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

Statement $- I :$ If the probabilities of solving a problem by $A$ and $B$ are $1/3$ and $1/4$ respectively,then the probability that the problem is solved is $7/12$.
Statement $- II :$ The events described above are independent events.

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 $P(A \cup B)$.

If $A$ and $B$ are mutually exclusive events,$P(A) = \frac{1}{2}$,$P(A \cup B) = \frac{3}{5}$,and $P(B') = p$,then $p = $ . . . . . . .

One of the two events must occur. If the chance of one is $\frac{2}{3}$ of the other,then the odds in favour of the other are

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: