Conditional probability. Targeting events 
Electric motors coming off two assembly lines are pooled for storage in a common stockroom, and the room contains an equal number of motors from each line. Motors are periodically sampled from that room and tested. It is known that 10% of the motors from line I are defective and 15% of the motors from line II are defective. If a motor is randomly selected from the stock-room and found to be defective, find the probability that it came from line I.

Here is my way to solve it. First it is a conditional probability. The formula is
$$P(A \mid B) = \frac{P (A\cap B) }{ P(B) }.$$
$P(B)$ = probability that it came from line 1 = $2 P_1$.
Now here is where it gets interesting. What would be $P(A\cap B)$ in that case? 
Is $P(A \cap B)=P(\text{came from line 1 * defective})$?

P(B) is not P(came from line 1) in this problem.  You are being asked to calculate P(came from line 1 | is defective) so B is "is defective" and A is "came from line 1".  You're right that P(AB) is "Came from line 1 and is defective", and if you know how to calculate P(B) correctly in this case then you're essentially doing the same thing as you would if you used Bayes' Theorem.
To calculate P(B) correctly: P(B) needs to be the probability that any given motor in the entire factory is defective, not just from line I.  Use the Law of Total Probability.
For P(AB): This is itself a conditional probability problem.  Consider P(is defective | came from line 1) = P(defective and from line 1)/P(came from line 1)
