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?

Question

9/8

5/8

7/8

2/8

Question 2
Part 2: Explain how you got your answer. Show your work and use math vocabulary to explain. (In kids words)

Answer 1

To find the probability of randomly selecting an employee who eats breakfast or lunch at the office, we need to add the number of employees who eat breakfast to the number of employees who eat lunch, and then subtract the employees who eat both breakfast and lunch (since we don't want to count them twice).

So, 20 employees eat breakfast, 50 employees eat lunch, and 10 employees eat both breakfast and lunch. If we add 20 and 50, we get 70 employees who eat either breakfast or lunch. Then, we subtract the 10 employees who eat both to avoid double counting, giving us 60 employees total who eat either breakfast or lunch.

Out of the 80 total employees, the probability of randomly selecting an employee who eats breakfast or lunch is 60/80 or 3/4, which is equal to 7/8 in simplified form.

So, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 7/8.

Answer 2

can you just do part 2

Answer 3

To find the probability of randomly selecting an employee who eat breakfast or lunch at the office, we first add the number of employees who eat breakfast to the number of employees who eat lunch: 20 + 50 = 70 employees.

However, this counts the employees who eat both breakfast and lunch twice, so we need to subtract them once: 70 - 10 = 60 employees.

Out of the total 80 employees, there are 60 employees who eat either breakfast or lunch.

Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 60/80, which simplifies to 3/4 or 7/8.