In a sample of 800800 U.S. adults, 174174 dine out at a resaurant more than once

Question

In a sample of 800800 U.S. adults, 174174 dine out at a resaurant more than once per week. TwoTwo 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) Which of the events can be considered unusual? Explain. Select all that apply. A. The event in part left parenthesis a right parenthesis is unusual because its probability is less than or equal to 0.05The event in part (a) is unusual because its probability is less than or equal to 0.05. B. None of these events are unusualNone of these events are unusual. C. The event in part (b) is unusual because its probability is less than or equal to 0.05. D. The event in part (c) is unusual because its probability is less than or equal to 0.05

Explanation / Answer

a) probability that both adults dine out more than once per week ==P(first one go)*P*(second one go)

=(174/800)*(173/799)=0.047

b) probability that neither adult dines out more than once per week =P(first one not go)*P*(second one not go)

=(1-174/800)*(1-174/799)=0.612

c)

probability that at least one of the two adults dines out more than once per week =1-P(neither go)

=1-0.612 =0.388

d)

option A is correct

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