Specific Probability Question So I had come up with this problem during my spare time and was wondering if my answers to the questions were correct? The problem is:
Let's say an individual does not go out that much from their house and lives in the basement, and so they decide to check the outside weather every time when that individual's friend/roommate goes out of the house. If the friend/roommate is wet when they go downstairs to the basement, then the weather is rainy. Moreover, if the friend/roommate is dry, then the weather is sunny. However, here is the catch:
The roommate/friend decides to flip a fair two-sided coin every time when he reaches back. If the coin reveals tails and the weather is rainy, then the roommate/friend decides to dry himself before going down to the basement, hence being dry. If the coin reveals tails and the weather is sunny, then the roommate/friend decides to take a shower before going down to the basement, hence being wet. Hence, the individual may predict the weather incorrectly if the coin lands on tails.
Now, what would the probability be such that the individual in the basement predicts that it is a rainy day and it is indeed rainy outside (finding P(Predicts rainy AND is actually rainy))? Also, what would the probability be such that the individual in the basement predicts that it is a rainy day given that it is actually rainy (P(Predicts rainy | actually rainy))? Using these probabilities, can we deduce if the individual's prediction of rain is independent of the weather actually being rainy that day?
My approach to this question was that we have 4 possible outcomes, which are:

*

*Individual predicts rainy, roommate lands on tails, weather is actually dry

*Individual predicts rainy, roommate lands on heads, weather is actually rainy

*Individual predicts sunny, roommate lands on tails, weather is actually rainy

*Individual predicts sunny, roommate lands on heads, weather is actually dry

Using this, P(Predicts rainy AND is actually rainy) = 1 outcome from 4, hence 0.25. And P(Predicts rainy | actually rainy) = 1 outcome from 2, hence 0.5. But what confuses me is finding the independence, since I would need the probability of the weather being rainy/sunny to decide the independence, but I have not included any probability within the question. So are my 3 answers to this problem correct? If not, what was wrong with my answer exactly? Thank you so much :)
 A: You are doing a great job doing a sanity check on the question and information and thinking about your answer.
Your answers are correct if you assume that there is an equal probability between it being sunny and rainy. In this case, the 4 outcomes you suggested are all equally likely.
However, if that is not the case, you will get different answers. For example, say there is a 10% chance of it raining. Then the probability of it being rainy will be greatly reduced, and instead of a uniform distribution over your 4 options, you would instead get something like this:

*

*45%

*5%

*5%

*45%

Hopefully you can see where I got these numbers. In this case, the answer to the second question is still 50%, but the first is instead 5%.
I hope that was helpful. Let me know if you have any questions.
A: Let the probability that it will rain on any given day is represented by $\mathbb P_R$. Similarly, let the probability that it will be sunny on any given day is represented by $\mathbb P_S = 1 - \mathbb P_R$. In the given scenario, we assume that the probability that it will rain on a given day is independent of whether or not it rains on the previous day or not. Then the following probability tree should give you answers to your four questions.

