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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 $P(A \cap B)=P(\text{came from line 1 * defective})$?

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When you are doing a conditional probability calculation, it is useful to write down explicitly what you mean by $A$ and $B$. In this case, you want to know the probability the motor came from line I given that it is defective. So $A$ is the event "came from line I" and $B$ is the event "is defective." Now everything should be straightforward. We have $P(A)=1/2$. For $P(B)$, defective happens if came from I and is defective or came from II and is defective. Thus P(B)=(1/2)(0.10)+(1/2)(0.15)$. – André Nicolas Jan 23 '12 at 6:07
pefect :) u made my day – WantIt Jan 23 '12 at 6:10
im confuse in P (AB). how de we get the intersection. (multiply a and b doesnt seem to work) – WantIt Jan 23 '12 at 6:13
ei last question i cant still figure out to find the intersection between a and b what i just do is 0.5 * .01 but cant get the logic behind it im trying to figure out by using venn diagram – WantIt Jan 23 '12 at 7:41
$P(AB)=P(A)P(B|A)$. $P(A)=1/2$, $P(B|A)=0.10$ (we were told that if from line I, probability defective is $0.10$). Or more concretely, think in terms of tree, $P(AB)$ is probability we picked from line I, times the probability a line I item is defective. Or more concretely still, $1000$ items in stock, $500$ from I, $500$ from I. Then approx. $50$ bad from I, $75$ bad from II. So $P(AB)=50/1000$. – André Nicolas Jan 23 '12 at 11:52
up vote 1 down vote accepted

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)

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Right. Try this: Imagine the factory has exactly 100 motors in it (giving 50 in each line). How many of them are defective? – Chad Miller Jan 23 '12 at 6:20
Can you please stop deleting your comments so I don't look like a madman talking to himself? lol – Chad Miller Jan 23 '12 at 6:21
any idea to get P(AB). my thought is finding intersection. but doesnt work – WantIt Jan 23 '12 at 6:28
ok sorry chad. my fault – WantIt Jan 23 '12 at 6:29
I think I made my last edit while you were making your comment (I don't have enough rep to post in the comment thread above). Tell me if that makes sense – Chad Miller Jan 23 '12 at 6:29

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