My question is about: if I am given simple prior probabilities, how do I calculate "complex" (events formed from the simple events in the simple prior probabilities) posterior probabilities? I am aware of Bayes' Rule, where I can go between the posterior and prior probabilities.
A specific question: Suppose there are three aircraft. We want to travel from our current location to a remote location, and back again. Let us call the remote location $d$ and the local location $l$. Thus, there are two "moves" (transits). The event $D$ means the aircraft arrives late to the remote location and the event $L$ means the aircraft arrives late to the local location. The probability of choosing aircraft $i$ is given.
$$ P(A_1) = 0.5, P(A_2) = 0.3, P(A_3) = 0.2 $$
We are given the probability of being late to the remote location if we are on an aircraft $i$. We are also given the probability of being late to the local location if we are riding on an aircraft $i$.
$$ P(D|A_1) = 0.3, P(L|A_1) = 0.1 $$ $$ P(D|A_2) = 0.25, P(L|A_2) = 0.2 $$ $$ P(D|A_3) = 0.4, P(L|A_3) = 0.25 $$
Then, how do I calculate the probability that I am on airline 1, given the observation that I am late on one and exactly one transit?
Note that this is not a simple event, it is the event $(D \cap L') \cup (D' \cap L)$, where $'$ means event negation. This is causing me difficulties in using Bayes' rule. If it were asking for probabilities involving simple events like $P(A_1 | D)$ or $P(A_1 | D')$, I could immediately use Bayes' rule.
My understanding is the probability we are looking for is $P(A_1 | (D \cap L') \cup (D' \cap L))$. I have already tried doing $ P(A_1 | D') \times P(A_1 | L) + P(A_1 | D) \times P(A_1 | L') $. This does not seem to match the expression $P(A_1 | (D \cap L') \cup (D' \cap L))$.
Also, the chance of a late arrival at L is unaffected by a late arrival at D. This question comes before the section on independence.
Also, there was a hint to use a tree diagram and label the branches in terms of number of late arrivals (0,1,2). I have tried this but I got stuck. Specifically, my first branch is: (late remote branch up, p = 0.305), (not late remote branch down, p = 0.695). These probabilities were calculated by enumerating all possible (late, airline) combinations and adding the appropriate probabilities.
Then, out of the first two branches, I can branch further by doing (aircraft $i$ given late remote, p = determined through Bayes' since they are simple events), (aircraft $i$ given not late remote, p = determined through Bayes' since they are simple events). At this stage, I have six branches with 2 levels, and I am unsure of how to continue. Also, the hint in the book seemed to imply we only needed 2 levels.
Any help would be appreciated, specifically on how to solve this type of problem.