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The following is an excerpt from OpenStax's Introductory Business Statistics text:

True random sampling is done with replacement. That is, once a member is picked, that member goes back into the population and thus may be chosen more than once. However for practical reasons, in most populations, simple random sampling is done without replacement. Surveys are typically done without replacement. That is, a member of the population may be chosen only once. Most samples are taken from large populations and the sample tends to be small in comparison to the population. Since this is the case, sampling without replacement is approximately the same as sampling with replacement because the chance of picking the same individual more than once with replacement is very low.

In a college population of 10,000 people, suppose you want to pick a sample of 1,000 randomly for a survey. For any particular sample of 1,000, if you are sampling with replacement,
• the chance of picking the first person is 1,000 out of 10,000 (0.1000);
• the chance of picking a different second person for this sample is 999 out of 10,000 (0.0999);
• the chance of picking the same person again is 1 out of 10,000 (very low).

If you are sampling without replacement,
• the chance of picking the first person for any particular sample is 1000 out of 10,000 (0.1000);
• the chance of picking a different second person is 999 out of 9,999 (0.0999);
• you do not replace the first person before picking the next person.

I do not understand some of the content in the bullet points. In the first bullet point, the authors claim that "the chance of picking the first person is 1,000 out of 10,000 (0.1000)". Who is this "first person" they are referring to and how was this probability computed?

Similarly, in the second bullet point, they assert that "the chance of picking a different second person for this sample is 999 out of 10,000 (0.0999)". Who is this "second person" and how was THAT probability computed?

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    $\begingroup$ That these reasonable questions are raised and not answered within the quotation, and the fact that some of its points are obviously wrong (e.g., with sampling without replacement, the chance of picking a different second person is $1$ by definition, not $0.0999$), suggest finding another source for learning these elementary concepts: there are many good selections available, especially in textbooks. $\endgroup$ – whuber Aug 12 at 12:58
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Welcome to Cross Validated!

Answer edited

The language used there is absolutely horrendous. What we are really talking about is unordered selection with and without replacement in standard combinatorics.

See if this is clearer.

With Replacement You are choosing 1,000 people at random from 10,000.

The probability of picking any single person is $10^{-4}$. If you choose with replacement, then the selections are completely independent, and the probability of choosing some particular set of 1,000 people is

$\sum_{n=1}^{1000} 10^{-4} = 1,000 \times 10^{-1} = 0.1$

If we want to make it complicated, we could break it down this way:

  1. We will pick 1000 people, p1, 2, ..., p1000
  2. The probability of picking some random individual as the first person, p1, is 1 in 10,000.
  3. When we pick the second person, p2, there are two things that can happen. We pick the same person again or a different person.
  4. Let's partition the 10,000 people into two groups: (1) The first person we picked (p1), and (2) everone else (p2, ..., p10000).
  5. The probability of p2 coming from group (1) is 1 in 10,000
  6. The probability of p2 coming from group (2) 9,999 in 10,000 respectively.
  7. If p2 comes from group (1), then p2 is the same person as p1, and the probability of choosing them is exactly 1. There is only one person (p1) in group (1).
  8. If the person comes from group (2), then the probability of selecting a particular person is 1 in 9,999.
  9. The probability of choosing the second person is, therefore,

$\frac{1}{10,000} \times 1 + \frac{9,999}{10,000} \times \frac{1}{9,999} = \frac{2}{10,000} = 2 \times 10^{-4}$ as we already knew.

If we keep going in this vein, we get $1,000 \times \frac{1}{10,000} = 0.1$.

I think that is what the text did, and I would not want to do it that way again.

If you sample without replacement, then the selections are not independent as you can no longer choose a person you have previously chosen. When you choose p1 you have 10,000 options. However, when you choose p2, you only have 9,999 remaining options.

This sort of selection is called unordered selection without replacement.

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    $\begingroup$ It is important to use clear, precise language. Descriptions like "picking any single person," "choosing some particular 1,000," and "picking the first person" have multiple possible interpretations. Indeed, that's precisely what is the matter with the original text! Could you edit your answer to clarify what you intended these phrases to mean? $\endgroup$ – whuber Aug 13 at 20:04
  • $\begingroup$ Thanks @whuber. I'll see what I can do. $\endgroup$ – abalter Aug 13 at 20:16
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I have read this passage about a dozen times and it doesn't make any sense to me. Only the author knows what they meant by 'the chance of picking the first person' etc.

I also have issues with the claim that 'true random sampling is with replacement'. With and without replacement are different and must each be analysed appropriately, but one is no more 'truly random' than the other.

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