Q.There are three events $A$, $B$ and $C$, one of which must, and only one can happen. The odds against the event $A$ are 7:4 and odds against event $B$ are 5:3. Find the odds against event $C$.

Odds against A, $7:4$, gives $P(A)=4/11$. Odds against B, $5:3$, gives $P(B)=3/8$. Since exactly one of A, B, C must happen, $P(A)+P(B)+P(C)=1$: $P(C)=1-\frac{4}{11}-\frac{3}{8}$. With LCM 88: $\frac{4}{11}=\frac{32}{88}$, $\frac{3}{8}=\frac{33}{88}$, sum $=\frac{65}{88}$, so $P(C)=1-\frac{65}{88}=\frac{23}{88}$. …

Convert given odds to probabilities, use the fact that exactly-one-of-three events are mutually exclusive and exhaustive so their probabilit …

Forgetting that A, B, C being mutually exclusive and exhaustive (exactly one happens) is what licenses adding their probabilities to …