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Math Help - Simple Random Sampling

  1. #1
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    Simple Random Sampling

    If you have a population of size n and a sample of size m what is the probability that two values x and y from the population both appear in the sample? So far, I have that each member of the population has probability m/n of being in the sample. Therefore I would guess that the probability that both x and y appear in the sample is (m/n)*(m/n)=m^2/n^2. However the answers show that the probability is (m^2)/n(n-1) but I don't understand why?
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    Quote Originally Posted by Dredito View Post
    If you have a population of size n and a sample of size m what is the probability that two values x and y from the population both appear in the sample? So far, I have that each member of the population has probability m/n of being in the sample. Therefore I would guess that the probability that both x and y appear in the sample is (m/n)*(m/n)=m^2/n^2. However the answers show that the probability is (m^2)/n(n-1) but I don't understand why?
    I'm not sure I understand your problem, so let me state a problem which I think is equivalent. You have a bowl containing n balls, 2 of which are red, the remainder white. You draw a sample of m balls without replacement.

    What is the probability that both red balls are in the sample?
    \frac{\binom{2}{2} \binom{n-2}{m-2}}{\binom{n}{m}} = \frac{m(m-1)}{n(n-1)}
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