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Part 1: A survey of 80 employees was conducted asking about which meals they regularly eat at the office. 20 employees eat breakfast at the office, 50 eat lunch, and 10 eat both breakfast and lunch. If an employee is randomly selected from this group, what is the probability of randomly selecting an employee who only eats breakfast or only eats lunch at the office?(2 points) Responses 98 9 eighths 58 5 eighths 78 7 eighths 28 2 eighths Question 2 Part 2: Explain how you got your answer. Show your work and use math vocabulary to explain. (2 points)

1 answer

To solve Part 1, we can use the information provided to find the probabilities we're interested in.

Let's denote the sets as follows:

  • Let \( A \) be the set of employees who eat breakfast.
  • Let \( B \) be the set of employees who eat lunch.

From the problem, we have:

  • \( |A| = 20 \) (employees who eat breakfast),
  • \( |B| = 50 \) (employees who eat lunch),
  • \( |A \cap B| = 10 \) (employees who eat both breakfast and lunch).

We need to find the number of employees who only eat breakfast and the number of employees who only eat lunch.

  1. Employees who only eat breakfast: \[ |A \text{ only}| = |A| - |A \cap B| = 20 - 10 = 10 \]

  2. Employees who only eat lunch: \[ |B \text{ only}| = |B| - |A \cap B| = 50 - 10 = 40 \]

Now, we add the two results to find the total number of employees who eat either only breakfast or only lunch: \[ \text{Total} = |A \text{ only}| + |B \text{ only}| = 10 + 40 = 50 \]

Now, to find the probability of randomly selecting an employee who only eats breakfast or only eats lunch, we can use the total number of employees surveyed (80): \[ P(\text{only breakfast or only lunch}) = \frac{\text{Number of employees who only eat breakfast or lunch}}{\text{Total number of employees}} = \frac{50}{80} = \frac{5}{8} \]

Thus, the answer to Part 1 is: \[ \frac{5}{8} \]

Part 2 Explanation:

In Part 2, we explain the math vocabulary and show our work:

  1. Understanding Set Theory: We used the concept of sets to categorize the employees based on their meal choices.

    • The employees who eat breakfast were represented by the set \( A \), and those who eat lunch by set \( B \).
    • The intersection \( |A \cap B| \) represents employees who belong to both sets: those who eat both breakfast and lunch.

  2. Calculating Exclusivity:

    • To find out how many employees eat only one type of meal, we subtracted the number of employees who eat both meals from each respective meal category. This gives us two exclusive counts: those who only eat breakfast and those who only eat lunch.

  3. Adding Exclusive Counts:

    • We then combined these two exclusive groups to get the total number of employees who only eat breakfast or only eat lunch.

  4. Calculating Probability:

    • Finally, we calculated the probability by taking the total exclusive count (50) and dividing it by the total number of employees surveyed (80). This results in a simplified fraction of \( \frac{5}{8} \).

Thus, the probability of selecting an employee who eats only breakfast or only lunch at the office is \( \frac{5}{8} \).