Conditional probability - Electronic components company 
A small company produces electronic components and receives a report that $87\%$ of their products are faulty. The company then decides to run a test on these products. The test gives back a positive result if the product is indeed faulty, and if no problem is found, returns a negative result. $15\%$ of all the test's results are positive. The company later found out that out of all these positive results, only $90\%$ were actual faulty products.
Let $F$ be the event of the test returning a positive result, and let $A$ be the event that the product isn't faulty and works perfectly.  Calculate:
(a) $P(A), P(F)$ and $P(A^c|F)$.
(b) $P(A|F)$, $P(F|A^c), P(A^c|F^c)$ and $P(A ∩ F^c)$ .

I constructed the following tree diagram:

The part I marked with blue is confusing me quite a lot, since the probability of the empty set has to be $0$, so something must be wrong in my diagram, I just can't find it.
For (a) I have:
$P(A) = 0.13, P(F) = P(F ∩ A^c) = 0.15$ and $P(A^c|F) = 0.9.$
I'm not sure how to proceed. Any help would be much appreciated.
 A: It might be clearer if at each level in the tree you condition on the events above it, rather than taking the intersection (actual the probabilities youve written on the tree are the conditional probabilities). The values on the last level dont make sense to me, but your work does, you're seeing that by partitioning the even space by $A^c$ again, you just end up back where you were. You're right, $A \cap A^c$ has probability 0.
The rest of your work is correct except $P(F\cap A^c) = 0.9 \times 0.15$.
The rest of the problem can be solved by applying bayes rule and total probability.
A: I think the issue here is that the original report (0.87 figure) should not be part of the tree. There are two steps: the first step is whether the test says "faulty" (event F), which it does with probability $0.15$, and then the second step is checking whether the item actually is faulty. You are told that in cases where the test says "faulty", the component is actually faulty with probability $0.9$. You are not told what the corresponding probability is when the test doesn't say "faulty". So your tree looks like this:

Now you need to work out the ?s, and to do this you can use the fact (assuming the report is reliable) that $P(A)=0.13$.
I think the wording of this question is very confusing.
