Bayes theorem - is it applicable in any case? I'm studying the Bayes' Theorem and I have a doubt. In this wikipedia page there's an example of application for the following events:
A = probability of raining on Sunday
B = probability of raining on Monday

but with the following example:
Suppose we have a marble bag with 3 red marbles and 2 blue marbles
A = probability of getting a blue marble first
B = probability of getting a blue marble second

I can calculate P(A and B) = P(B|A) * P(A).
My question is: does it make sense to say, for Bayes' theorem, that
P(A and B) = P(A|B) * P(B) ?
P(A|B) seems temporally illogical.. and P(B) seems impossible to calculate with a fixed value (since it depends on A).
Where am I getting wrong? Am I missing something?
 A: If we don't distinguish between the different marbles of the same color, then in our experiment we have four possible outcomes: RR,RU,UR,UU. (I use "U" for blue to avoid confusing it with the event $B$). Conditioning on the event you're calling $B$ amounts to assuming that the second outcome is U, that is, restricting attention to the second and fourth outcomes, and updating the distribution accordingly.
Bayes' rule and the total probability formula are the main tools here. For instance you mention that the distribution of the second draw depends on the outcome of the first draw. But that doesn't mean that you can't calculate $P(B)$, it just means you have to consider all the possibilities leading into the second draw in order to calculate $P(B)$. Specifically, you get a red with probability $3/5$, in which case you have probability $1/2$ to get a blue the second time; or you get a blue with probability $2/5$, in which case you have probability $1/4$ to get a blue the second time. So $P(B)=\frac{3}{5} \frac{1}{2} + \frac{2}{5} \frac{1}{4} = \frac{8}{20} = \frac{2}{5}$.
(Here I assume from context that the draws are without replacement.)
