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A manufacturer of computer keyboards has assembly plants at two different locations, one in Iowa and the other in South Carolina. The keyboard faults in general are classified into three different categories depending on the location: fault related to a letter key, a number key, or other function key. The summary of last year’s inspection of production estimated the fault rates as follows:

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If a keyboard from this manufacturer is chosen at random, are the events “faulty letter key” and “produced in South Carolina facility” independent of each other?

I just want to ask and verify if my reasoning is correct. First of all, to check if two event are independent, you simply use $P(A \mid B) = P(A) P(B)$ (where $A$ and $B$ are the events you want to check for independence. If you multiply those things, and when they are equal, then they are independent.

My $P(A)$ is the fauly letter key and its $90/250$ and $P(B)$ (from South Carolina) is $150/250$. Are those numbers correct? I just simply count the number it matches in the table over the total items. How about $P(A \mid B)$? How do we get it? Thanks.

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Two events are independent when $P(A \cap B) = P(A)P(B)$. If $P(B) \neq 0$, this is the same as saying $P(A | B) = P(A)$. The conditional probability is defined as $P(A | B) = \frac{P(A \cap B)}{P(B)}$. – Mikko Korhonen Jan 26 '12 at 14:09
@m.k. what would be P(A | B) according to the table? – WantIt Jan 26 '12 at 14:20
are the events “faulty letter key” and “produced in South Carolina facility” independent of each other? – WantIt Jan 26 '12 at 14:29
People aren't going to just do your homework problem for you. What ideas do you have to try to find those probabilities? – Chad Miller Jan 26 '12 at 15:36
There's a little glitch. You've given us the number of defective items. But, out of how many items surveyed, are so many defective. Then, there's some hope of an answer! – user21436 Jan 26 '12 at 15:39

As the first comment points out, the actual formula you want to test is:

P(A and B) = P(A)P(B)

If that equation is true, you have independence.

So the next step is to calculate all those probabilities. You are on the right track for P(A) and P(B), so use identical reasoning for P(A and B).

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