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)

Question

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. (2 points)

Answer 1

To find the probability of randomly selecting an employee who eats breakfast or lunch at the office, we need to find the total number of employees who eat either breakfast or lunch, and then divide that by the total number of employees surveyed.

We know that 20 employees eat breakfast, 50 eat lunch, and 10 eat both breakfast and lunch. However, we need to make sure we do not double count those who eat both breakfast and lunch.

So, the total number of employees who eat either breakfast or lunch = (number of employees who eat breakfast) + (number of employees who eat lunch) - (number of employees who eat both breakfast and lunch)
= 20 + 50 - 10
= 60

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

To explain this, we used the concept of probability which is the likelihood of an event occurring. The probability is calculated by taking the number of favorable outcomes (in this case, employees who eat breakfast or lunch) and dividing it by the total number of possible outcomes (total number of employees surveyed).