Let $A$ and $B $ be two events such that  $P\left( {\overline {A \cup B} } \right) = \frac{1}{6}\;,P\left( {A \cap B} \right) = \frac{1}{4}$ and $P\left( {\bar A} \right) = \frac{1}{4}$ where $\bar A$ stands for the complement of the event $A$. Then the events $A$ and$B$ are

Let $A$ and $B$ are two events and $P(A') = 0.3$, $P(B) = 0.4,\,P(A \cap B') = 0.5$, then $P(A \cup B')$ is

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) = $

Let $\mathrm{E}$ and $\mathrm{F}$ be events with $\mathrm{P}(\mathrm{E})=\frac{3}{5}, \mathrm{P}(\mathrm{F})$ $=\frac{3}{10}$ and $\mathrm{P}(\mathrm{E} \cap \mathrm{F})=\frac{1}{5} .$ Are $\mathrm{E}$ and $\mathrm{F}$ independent ?

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) = $

Given two independent events $A$ and $B$ such $P(A)=0.3,\, P(B)=0.6 .$ Find $P(A $ and not $B)$