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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. 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?

Step by step solution

01

Identify the Probability of Exceeding Cost

We are told that one-third of the restaurants have a meal cost exceeding $50. Since 15 restaurants are surveyed, the probability that a randomly selected restaurant exceeds the cost is \( p = \frac{1}{3} \).

02

Define the Probability of Not Exceeding Cost

The probability that a randomly selected restaurant does not exceed the cost is \( 1 - p = 1 - \frac{1}{3} = \frac{2}{3} \).

03

Calculate Probability of None Exceeding Cost

To find the probability that none of the three chosen meals will exceed $50, use \( (\frac{2}{3})^3 \): \[ P(0 \text{ exceed }) = \left( \frac{2}{3} \right)^3 = \frac{8}{27} \].

04

Calculate Probability of One Exceeding Cost

Use the binomial probability formula \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \) where \( n = 3 \), \( k = 1 \), \( p = \frac{1}{3}\):\[ P(1 \text{ exceed }) = \binom{3}{1} \left( \frac{1}{3} \right)^1 \left( \frac{2}{3} \right)^2 \ = 3 \times \frac{1}{3} \times \frac{4}{9} = \frac{12}{27} = \frac{4}{9} \].

05

Calculate Probability of Two Exceeding Cost

Using the same formula with \( k = 2 \):\[ P(2 \text{ exceed }) = \binom{3}{2} \left( \frac{1}{3} \right)^2 \left( \frac{2}{3} \right)^1 \ = 3 \times \frac{1}{9} \times \frac{2}{3} = \frac{6}{27} = \frac{2}{9} \].

06

Calculate Probability of All Three Exceeding Cost

If all three meals exceed the cost, use \( (\frac{1}{3})^3 \):\[ P(3 \text{ exceed }) = \left( \frac{1}{3} \right)^3 = \frac{1}{27} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Binomial Distribution

The binomial distribution is a fundamental concept in probability theory. It describes the number of successes in a fixed number of independent experiments, where each experiment has the same probability of success. For this problem, think of each restaurant visit as an independent trial, where success is the meal exceeding $50. We need to determine the likelihood of various scenarios (none, one, two, or all meals exceeding $50) which are perfect applications for binomial probabilities. Here, the number of trials is 3 (the number of restaurants), and the probability of success p is 1/3 (the fraction of restaurants exceeding $50). This sets up a binomial distribution problem that allows us to use its formula to find each probability exactly.

The Role of Random Selection

Random selection is crucial in probability because it ensures each item in a set has an equal chance of being chosen. When we select 3 restaurants out of 15 randomly, it means no bias influences which restaurants are selected. This randomness is vital for applying pure probability calculations like the binomial distribution. It gives us confidence that the probability of exceeding $50 is truly represented by 1/3, as all restaurants have equal opportunity to be chosen in the sample.

Conducting Cost Analysis

Cost analysis in this context involves evaluating the likelihood that the meal expenses, reimbursed by your company, do not exceed the set budget (\(50 per meal).

• None Exceeding Cost: We calculate the probability using \( (\frac{2}{3})^3 \), resulting in a probability of 8/27 that none of the meals will exceed \)50.
• One Meal Exceeding Cost: With the binomial formula, we find there's a 4/9 chance for this scenario.
• Two Meals Exceeding Cost: Again using the formula, there's a probability of 2/9.
• All Three Exceeding Cost: Lastly, the chance is 1/27 that all exceed the budget.

These analyses provide useful insights, helping you plan financially for your trip by understanding potential out-of-pocket costs.

Approaches to Statistical Problem Solving

Statistical problem solving requires identifying the correct method to analyze data and draw conclusions. Here, we use a structured approach:

• Key Information Identification: Recognize the probabilities given (1/3 of restaurants over budget) and the scenario (3 restaurants are chosen).
• Determination of Probabilities: Understand the context involves independent events, pointing to the binomial distribution as the tool to calculate probabilities.
• Mathematical Calculation: Utilize binomial probability formulas to calculate the likelihoods of different outcomes (none, one, two, and all restaurants exceeding $50).
• Interpretation of Results: Analyze and explain the results contextually, providing a clear understanding of financial expectations.

Such a methodological approach helps to break down complex statistical scenarios, making it easier to comprehend and apply solutions effectively.