Question 1043349: In a sample of 1200 U.S. adults, 188 dine out at a restaurant more than once per week. Two U.S. adults are selected at random from the population of all U.S. adults without replacement. Assuming the sample is representative of all U.S. adults, complete parts (a) through (d).
(a) Find the probability that both adults dine out more than once per week.
The probability that both adults dine out more than once per week is
nothing.
(Round to three decimal places as needed.)
(b) Find the probability that neither adult dines out more than once per week.
The probability that neither adult dines out more than once per week is
nothing.
(Round to three decimal places as needed.)
(c) Find the probability that at least one of the two adults dines out more than once per week.
The probability that at least one of the two adults dines out more than once per week is
nothing.
(Round to three decimal places as needed.)
(d) Is this considered unusual? If yes explain
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Probability one adult will dine out more than once a week is 188/1200
Probability both selected will dine out more than once a week is (188/1200)(187/1199)=0.024
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Probability neither dines out is (1012/1200)*(1011/1199)=0.711
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Probability at least one should be the complement of 1-0711=0.289
This can be checked by the ways one can occur
first and not second is (188/1200)(1012/1199)=0.132
not first but second is (1012/1200)(188/1199), same product just different order=0.132
both (since at least one) and that is 0.024
Those three add to 0.288, and rounding error is reason it isn't the complement exactly.
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As for the last part, unusual vs. usual is not a statistical question. The answers are what they are. If someone wishes to state whether it were surprising, it would be incumbent upon them to state what they thought and why.
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