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

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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