For two events $A$ and $B$,if $P(A \cup B) = \frac{3}{4}$,$P(A \cap B) = \frac{1}{4}$,and $P(A') = \frac{2}{3}$,then find $P(A' \cap B)$.

The probability that Ajay will not appear in the $JEE$ exam is $p = \frac{2}{7}$,while the probability that both Ajay and Vijay will appear in the exam is $q = \frac{1}{5}$. Then the probability that Ajay will appear in the exam and Vijay will not appear is:

$A$ problem in mathematics is given to $4$ students whose chances of solving it individually are $\frac{1}{2}, \frac{1}{3}, \frac{1}{4},$ and $\frac{1}{5}.$ The probability that the problem will be solved by at least one student is

Two students appeared simultaneously for an entrance exam. If the probability that the first student gets qualified in the exam is $\frac{1}{4}$ and the probability that the second student gets qualified in the same exam is $\frac{2}{5}$,then the probability that at least one of them gets qualified in that exam is

The 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:

If $P(A)=0.59, P(B)=0.30$ and $P(A \cap B)=0.21$,then $P(A^{\prime} \cap B^{\prime})$ is equal to