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The 2003 Zagat Restaurant Survey provides food, décor, and service ratings for some of the top restaurants
across the United States. For 15 top-ranked restaurants located in Boston, the average price of a dinner
including, one drink and tip, was $48.60. You are leaving for a business trip to Boston 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 restaurants for dinner.

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?
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Answer #3

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Answer #1
Since the problem states that you will eat at 3 of these restaurants then there is no replacement.
5 restaurants' meals are over $50. and 10 are not.
a)10C3/15C3 = 120/455 = .2637


b)(10C2 * 5C1)/15C3 = 225/455 = .4945
c)(10C1 * 5C2)/15C3 = 100/455 = .2198
d) 5C3/15C3 = 10/455 = .0220
answered by: Sherrie
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Answer #2

Compute the probability of two occurrences in two time periods (to 4 decimals).

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Answer #4

source: SLADER
answered by: mae
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