The probabilities of three events $A , B$ and $C$ are given by $P ( A )=0.6, P ( B )=0.4$ and $P ( C )=0.5$ If $P ( A \cup B )=0.8, P ( A \cap C )=0.3, P ( A \cap B \cap$ $C)=0.2, P(B \cap C)=\beta$ and $P(A \cup B \cup C)=\alpha$ where $0.85 \leq \alpha \leq 0.95,$ then $\beta$ lies in the interval

Given two independent events $A$ and $B$ such $P(A)=0.3,\,P(B)=0.6 .$ Find  $P($ neither $A$or $B)$

Let $E$ and $F$ be two independent events. The probability that both $E$ and $F$ happens is $\frac{1}{{12}}$ and the probability that neither $E$ nor $F$ happens is $\frac{1}{2},$ then

If $A$ and $B$ are two events such that $P\left( {A \cup B} \right) = P\left( {A \cap B} \right)$, then the incorrect statement amongst the following statements is

The probability that at least one of $A$ and $B$ occurs is $0.6$. If $A$ and $B$ occur simultaneously with probability $0.3$, then $P(A') + P(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 $P \left( B \cap A ^{\prime}\right)$.