Three persons $P, Q$ and $R$ independently try to hit a target . If the probabilities of their hitting the target are $\frac{3}{4},\frac{1}{2}$ and $\frac{5}{8}$ respectively, then the probability that the target is hit by $P$ or $Q$ but not by $R$ is

Let $A$,$B$ and $C$ be three events such that $P\left( {A \cap \bar B \cap \bar C} \right) = 0.6$, $P\left( A \right) = 0.8$ and $P\left( {\bar A \cap B \cap C} \right) = 0.1$, then the value of $P$(atleast two among $A$,$B$ and $C$ ) equals

The probabilities of occurrence of two events are respectively $0.21$ and $0.49$. The probability that both occurs simultaneously is $0.16$. Then the probability that none of the two occurs is

If $A$ and $B$ are any two events, then the probability that exactly one of them occur is

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

Given that the events $A$ and $B$ are such that $P(A)=\frac{1}{2}, P(A \cup B)=\frac{3}{5}$ and $P(B)=p .$ Find $p$ if they are independent.