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

A die is loaded in such a way that each odd number is twice as likely to occur as each even number. If $E$ is the event that a number greater than or equal to $4$ occurs on a single toss of the die then $P(E)$ is equal to

If $A$ and $B$ are any two events, then $P(\bar A \cap B) = $

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

Let $A$ and $B$ be independent events with $P(A)=0.3$ and $P(B)=0.4$. Find  $P(A \cap B)$

If $A$ and $B$ are two events such that $P\,(A \cup B) = P\,(A \cap B),$ then the true relation is