting a sum that is a multiple of $3$,when a pair of dice is thrown,is:

It is $5:2$ against a husband who is $65$ years old living till he is $85$ and $4:3$ against his wife who is now $58$,living till she is $78$. If the probability that at least one of them will be alive for $20$ years is $k$,then the value of $49k$ is: