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Really no idea how to go about this. I thought about using a uniform normal distribution law but the answers I got made no sense.

In a country that has a population between 1500000 and 3000000 people, there are 2262 legal first names and 1030 legal last names. A statistician picks a sample of 225 000 people to try to compute the name distribution. However, he is informed that although the name distribution is unknown, there is necessarily two people with the same First name + last name combination.

What is the probability that the sample represents between 7.4% and 8.6% ?

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The fact that there must be a duplicate first name/last name pair means there are at least ???? people in the population. For how big a population would $225,000$ be $7.4\%$? For how big a population would $225,000$ be $8.6\%$? I think you are supposed to find the interval of population that meets the criteria, assume that all values are equally likely (!?!?) and find what fraction of the values your sample is within the range $7.4\%$ to $8.6\%$. A badly worded question.

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  • $\begingroup$ Thank you for your reply. I tried to translate from French as possibly as I could but even in the original language the question is not very clear. "The fact that there must be a duplicate first name/last name pair means there are at least ???? people in the population.". Could you elaborate on this ? I would assume that given the number of possible combinations, that number of people in the population would be far below the given limits (1.5M and 3.0M). $\endgroup$ – Jack Jan 26 '16 at 16:41
  • $\begingroup$ No, given the number of first names and last names, how many pairs are there? It is in the range of interest. For each choice of first name, you have (how many?) choices of last name $\endgroup$ – Ross Millikan Jan 26 '16 at 16:46
  • $\begingroup$ There are 2262*1030 = 2329860 combinations. From what I'm understanding, you're saying that since there MUST be two people with the same name, that the population size has to be 2329860 + 1 or higher ? Is that a safe assumption ? $\endgroup$ – Jack Jan 26 '16 at 16:49
  • $\begingroup$ Yes, that is what I was saying. We have now reduced the range of possible populations to $ [2329861,3000000]$. This is the interval I was talking about. $\endgroup$ – Ross Millikan Jan 26 '16 at 16:51
  • $\begingroup$ Thank you. I just tried a counter-example. I assumed that the population size is 1.5M, and calculated the probability of having exclusively unique names. The probability is (2329860!/(2329860 - 1.5M)!)/(2329860^1.5M). I used a formula similar to the classic "birthday problem" formula. The result comes to 3.0 * 10^(-279395), which can be safely approximated as 0. $\endgroup$ – Jack Jan 26 '16 at 16:56

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