Two events $A$ and $B$ will be independent, if
$A$ and $B$ are mutually exclusive
$P\left(A^{\prime} B^{\prime}\right)=[1-P(A)][1-P(B)]$
$P(A)=P(B)$
$P(A)+P(B)=1$
$A$ and $B$ are two events such that $P(A)=0.54$, $P(B)=0.69$ and $P(A \cap B)=0.35.$ Find $P \left( A ^{\prime} \cap B ^{\prime}\right)$.
$A$ and $B$ are events such that $P(A)=0.42$, $P(B)=0.48$ and $P(A$ and $B)=0.16 .$ Determine $P ($ not $A ).$
For an event, odds against is $6 : 5$. The probability that event does not occur, 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 probability of happening at least one of the events $A$ and $B$ is $0.6$. If the events $A$ and $B$ happens simultaneously with the probability $0.2$, then $P\,(\bar A) + P\,(\bar B) = $