In a sample of 1200 U S adults, 158 dine out at a restaurant more than once per

Question

In a sample of 1200 U S adults, 158 dine out at a restaurant more than once per week. Two U.S adults are selected at random from the population of at U.S adults, without replacement. Assuming the sample is of all U.S adults complete parts (a) trough (d). 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 A. The event in, part (c) is unusual because its probably is less than or equal to 0.05 B. The event in part (a) is unusual because its probably is less than or equal to 0.05. C. None of these events are unusual D. The event in part (b) is unusual because its probability is less than or equal to 0.05

Explanation / Answer

Let k be number if adult dine out more than once per a week

Combinations of 1100 U.S adult out more than once per a week = 1200C2

Combinations of 158 U.S dine out more than once per a week = 158C2

a) Probability that both adults dine out more than once per a week

P(k =2) = 158C2 / 1200C2 = 0.0172

b) Probability of neither adults dine out more than once per a week (1200-158 =1042)

P(k=0) = 1042C2 / 1200C2 =0.7539

c) Probability of atleast one of two adults dine out more than once per a week

P(k 1) = 1- P(k = 0) =1-1.7539 =0.2461

d) Option B is correct

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.