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"?
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.
$$P(\text{win}) = P(\text{rain})\cdot P(\text{win given rain})+P(\text{no rain})\cdot P(\text{win given no rain})$$
$$P(\text{win}) = 0.4 \cdot 0.3 + 0.6 \cdot 0.55 = \boxed{0.45}=\frac{9}{20}$$