Question

The 2003 Zagat Restaurant Survey provides food, decor, and service ratings for some of the top restaurants across the United States. For 15 top-ranking restaurants located in Boston, the average price of a dinner, including one drink and tip, was $48.60. You are leaving on 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 the restaurants have told you that the meal cost at 5 of the restaurants will exceed $50. Suppose that you randomly select three of these restaurants for dinner.
a. What is the probability that none of the meals will exceed the cost covered by your company (to 4 decimals)?
b. What is the probability that one of the meals will exceed the cost covered by your company (to 4 decimals)?
c. What is the probability that two of the meals will exceed the cost covered by your company (to 4 decimals)?
d. What is the probability that all three of the meals will exceed the cost covered by your company (to 4 decimals)?

Answer 1

Answer:

See explanation

Step-by-step explanation:

5 of 15 top-ranking restaurants offer dinner for more than $50 and 10 offer dinner for less than $50.

You will eat dinner at three of these restaurants.

a) The probability that none of the meals will exceed the cost covered by your company is

$\dfrac{C^{10}_3}{C^{15}_3}=\dfrac{\frac{10!}{3!(10-3)!}}{\frac{15!}{3!(15-3)!}}=\dfrac{10!\cdot 12!}{7!\cdot 15!}=\dfrac{8\cdot 9\cdot 10}{13\cdot 14\cdot 15}\approx 0.2637$

b) The probability that one of the meals will exceed the cost covered by your company is

$\dfrac{C^{10}_2\cdot C^5_1}{C^{15}_3}=\dfrac{\frac{10!}{2!(10-2)!}\cdot \frac{5!}{1!(5-1)!}}{\frac{15!}{3!(15-3)!}}=\dfrac{10!\cdot 5!\cdot 3!\cdot 12!}{2!\cdot 8!\cdot 4!\cdot 15!}=\dfrac{ 9\cdot 10\cdot 5\cdot 3}{\cdot 13\cdot 14\cdot 15}\approx 0.4945$

c) The probability that two of the meals will exceed the cost covered by your company is

$\dfrac{C^{10}_1\cdot C^5_2}{C^{15}_3}=\dfrac{\frac{10!}{1!(10-1)!}\cdot \frac{5!}{2!(5-2)!}}{\frac{15!}{3!(15-3)!}}=\dfrac{10!\cdot 5!\cdot 3!\cdot 12!}{9!\cdot 2!\cdot 3!\cdot 15!}=\dfrac{10\cdot 3\cdot 4\cdot 5}{13\cdot 14\cdot 15}\approx 0.2198$

d) The probability that all three of the meals will exceed the cost covered by your company is

$\dfrac{C^{5}_3}{C^{15}_3}=\dfrac{\frac{5!}{3!(5-3)!}}{\frac{15!}{3!(15-3)!}}=\dfrac{5!\cdot 12!}{2!\cdot 15!}=\dfrac{3\cdot 4\cdot 5}{13\cdot 14\cdot 15}\approx 0.0220$