Expected values for random variables

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Homework Help Overview

The problem involves sampling items from a lot where some are defective. The inspector tests items in a random order until a defective item is found, and the variable Y represents the number of items tested before finding a defective one. Participants are discussing the probability distribution of Y, specifically the calculations for p(1), p(2), and p(3).

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to understand the probabilities associated with the sampling process, particularly how to calculate p(2) given the conditions of the problem. Questions are raised about the interpretation of p(2) and the implications of testing non-defective items first.

Discussion Status

The discussion is ongoing, with participants exploring the definitions and calculations of the probabilities involved. Some have provided clarifications on the meaning of p(2) and the conditions under which it is calculated, but no consensus has been reached on the exact calculations.

Contextual Notes

Participants are working under the assumption that the inspector does not know the defectiveness of the items and that the testing stops upon finding a defective item. There is a focus on understanding the implications of the sampling without replacement.

cue928
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I am stuck on the following problem: Five items are to be sampled from a large lot of samples. The inspector doesn't know that three of the five sampled items are defective. They will be tested in randomly selected order until a defective item is found, at which point the entire lot is rejected. Y is the number of firing pins the inspector must test. Find, graph the probability distribution of Y.

I understand that p(1) = 3/5. But it is saying that p(2) =3/10 and p(3)=1/10, I do not see how they are calculating that.
 
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So p(2) is the probability that the second sample tested is defective. That means that you are looking at the case where the first one is non-defective (otherwise testing would stop there!) and the second one is broken.
What are the probabilities for each of those? Then how do you calculate p(2)?
 
That's what I can't figure out. Is p(2) asking what the chance of the second draw being defective?
 
It says "Y is the number of firing pins the inspector must test." and that they will stop testing the moment a defective item is found.

p(2) is "sloppy" shorthand for P(Y = 2), that is: when you do the experiment, what is the probabillity of finding the value 2 for Y.
 
cue928 said:
That's what I can't figure out. Is p(2) asking what the chance of the second draw being defective?

After finding that the first item is non-defective, how many items are left to test? How many of those are defective?

RGV
 

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