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

Probability of solving specific problem independently by $A$ and $B$ are $\frac{1}{2}$ and $\frac{1}{3}$ respectively. If both try to solve the problem independently, find the probability that exactly one of them problem

If $A$ and $B$ are two independent events, then $P\,(A + B) = $

If $E$ and $F$ are events such that $P(E)=\frac{1}{4}$,  $P(F)=\frac{1}{2}$ and $P(E$ and $F )=\frac{1}{8},$ find $:$ $P($ not $E$ and not $F)$.

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

Let $A$ and $B$ be independent events with $P(A)=0.3$ and $P(B)=0.4$. Find $P(A \cup B)$