If $A$ and $B$ are arbitrary events, then
$P(A \cap B) \ge P(A) + P(B)$
$P(A \cup B) \le P(A) + P(B)$
$P(A \cap B) = P(A) + P(B)$
None of these
Let $A$ and $B$ be independent events with $P(A)=0.3$ and $P(B)=0.4$. Find $P(A \cup B)$
If $P\,({A_1} \cup {A_2}) = 1 - P(A_1^c)\,P(A_2^c)$ where $c$ stands for complement, then the events ${A_1}$ and ${A_2}$ are
The probability that a student will pass the final examination in both English and Hindi is $0.5$ and the probability of passing neither is $0.1$. If the probability of passing the English examination is $0.75$, what is the probability of passing the Hindi examination?
If $A$ and $B$ are two events of a random experiment, $P\,(A) = 0.25$, $P\,(B) = 0.5$ and $P\,(A \cap B) = 0.15,$ then $P\,(A \cap \bar B) = $
Given two independent events $A$ and $B$ such $P(A)=0.3,\, P(B)=0.6 .$ Find $P(A $ and not $B)$