Question

The Zagat Restaurant Survey provides food, decor, and service ratings for some of the top restaurants across the United States. For 18 restaurants located in a certain city, the average price of a dinner, including one drink and tip, was $48.60. You are leaving on a business trip to this city and will eat dinner at three of these restaurants. Your company will reimburse you for a maximum of $50 per dinner. Business associates familiar with these restaurants have told you that the meal cost at one-third of these restaurants will exceed $50. Suppose that you randomly select three of these restaurants for dinner. (Round your answers to four decimal places.)

(a)

What is the probability that none of the meals will exceed the cost covered by your company?

(b)

What is the probability that one of the meals will exceed the cost covered by your company?

(c)

What is the probability that two of the meals will exceed the cost covered by your company?

(d)

What is the probability that all three of the meals will exceed the cost covered by your company?

Answer 1

P(Exceeds) = 1/3

So P(Does not exceed) = 1 - 1/3 = 2/3

Therefore,

a) P(None) = (2/3)³ = 8/27

b) P(One) = $\binom{3}{1}(1/3)^1(2/3)^2$ = 4/9

c) P(Two) = $\binom{3}{2}(1/3)^2(2/3)^1$ = 2/9

d) P(Three) = (1/3)³ = 1/27