If $P(A \cup B) = 0.8$ and $P(A \cap B) = 0.3,$ then $P(\bar A) + P(\bar B) = $

Let $A$ and $B$ are two independent events. The probability that both $A$ and $B$ occur together is $1/6$ and the probability that neither of them occurs is $1/3$. The probability of occurrence of $A$ is

If $A$ and $B$ an two events such that $P\,(A \cup B) = \frac{5}{6}$,$P\,(A \cap B) = \frac{1}{3}$ and $P\,(\bar B) = \frac{1}{3},$ then $P\,(A) = $

If $P\,(A) = 0.4,\,\,P\,(B) = x,\,\,P\,(A \cup B) = 0.7$ and the events $A$ and $B$ are independent, then $x =$

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 : $P ( E$ or  $F )$

If $E$ and $F$ are independent events such that $0 < P(E) < 1$ and $0 < P\,(F) < 1,$ then