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In a sample of 1000 U.S. adults, 214 dine out at a resaurant more than once per

ID: 3294890 • Letter: I

Question

In a sample of 1000 U.S. adults, 214 dine out at a resaurant 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 (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 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 Round to three decimal places as needed.) (d) Which of the events can be considered unusual? Explain. Select all that apply

Explanation / Answer

A)Probability one adult will dine out more than once a week is 214/1000=0.214

Probability both selected will dine out more than once a week is (214/1000)(213/999)=0.046
B)Probability neither dines out is (786/1000)*(785/999)=0.618
C) Probability at least one should be the complement of

1-0.618=0.382

This can be checked by the ways one can occur
first and not second is (214/1000)(786/999)=0.169
not first but second is (786/1000)(214/999), same product just different order=0.169

Those two add to 0.338, and rounding error is reason it isn't the complement exactly

D)the answer is "A' The event with the lowest probability can be considered the most unusual. Out of the three above, that would be the case that both adults dine out more than once per week.

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