$A, B, C$ are three events, one of which must and only one can happen the odds in favour of $A$ are $4: 6$,…
$A, B, C$ are three events, one of which must and only one can happen the odds in favour of $A$ are $4: 6$, odds against $B$ are $7: 3$, then odds against $C$ are
7:3
3:7
6:4
4:6
Solution
$P(A)=\frac{4}{4+6}=\frac{2}{5}, P(B)=\frac{3}{3+7}=\frac{3}{10}$
$\because A, B$ and $C$ are mutually exclusive and exhaustive
$\begin{aligned}
& \Rightarrow P(A)+P(B)+P(C)=1 \\
& \Rightarrow \frac{2}{5}+\frac{3}{10}+P(C)=1 \\
& \Rightarrow P(C)=\frac{3}{10}
\end{aligned}$
Now, odds against to the event $C=\frac{1-P(C)}{P(C)}=\frac{1-\frac{3}{10}}{\frac{3}{10}}=\frac{7}{3}$