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 eat breakfast or lunch at the office?(2 points)

Responses

9/8
9 eighths

5/8
5 eighths

7/8
7 eighths

2/8
2 eighths
Question 2
Part 2: Explain how you got your answer. Show your work and use math vocabulary to explain. in one sentence (2 points)

To calculate the probability of randomly selecting an employee who eats breakfast or lunch at the office, I added the number of employees who eat breakfast and lunch separately (20 + 50 = 70) and subtracted the number of employees who eat both meals (10), resulting in 70 - 10 = 60 employees who eat breakfast or lunch. Therefore, the probability is 60 out of 80, or 60/80, which simplifies to 3/4, or 3 quarters.