$\mathrm{A}, \mathrm{B}, \mathrm{C}$ are three events, one of which must and only one can happen. The odds…
$\mathrm{A}, \mathrm{B}, \mathrm{C}$ 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
$7:3$
$4:6$
$6:4$
$3:7$
Solution
Odd in favor of $\mathrm{A}$ is $4: 6$.
$\therefore \quad \mathrm{P}(\mathrm{A})=\frac{4}{10}$
Odd against $\mathrm{B}$ is $7: 3$
$\therefore \quad \mathrm{P}(\mathrm{B})=\frac{3}{10}$
Since only one of the events A, B and C can happen, we get
$\begin{array}{ll}
& \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{C})=1 \\
\therefore & \frac{4}{10}+\frac{3}{10}+\mathrm{P}(\mathrm{C})=1 \\
\therefore & \mathrm{P}(\mathrm{C})=\frac{3}{10} \\
\therefore \quad & \mathrm{P}\left(\mathrm{C}^{\prime}\right)=\frac{7}{10}
\end{array}$
$\therefore \quad$ odds against the event $\mathrm{C}$ are $\mathrm{P}\left(\mathrm{C}^{\prime}\right): \mathrm{P}(\mathrm{C})$
$\begin{aligned}
& =\frac{7}{10}: \frac{3}{10} \\
& =7: 3
\end{aligned}$