Probability of the horse winning, given the chance of rain Here's the question:

In the past two racing seasons Seahorse has won 55% of the time if the track is 
     dry. On rainy days when the track is muddy he won only 30% of the time. For the
   next race day there is a 40% chance of rain. What is the probability that
   Seahorse will win the next race?

This what I attempted:
$$
P(Dry) = \frac{55}{100} = \frac{11}{20}\\
P(Muddy) = \frac{30}{100}=\frac{3}{10}\\
P(Rain) = \frac{40}{100} = \frac{2}{5}
$$
Since $P(Muddy)$ and $P(Rain)$ are dependent events, I did the following:
$$
P(Muddy ∩ Rain) = \frac{3}{10} \times \frac{2}{5} = \frac{3}{25}
$$
Does anybody know if I'm right or am I totally on the wrong "track"?
 A: The way I read the question, rainy days are exactly the days on which the track is muddy, so there shouldn't be two separate probabilities for that. 
$$P(\text{win}) = P(\text{rain})\cdot P(\text{win given rain})+P(\text{no rain})\cdot P(\text{win given no rain})$$
$$P(\text{win}) = 0.4 \cdot 0.3 + 0.6 \cdot 0.55 = \boxed{0.45}=\frac{9}{20}$$
A: You are sort of on the right track but you have run only half the race.  You also have to add the probability of winning of a driy track which is 
$$
\frac{11}{20} \times \frac{3}{5} == 0.33
$$
Another helpful hint:  $\frac{30}{100}$ is not quite $\frac{1}{3}$.
A: For clarity it is best to simply identify the two events as simply Winning and Raining.   Don't muddy the waters by giving the conditional events confusing names.   Always label things as clearly as possible, because it then often becomes obvious what you need to do.
For brevity let's use $W$ for Wins and $R$ for Rains (and $\neg R$ for not rains).

In the past two racing seasons Seahorse has won 55% of the time if the track is 
   dry.

This is the conditional probability that Seahorse wins when it does not rain.
$$\mathsf P(W\mid \neg R) = 55\%$$

On rainy days when the track is muddy he has won only 30% of the time. 

This is the conditional probability of winning when it rains:
$$\mathsf P(W\mid R) = 30\%$$

For the next race day there is a 40% chance of rain. 

This is what it sais:
$$\mathsf P(R) = 40\%$$

What is the probability that Seahorse will win the next race?

This is obtained by using the Law of Total Probability:
$$\begin{align}\mathsf P(W) & = \mathsf P(W\mid\neg R)\mathsf P(\neg R)+\mathsf P(W\mid R)\mathsf P(R)
\\ & = 0.55\times(1-0.40) + 0.30\times 0.40
\\ & =
\end{align}$$
