Suppose there are 2 defective among five items

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The discussion centers on a probability problem involving five items, two of which are defective. It concludes that identifying the second defective item requires at most three tests. The probability of finding the first defective item on the first test, given that the second defective item is found on the third test, can be calculated using Bayes' Theorem. Participants emphasize the importance of considering all possible combinations of defective items to accurately determine probabilities.

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shinkansenfan
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Homework help:

suppose there are 2 defective among five items, you test one item at a time until you identify the second defective. find:

1. you need at most 3 tests.
2.given the second defective item found at the 3rd test, find the Probability that you found the first defective at the 1st test.
 
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One, I don't think this is the right section for your question.

Unless you mean you expect to only check 3 times I don't think that is right.
Let's say I check one and it's fine, the second one I check is defective and the third one I check is defective. I would need one more test to see which one out of the two left is defective. And that still leaves the question of do I have to check the last one to ensure it's defective for this problem or can we assume that we can use elimination and not have to check the one we know to be defective?
The last part is just a conditional probability, try Bayes Theorem.
 
Hi shinkansenfan. jwatts is right that you should be using the separate forums for homework questions.

But since you're here: If you are asking for the probabilities of the events in your questions, think about all the possible ways that two of five ordered items can be defective.

Consider a simpler example: one of two items is defective. There are two possibilities in this simple case:
A. Item 1 is defective, Item 2 is not.
B. Item 1 is not, Item 2 is defective.

What is the probability Item 1 is defective? Just count! In one of the two cases, Item 1 is defective, so the probability is 0.5.

If two of three items are defective,
A. I1 defective, I2 defective, I3 not.
B. I1 defective, I2 not, I3 defective.
C. I1 not, I2 defective, I3 defective.

What is the probability that Item 1 is defective? 2/3. Can you answer versions of your two questions in this simpler case?
What is the probability you need at most two tests to find the second defective item?
Given that you find the second defective item is Item 3, what is the probability you found Item 1 to be the first defective item?

Use the same logic to solve the full question that you asked, writing out all 10 possibilities if you really have to. (Hint: write 11000 for Items 1 and 2 defective, Items 3-5 not).
 
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