Suppose there are 2 defective among five items

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The discussion revolves around determining the maximum number of tests needed to identify two defective items among five. It is established that at most three tests are required, but clarification is sought on whether further testing is needed to confirm the last item’s status. The conversation also touches on calculating the probability of finding the first defective item during the testing process, suggesting the use of Bayes Theorem and counting possible outcomes. Simplified examples are provided to illustrate the probability calculations, emphasizing the importance of considering all combinations of defective items. The overall focus is on understanding the testing strategy and probability assessment in this scenario.
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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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I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...

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