MATH SOLVE

4 months ago

Q:
# The Zagat Restaurant Survey provides food, decor, and service ratings for some of the top restaurants across the United States. For 15 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 of these restaurants for dinner.What is the probability that none of the meals will exceed the cost covered by your company?What is the probability that one of the meals will exceed the cost covered by your company?What is the probability that two of the meals will exceed the cost covered by your company?What is the probability that all three of the meals will exceed the cost covered by your company?

Accepted Solution

A:

Answer:Probability that none of the meals will exceed the cost covered by your company=0.2637
Probability that one of the meals will exceed the cost covered by your company=0.4945Probability that two of the meals will exceed the cost covered by your company=0.2197
Probability that three of the meals will exceed the cost covered by your company=0.02197
Step-by-step explanation:15 restaurants located in Boston is relevant for this question.
One third of 15 means, 5 restaurants will always exceed 50$
So,
We can use combinations, as we donβt know the exact 5 restaurants out of the 15
None of the meals will exceed,
10 restaurants will not exceed the rate and we will surely have to go to 3 restaurants so
10C3/15C3
= (10*9*8/1*2*3)/ (15*14*13/1*2*3)
=120/455
=0.2637
One of the meals will exceed,
We have to visit 3 restaurants but one will exceed.
So,
(10C2*5C1)/15C3
= ((10*9/1*2) *5)/ (15*14*13/1*2*3)
=45*5/455
=225/455
=0.4945
Two of the meals will exceed,
We visit 3 restaurants but two will exceed.
So,
(10C1*5C2)/15C3
= ((10*5*4/1*2))/ (15*14*13/1*2*3)
=100/455
=0.2197
All 3 meals will exceed.
So,
5C3/15C3
= (5*4*3/1*2*3)/ (15*14*13/1*2*3)
=10/455
=0.02197