
Re: Conditional Probability Question
Posted:
Sep 25, 2001 1:50 PM


Aaron Davies wrote: > > Hi, I'm a senior at Columbia University majoring in Computer > Engineering, and I'm taking my first course on probability this > semester. The following problem came up in class during the past two > sessions and was the subject of much debate. > > There is a bucket containing 100 machine parts, one of which is known to > be defective. You draw parts from the bucket one at a time, examining > them for the defect, and then set them aside (ie no replacement). What > is the probability that the ith part you examine will be defective? > > The professor is of the opinion that the probability is always 1/100 and > is thus independant of i. Most of the students are of the opinion that > it is 1/(101i). Which is correct? Yesterday, the professor presented > some math he'd worked out to support his answer; if anyone's interested, > I'll post that too.
It could be a question of what probability is being asked for. Here is the question you posed:
> What is the probability that the ith part you examine > will be defective?
Interpretation #1: I pull all of the parts out of the bucket and lay them down in a row. What is the probability that the defective part is the ith one in the row? Answer: 1/100. Exactly 1/100 of all possible arrangements have the bad part in position i.
Interpretation #2: Given that I have already pulled out (i1) parts and found them to be good, what is the probability that the next part is bad? In that case, the answer is 1/(100  (i1)) = 1/(101i).
Note that the second one is a conditional probability, and only counts from among those experiments in which you have already found (i1) good parts. I'd vote with your professor, since that is the unconditional probability from among all testing scenarios.
You would not be the first person to be confused about when to use a conditional probability. There are some particularly devious "paradoxes" based on conditional probability. Some of them lead to quite heated debate here.
 Randy

