In a horse race,the odds in favour of three horses are $1:2$,$1:3$,and $1:4$. The probability that one of the horses will win the race is

Let $A, B,$ and $C$ be events such that $P(A) = P(B) = P(C) = \frac{1}{4}, P(AB) = P(CB) = 0,$ and $P(AC) = \frac{1}{8}.$ Then $P(A \cup B) = \dots$

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:

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

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

The probability that a leap year selected at random contains either $53$ Sundays or $53$ Mondays is: