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
Independent but not equally likely
Mutually exclusive and independent
Equally likely and mutually exclusive
Equally likely but not independent
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 the probability that exactly one of them occurs is $\frac{2}{5}$ and the probability that $A$ or $B$ occurs is $\frac{1}{2}$ then the probability of both of them occur together is
Fill in the blanks in following table :
$P(A)$ | $P(B)$ | $P(A \cap B)$ | $P (A \cup B)$ |
$0.5$ | $0.35$ | ......... | $0.7$ |
If $A$ and $B$ are any two events, then $P(\bar A \cap B) = $
$A$ and $B$ are events such that $P(A)=0.42$, $P(B)=0.48$ and $P(A$ and $B)=0.16 .$ Determine $P ($ not $A ).$