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 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

To solve this problem, we need to use the concept of probability in a scenario involving a binomial distribution, as each selected restaurant either exceeds the $50 limit or does not.

Let's define our variables and steps:

1. **Probability of a Restaurant Exceeding $50**: We know that one-third of the restaurants exceed the $50 threshold. Thus:
   \[
   p = \frac{1}{3} = 0.333
   \]
   This is the probability that any randomly chosen restaurant will cost more than $50.

2. **Number of Trials**: Since you will dine at three randomly chosen restaurants, the number of trials $ n = 3 $.

3. **Binomial Probability Formula**: For a binomial distribution, the probability of having exactly $ k $ successes (in this case, meals over $50) in $ n $ trials is given by:
   \[
   P(X = k) = \binom{n}{k} p^k (1 - p)^{n - k}
   \]
   where $ \binom{n}{k} = \frac{n!}{k!(n - k)!} $ is the binomial coefficient.

### Part (a): Probability that None of the Meals Will Exceed $50
Here, $ k = 0 $, so we want $ P(X = 0) $.
\[
P(X = 0) = \binom{3}{0} (0.333)^0 (1 - 0.333)^3
\]
Calculating each term:
- $ \binom{3}{0} = 1 $
- $ (0.333)^0 = 1 $
- $ (1 - 0.333)^3 = (0.667)^3 \approx 0.2963 $

So:
\[
P(X = 0) \approx 1 \times 1 \times 0.2963 = 0.2963
\]

### Part (b): Probability that One Meal Will Exceed $50
For this part, $ k = 1 $, so we want $ P(X = 1) $.
\[
P(X = 1) = \binom{3}{1} (0.333)^1 (1 - 0.333)^2
\]
Calculating each term:
- $ \binom{3}{1} = 3 $
- $ (0.333)^1 = 0.333 $
- $ (1 - 0.333)^2 = (0.667)^2 \approx 0.445 $

So:
\[
P(X = 1) \approx 3 \times 0.333 \times 0.445 \approx 0.444
\]

### Part (c): Probability that Two Meals Will Exceed $50
For this part, $ k = 2 $, so we want $ P(X = 2) $.
\[
P(X = 2) = \binom{3}{2} (0.333)^2 (1 - 0.333)^1
\]
Calculating each term:
- $ \binom{3}{2} = 3 $
- $ (0.333)^2 \approx 0.111 $
- $ (1 - 0.333)^1 = 0.667 $

So:
\[
P(X = 2) \approx 3 \times 0.111 \times 0.667 \approx 0.222
\]

### Part (d): Probability that All Three Meals Will Exceed $50
For this part, $ k = 3 $, so we want $ P(X = 3) $.
\[
P(X = 3) = \binom{3}{3} (0.333)^3 (1 - 0.333)^0
\]
Calculating each term:
- $ \binom{3}{3} = 1 $
- $ (0.333)^3 \approx 0.037 $
- $ (1 - 0.333)^0 = 1 $

So:
\[
P(X = 3) \approx 1 \times 0.037 \times 1 = 0.037
\]

### Summary of Results
The probabilities for each scenario are:
1. **No meals exceed $50**: $ P(X = 0) \approx 0.2963 $
2. **One meal exceeds $50**: $ P(X = 1) \approx 0.444 $
3. **Two meals exceed $50**: $ P(X = 2) \approx 0.222 $
4. **All three meals exceed $50**: $ P(X = 3) \approx 0.037 $

Would you like more detailed explanations of any specific step?

#### Related Questions:
1. What is the probability that at least one meal will exceed $50?
2. How would the probabilities change if two-thirds of restaurants exceeded $50?
3. What assumptions are we making about meal costs and randomness in this scenario?
4. How does increasing the number of restaurants chosen affect these probabilities?
5. Can we apply the same approach if the number of trials (dinners) is unknown?

#### Tip:
In probability problems, clearly defining each variable and the conditions before starting calculations helps simplify complex scenarios.

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? 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?

Explore a probability problem involving meal costs where the binomial distribution is used to calculate the likelihood that meals exceed a company budget. Learn the step-by-step solution, including calculations for scenarios where 0, 1, 2, or all 3 meals go over the $50 limit. This example illustrates concepts of binomial probability and is suitable for Grades 11-12.

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