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

Odd in favor of \(\mathrm{A}\) is \(4: 6\).
\(\therefore \mathrm{P}(\mathrm{A})=\frac{4}{10}\)
Odd against \(\mathrm{B}\) is \(7: 3\)
\(\therefore \mathrm{P}(\mathrm{B})=\frac{3}{10}\)
Since only one of the events A, B and C can happen, we get
\(\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{C})=1 \)
\( \therefore \frac{4}{10}+\frac{3}{10}+\mathrm{P}(\mathrm{C})=1 \)
\( \therefore \mathrm{P}(\mathrm{C})=\frac{3}{10} \)
\( \therefore \mathrm{P}\left(\mathrm{C}^{\prime}\right)=\frac{7}{10}\)
\(\therefore\) odds against the event \(\mathrm{C}\) are \(\mathrm{P}\left(\mathrm{C}^{\prime}\right): \mathrm{P}(\mathrm{C})\)
\(=\frac{7}{10}: \frac{3}{10}\)
\(=7: 3\)