Questions: 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 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 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. 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)?

Solution Steps

We are analyzing the probability of meal costs at three randomly selected restaurants in Boston, where the probability that a meal exceeds the reimbursement limit of \$50 is \( p = \frac{1}{3} \). Consequently, the probability that a meal does not exceed this limit is \( q = 1 - p = \frac{2}{3} \).

To find the probability that none of the meals exceed \$50, we calculate \( P(X = 0) \) using the binomial probability formula:

\[ P(X = x) = \binom{n}{x} \cdot p^x \cdot q^{n-x} \]

For \( n = 3 \) and \( x = 0 \):

\[ P(X = 0) = \binom{3}{0} \cdot \left(\frac{1}{3}\right)^0 \cdot \left(\frac{2}{3}\right)^{3} = 1 \cdot 1 \cdot \left(\frac{8}{27}\right) = 0.2963 \]

Thus, the probability that none of the meals will exceed \$50 is \( 0.2963 \).

Next, we calculate the probability that exactly one meal exceeds \$50, \( P(X = 1) \):

\[ P(X = 1) = \binom{3}{1} \cdot \left(\frac{1}{3}\right)^1 \cdot \left(\frac{2}{3}\right)^{2} = 3 \cdot \left(\frac{1}{3}\right) \cdot \left(\frac{4}{9}\right) = 0.4444 \]

Thus, the probability that one of the meals will exceed \$50 is \( 0.4444 \).

Now, we calculate the probability that exactly two meals exceed \$50, \( P(X = 2) \):

\[ P(X = 2) = \binom{3}{2} \cdot \left(\frac{1}{3}\right)^2 \cdot \left(\frac{2}{3}\right)^{1} = 3 \cdot \left(\frac{1}{9}\right) \cdot \left(\frac{2}{3}\right) = 0.2222 \]

Thus, the probability that two of the meals will exceed \$50 is \( 0.2222 \).

Final Answer