The probability that a man will be alive in $20$ years is $\frac{3}{5}$ and the probability that his wife will be alive in $20$ years is $\frac{2}{3}$. Then the probability that at least one will be alive in $20$ years is:

$P(A \cup B) = P(A \cap B)$ if and only if the relation between $P(A)$ and $P(B)$ is

Two candidates $A$ and $B$ have attended an interview conducted by a recruitment board for two jobs. If the probability that candidate $A$ will get the job is $0.8$ and the probability that candidate $B$ will get the job is $0.7$,then the probability that at least one of them will get the job is

For two events $A$ and $B$,$P(A)=0.5$,$P(A \cup B)=0.6$ and $P(B)=K$ are given. If $A$ and $B$ are independent events,then $K=$ . . . . . . .

Fill in the blanks in the following table:

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

For an event,the odds against are $6 : 5$. The probability that the event does not occur is