How to find the probability that the next three are not flawed?

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Homework Help Overview

The discussion revolves around finding the probability that the next three items chosen from a sample of size n=20, with 5 flawed items, will be flawless. Participants are exploring the implications of sampling methods and the definitions of probability in this context.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants are questioning the meaning of "next" in the context of the problem and whether it refers to items 21, 22, and 23. There is discussion about whether the sampling is done with or without replacement and how that affects the probability calculations. Some participants suggest considering the size of the overall population from which the sample is drawn.

Discussion Status

There is an ongoing exploration of the problem's assumptions and definitions. Some participants have offered insights into the implications of sampling methods, while others express confusion about the requirements of the problem and the relationship between the Maximum Likelihood Estimator and the probability of selecting flawless items.

Contextual Notes

Participants note the lack of clarity regarding the total population size and the implications of sampling without replacement. There is also mention of a previous part of the problem that relates to finding the Maximum Likelihood Estimator, which adds complexity to the current question.

nontradstuden
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Homework Statement



If you have a sample of size n=20, where X=the number among the n that are flawed=5, how do you go about finding the probability that the next three chosen will be flawless?

do I say (15*14*13*12*11)/ (20*19*18*17*16) .

Or is it (15/20)^5?


How do I go about solving this?
 
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nontradstuden said:

Homework Statement



If you have a sample of size n=20, where X=the number among the n that are flawed=5, how do you go about finding the probability that the next three chosen will be flawless?

do I say (15*14*13*12*11)/ (20*19*18*17*16) .

Or is it (15/20)^5?


How do I go about solving this?

I don't understand the question. You say we have 20 items among which 5 are flawed. OK so far? Now you ask about the next three. What does "next" mean here? Do you mean items 21, 22 and 23, or something else? If you mean items 21-23, you need to know something about the probability that a randomly-chosen item is flawed; also, you need to specify the size of the whole population. That is, did we choose 20 from a population of size 24 or size 2400 or size ∞, or what? (It makes a big difference.)

RGV
 
If this were "sampling with replacement" or if we were taking sample from a much, much larger "universe", then the probability of any three in a row (or any three) would be the same. But with this problem, the probability the first three will be flawed may be quite different from the probability of the last three. That is why we need to know what you mean by "the next three". Are you assuming that some number have already been chosen?
 
Hi folks. Sorry for the late reply.

I really don't know. This question is part three of my problem. I'm supposed to find the Maximum Likelihood estimator of (1-p)^3. I'm told to remember that this is both a probability and a parameter.

I'm confused. The first part asks me to find the Maximum Likelihood Estimator of p for n=20 and X= the number among the n that are flawed. So, I found that to be p-hat= X/n.

This part asks me to find the Maximum Likelihood estimator of (1-p)^3 given n=20 and X=5. Also, to find the probability that the next three are flawless.

This is all I'm given. I don't know where to go with it. I though I would just have to make (1-p) hat= 15/20 and then use sampling without replacement, but I'm grasping at straws.
 
I was searching for a way to edit my last response, but I don't see one. anywho, I've solved it. Forgot about assuming independence and the invariance principle. This can be deleted to cyber space because it is not contributing to the site, at all. :/
 

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