In a sample of 1000 adults, 186 dine out at a restaurant more than once per week

Question

In a sample of 1000 adults, 186 dine out at a restaurant more than once per week. To US adults are selected at random from the population of all US adults, without replacement. Assuming the sample is representative of all US adults.
A)Find the probability that both adults done out more than once a week (round three decimal of places)
B)Find the probability that neither adult dines out more than once per week. (Round three decimal places)
C)Find the probability that at least one of the two adult stands out more than once per week (three decimal places)
D)Which event can be considered unusual? Part A,B,C, none of the events are unusual In a sample of 1000 adults, 186 dine out at a restaurant more than once per week. To US adults are selected at random from the population of all US adults, without replacement. Assuming the sample is representative of all US adults.
A)Find the probability that both adults done out more than once a week (round three decimal of places)
B)Find the probability that neither adult dines out more than once per week. (Round three decimal places)
C)Find the probability that at least one of the two adult stands out more than once per week (three decimal places)
D)Which event can be considered unusual? Part A,B,C, none of the events are unusual
A)Find the probability that both adults done out more than once a week (round three decimal of places)
B)Find the probability that neither adult dines out more than once per week. (Round three decimal places)
C)Find the probability that at least one of the two adult stands out more than once per week (three decimal places)
D)Which event can be considered unusual? Part A,B,C, none of the events are unusual

Explanation / Answer

A) p [2] = (186/1000)2 = 0.035

B) p [0] = (814/1000)2 = 0.663

C) p [>=1] = p [1] + p [2]

= 2 * (186/1000) (814/1000) + (186/1000)2

= 0.337

D) part A is unusual, Since p value < 0.05

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