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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.

Thanks.

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2 Answers 2

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Call $E = (D\cap L') \cup (D' \cap L)$ Then, assuming arrivals and departings are independent, $P(E) = P(D)(1-P(L)) + P(L)(1-P(D))$. The same applies (with the same assumption!) for the conditioned events (on A, B or C).

So, you can compute $P(E | A) = P(D|A)(1-P(L|A)) + P(L|A)(1-P(D|A))$.

And from that you can apply Bayes formula to get $\displaystyle P(A | E) = \frac{P(E|A) P(A)}{P(E)} $

The denominator, of course, if computed by summing the numerator for all aircrafts (A, B ,C).

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For the purpose of learning, if $A,B$ are independent events given an event $C$ occurring, $P(A \cap B | C) = P(A|C) \times P(B|C)$. The formal term for this is "conditional independence". In this problem, knowing that we are taking aircraft 1, knowledge of being late to the remote location does not affect our knowledge of being late to the local location, and vice versa. –  jrand Jun 6 '11 at 17:03
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Instead of trying to break up $P(A_1|(D\cap L') \cup (D' \cap L))$, use Bayes rule to reduce this to calculating $P((D\cap L') \cup (D' \cap L))$ and $P((D\cap L') \cup (D' \cap L)|A_1)$. Now this can be broken up since the events $D \cap L'$ and $D' \cap L$ are disjoint. Finally, since $D$ and $L$ are independent, you can calculate the latter as $P(D \cap L') = P(D) P(L')$ and so on. Same applies for the probabilities conditioned on $A_1$.

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