Two events $A$ and $B$ will be independent, if
$A$ and $B$ are mutually exclusive
$P\left(A^{\prime} B^{\prime}\right)=[1-P(A)][1-P(B)]$
$P(A)=P(B)$
$P(A)+P(B)=1$
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
If $P(A \cup B) = 0.8$ and $P(A \cap B) = 0.3,$ then $P(\bar A) + P(\bar B) = $
If $A$ and $B$ are two independent events such that $P\,(A) = 0.40,\,\,P\,(B) = 0.50.$ Find $P$ (neither $A$ nor $B$)
$A$ and $B$ are two events such that $P(A)=0.54$, $P(B)=0.69$ and $P(A \cap B)=0.35.$ Find $P \left( B \cap A ^{\prime}\right)$.
Let $A$ and $B$ be two events such that $P\,(A) = 0.3$ and $P\,(A \cup B) = 0.8$. If $A$ and $B$ are independent events, then $P(B) = $