Four persons can hit a target correctly with probabilities $\frac{1}{2},\frac{1}{3},\frac{1}{4}$ and $\frac {1}{8}$ respectively. If all hit at the target independently, then the probability that the target would be hit, is

If $A$ and $B$ are two events such that $P\,(A \cup B) = P\,(A \cap B),$ then the true relation is

Let $A$ and $B$ be two events such that $P\overline {(A \cup B)} = \frac{1}{6},P(A \cap B) = \frac{1}{4}$ and $P(\bar A) = \frac{1}{4},$ where $\bar A$ stands for complement of event $A$. Then events $A$ and $B$ are

For any two events $A$ and $B$ in a sample space

$A$ and $B$ are two events such that $P(A)=0.54$, $P(B)=0.69$ and $P(A \cap B)=0.35.$ Find  $P \left( A ^{\prime} \cap B ^{\prime}\right)$.

For three events $A,B $ and $C$  ,$P ($ Exactly one of $A$ or $B$ occurs$)\, =\, P ($ Exactly one of $C$ or $A$ occurs $) =$ $\frac{1}{4}$ and $P ($ All the three events occur simultaneously $) =$ $\frac{1}{16}$ Then the probability that at least one of the events occurs is :