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 $P(A \cup B) = 0.8$ and $P(A \cap B) = 0.3,$ then $P(\bar A) + P(\bar B) = $

There are three events $A, B, C$,one of which must and only one can happen. The odds are $8:3$ against $A$,$5:2$ against $B$,and the odds against $C$ are $43:17k$. Then the value of $k$ is:

It is $5:2$ against a husband who is $65$ years old living till he is $85$ and $4:3$ against his wife who is now $58$,living till she is $78$. If the probability that at least one of them will be alive for $20$ years is $k$,then the value of $49k$ is:

The odds in favour of getting a sum that is a multiple of $3$,when a pair of dice is thrown,is:

One card is drawn from a pack of $52$ cards. The probability that it is a king or a diamond is