SOLUTION: In a large group of Americans in 2001-2002, the probability of visiting a doctor's office in the past 12 months was 0.80. (a) For two randomly chosen Americans in 2001-2002, what

Question 1000844: In a large group of Americans in 2001-2002, the probability of visiting a doctor's office in the past 12 months was 0.80.
(a) For two randomly chosen Americans in 2001-2002, what is the probability of at least one of them (either the first and not the second, or the second and not the first, or both) visiting a doctor that year? (Round your answer to two decimal places.)
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
In a large group of Americans in 2001-2002, the probability of visiting a doctor's office in the past 12 months was 0.80.
(a) For two randomly chosen Americans in 2001-2002, what is the probability of at least one of them (either the first and not the second, or the second and not the first, or both) visiting a doctor that year? (Round your answer to two decimal places.)
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P(x >= 1) = 1 - P(x = 0)
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= 1 - 0.2^2
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= 1 - 0.04
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= 0.96
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Cheers,
Stan H.
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