The probability of this happening unlucky is [tex]\frac{1}{1000}[/tex].
Given that
You are a member of a population of 10,000 people.
You happen to be chosen for a simple random sample of size 1,000 this year.
Then you are chosen for another simple random sample of size 100 (drawn independently of the first sample) next year.
P(getting in sample 1st year) = [tex]\frac{(\frac{9999}{99})(\frac{1}{1} ) }{(\frac{10000}{1000} )}[/tex]
= [tex]\frac{1000}{10000} = \frac{1}{10}[/tex]
P(getting in sample 2nd year) = [tex]\frac{(\frac{9999}{99}) }{(\frac{10000}{1000} )}[/tex]
= [tex]\frac{100}{10000} = \frac{1}{100}[/tex]
P(getting in both samples) = [tex]\frac{1}{10} * \frac{1}{100} = \frac{1}{1000}[/tex]
Hence the answer is The probability of this happening unlucky is [tex]\frac{1}{1000}[/tex].
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