For any two independent events ${E_1}$ and ${E_2},$ $P\,\{ ({E_1} \cup {E_2}) \cap ({\bar E_1} \cap {\bar E_2})\} $ is
$ < \frac{1}{4}$
$ > \frac{1}{4}$
$ \ge \frac{1}{2}$
None of these
$A$ and $B$ are two independent events. The probability that both $A$ and $B$ occur is $\frac{1}{6}$ and the probability that neither of them occurs is $\frac{1}{3}$. Then the probability of the two events are respectively
If $P(A \cup B) = 0.8$ and $P(A \cap B) = 0.3,$ then $P(\bar A) + P(\bar B) = $
Given two independent events $A$ and $B$ such that $P(A) $ $=0.3, \,P(B)=0.6$ Find $P(A$ and $B)$.
$A, B, C$ are any three events. If $P (S)$ denotes the probability of $S$ happening then $P\,(A \cap (B \cup C)) = $
$A$ and $B$ are events such that $P(A)=0.42$, $P(B)=0.48$ and $P(A$ and $B)=0.16 .$ Determine $P (A$ or $B).$