$$A$$ is one of $$6$$ horses entered for a race, and is to be ridden by one of two jockeys $$B$$ and $$C$$. It is $$2$$ to $$1$$ that $$B$$ rides $$A$$, in which case all the horses are equally likely to win. If $$C$$ rides $$A$$, his chance of winning is trebled. What are the odds against winning of $$A\,?$$

Solution :
Let
$$E = $$  the event that horse $$A$$ wins
$${E_1} = $$  the event that jockey $$B$$ rides horse $$A$$
$${E_2} = $$  the event that jockey $$C$$ rides horse $$A$$
According to question odds in favour of $${E_1} = 2:1$$
$$ \therefore \,P\left( {{E_1}} \right) = \frac{2}{3}{\text{ and }}P\left( {\frac{E}{{{E_1}}}} \right) = \frac{1}{6}$$
(Since, when $$B$$ rides $$A$$, all six horses are equally likely to win)
$$\eqalign{ & P\left( {{E_2}} \right) = 1 - P\left( {{E_1}} \right) = 1 - \frac{2}{3} = \frac{1}{3} \cr & {\text{and }}P\left( {\frac{E}{{{E_2}}}} \right) = 3P\left( {\frac{E}{{{E_1}}}} \right) = \frac{1}{2} \cr & {\text{Let }}{A_1} = {E_1} \cap E{\text{ and }}{A_2} = {E_2} \cap E \cr & {\text{Now, required probability}} \cr & P\left( E \right) = P\left( {{A_1}} \right) + P\left( {{A_2}} \right) \cr & = P\left( {{E_1} \cap E} \right) + P\left( {{E_2} \cap E} \right) \cr & = P\left( {{E_1}} \right)P\left( {\frac{E}{{{E_1}}}} \right) + P\left( {{E_2}} \right)P\left( {\frac{E}{{{E_2}}}} \right) \cr & = \frac{2}{3}.\frac{1}{6} + \frac{1}{3}.\frac{1}{2} \cr & = \frac{5}{{18}} \cr & {\text{So, that odds against winning of }}A{\text{ are }}13:5 \cr} $$