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 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.

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 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.

Answer 1

To find the probability of randomly selecting an employee who eats breakfast or lunch at the office, we can use the formula:

P(Breakfast or Lunch) = P(Breakfast) + P(Lunch) - P(Breakfast and Lunch)

We are given that 20 employees eat breakfast at the office, 50 eat lunch, and 10 eat both breakfast and lunch.

P(Breakfast) = 20/80 = 1/4
P(Lunch) = 50/80 = 5/8
P(Breakfast and Lunch) = 10/80 = 1/8

Therefore, P(Breakfast or Lunch) = 1/4 + 5/8 - 1/8 = 2/8 + 5/8 - 1/8 = 6/8 = 3/4

So, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 3/4 or 75%.