Fill in the blanks in following table :

$P(A)$	$P(B)$	$P(A \cap B)$	$P (A \cup B)$
$0.35$	...........	$0.25$	$0.6$

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

Two cards are drawn at random and without replacement from a pack of $52$ playing cards. Finds the probability that both the cards are black.

The chances to fail in Physics are $20\%$ and the chances to fail in Mathematics are $10\%$. What are the chances to fail in at least one subject ............ $\%$

If $P(A) = \frac{1}{2},\,\,P(B) = \frac{1}{3}$ and $P(A \cap B) = \frac{7}{{12}},$ then the value of $P\,(A' \cap B')$ is

If $\mathrm{A}$ and $\mathrm{B}$ are two events such that $\mathrm{P}(\mathrm{A})=\frac{1}{4}, \mathrm{P}(\mathrm{B})=\frac{1}{2}$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{8}$ find $\mathrm{P}$ $($ not $\mathrm{A}$ and not $\mathrm{B})$