For two events $A$ and $B$,the probability that at least one of them occurs is $0.6$ and the probability that both occur is $0.2$. Then $P(A) + P(B) = \dots$

$A$ coin is tossed twice. If events $A$ and $B$ are defined as: $A = \text{head on first toss}$,$B = \text{head on second toss}$. Then the probability of $A \cup B = $

Statement $- I :$ If the probabilities of solving a problem by $A$ and $B$ are $1/3$ and $1/4$ respectively,then the probability that the problem is solved is $7/12$.
Statement $- II :$ The events described above are independent events.

The probability of an event happening is the square of the probability of a second event,but the odds against the first are the cube of the odds against the second. The probabilities of the events are:

If $A$ and $B$ are independent events such that odds in favour of $A$ is $2:3$ and odds against $B$ is $4:5$,then $P(A \cap B)=$

If a number is selected from the first $30$ natural numbers, then the probability that the number selected is divisible by $4$ or $7$ is: