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 $K=$ . . . . . . .

If $P(A) = \frac{1}{4}$,$P(B) = \frac{5}{8}$ and $P(A \cup B) = \frac{3}{4}$,then $P(A \cap B) = $

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

If $A$ and $B$ are mutually exclusive events,then the value of $P(A \cup B)$ is

$A, B, C$ are three events,one of which must and only one can happen. The odds in favour of $A$ are $4:6$,and the odds against $B$ are $7:3$. Then,the odds against $C$ are:

Given two mutually exclusive events $A$ and $B$ such that $P(A) = 0.45$ and $P(B) = 0.35,$ then $P(A \text{ or } B) =$