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

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To find the probability that the next three items are flawless from a sample of 20, where 5 are flawed, one must clarify the meaning of "next" in the context of the sampling method. If sampling without replacement is assumed, the probability changes based on the remaining items, and the correct calculation involves the product of probabilities for each selection. The Maximum Likelihood Estimator for the probability of a flawless item can be calculated as (1-p), where p is the proportion of flawed items. Understanding whether the items are drawn from a larger population or if they are the next sequential items is crucial for accurate probability assessment. The discussion highlights the importance of defining the sampling context to solve the problem correctly.
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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. :/
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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