Question

Question 1
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)
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Answer 1

To solve this problem, we can use the principle of set theory and the concept of probability.

### Part 1: Calculate the Probability

Let:
- \( B \) = the set of employees who eat breakfast
- \( L \) = the set of employees who eat lunch

From the survey:
- \( |B| = 20 \) (employees who eat breakfast)
- \( |L| = 50 \) (employees who eat lunch)
- \( |B \cap L| = 10 \) (employees who eat both breakfast and lunch)

To find the number of employees who only eat breakfast, we can use the formula:
\[
|B \text{ only}| = |B| - |B \cap L|
\]
Substituting in the values:
\[
|B \text{ only}| = 20 - 10 = 10
\]

To find the number of employees who only eat lunch:
\[
|L \text{ only}| = |L| - |B \cap L|
\]
Substituting in the values:
\[
|L \text{ only}| = 50 - 10 = 40
\]

Now, we can add the number of employees who only eat breakfast and those who only eat lunch:
\[
\text{Total who eat only breakfast or only lunch} = |B \text{ only}| + |L \text{ only}|
\]
\[
= 10 + 40 = 50
\]

The probability of randomly selecting an employee who only eats breakfast or only eats lunch is given by the ratio of the favorable outcomes to the total number of employees:
\[
P(\text{only breakfast or only lunch}) = \frac{\text{number of employees who only eat breakfast or lunch}}{\text{total number of employees}}
\]
\[
= \frac{50}{80}
\]

Now, simplifying the fraction:
\[
= \frac{5}{8}
\]

Therefore, the answer to Part 1 is **5 eighths**.

### Part 2: Explanation of the Solution

To find the probability of selecting an employee who only eats breakfast or only eats lunch, I first identified the relevant sets based on the survey results.

1. **Set Definitions**: I designated the set \( B \) for breakfast eaters and \( L \) for lunch eaters.
2. **Counts**: I used the provided numbers to establish:
- Total breakfast eaters: 20
- Total lunch eaters: 50
- Employees eating both meals: 10.

3. **Exclusive Groups Calculation**: By computing the exclusive categories:
- For employees who only eat breakfast, I subtracted those who eat both meals: \( 20 - 10 = 10 \).
- For employees who only eat lunch, I similarly performed: \( 50 - 10 = 40 \).

4. **Total Count of Exclusive Eaters**: I combined these exclusive counts to get \( 10 + 40 = 50 \) employees who eat only one of the two meals.

5. **Calculating Probability**: Finally, I computed the probability as the ratio of favorable outcomes (50) to total outcomes (80), which simplifies to \( \frac{5}{8} \).

Using set theory allowed for clear differentiation of overlapping groups, ensuring accuracy in determining the exclusive counts and probability calculation.