If $P(A)=\frac{6}{11}, P(B)=\frac{5}{11}$ and $P(A \cup B)=\frac{7}{11},$ find $P(A \cap B).$

If the probability of an event occurring is $3:8$,what is the probability of the event not occurring?

The odds against a certain event is $5 : 2$ and the odds in favour of another event is $6 : 5$. If both the events are independent,then the probability that at least one of the events will happen is

If $A$ and $B$ are events such that $P(A \cup B) = \frac{5}{6}$, $P(\bar{A}) = \frac{1}{4}$, and $P(B) = \frac{1}{3}$, then $A$ and $B$ are

$A$ die is thrown. Let $A$ be the event that the number obtained is greater than $3$. Let $B$ be the event that the number obtained is less than $5$. Then $P(A \cup B)$ is

Fill in the blanks in the following table:

$P(A)$	$P(B)$	$P(A \cap B)$	$P(A \cup B)$
$0.5$	$0.35$	$.........$	$0.7$