Four persons independently solve a certain problem correctly with probabilities $\frac{1}{2}, \frac{3}{4}, \frac{1}{4}, \frac{1}{8}$. Then the probability that the problem is solved correctly by at least one of them is

In a competition,$A, B$,and $C$ are participating. The probability that $A$ wins is twice that of $B$,and the probability that $B$ wins is twice that of $C$. Then,the probability that $A$ loses is:

If $P(A) = \frac{1}{2}$,$P(B) = \frac{1}{3}$ and $P(A \cap B) = \frac{1}{4}$,then the value of $P(A' \cap B')$ is

An event $A$ is independent of itself if $P(A) = \dots$

Two students Anil and Ashima appeared in an examination. The probability that Anil will qualify the examination is $0.05$ and that Ashima will qualify the examination is $0.10$. The probability that both will qualify the examination is $0.02$. Find the probability that both Anil and Ashima will not qualify the examination.

The probability of at least one of the events $A$ and $B$ occurring is $0.6$. If the probability of both $A$ and $B$ occurring simultaneously is $0.3$,then $P(A') + P(B') = \dots$