"Mike races with another cyclist, the probability that Mike wins is 0.8 when it is a rainy day, and the probability that Mike wins on a dry day is 0.3. The probability that it is a rainy day is 0.3
-Find the probability that Mike wins the race"
From the example, it drew a tree diagram and showed that the answer would be 0.3 x 0.8 + 0.7 x 0.3 = 0.45.
But from my logical point of view, this should also give the right answer: P(M|R) + P(M|D), where the first is the probability of him wining when its raining and the probability of him wining when its given that its not rainy (dry).
So, it should be: Intersection of M and R divided by P(R) + Intersection of M and D divided by P(D), and it is given that the intersection when it rains is 0.8 and of when its dry is 0.3, so I dont understand how this would be wrong: 0.8/0.3 + 0.3/0.7.