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)

9/8

5/8

7/8

2/8

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

The answer is 5/8. Only answer part 2

Answer 1

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

P(A or B) = P(A) + P(B) - P(A and B)

In this case, the probability of selecting an employee who eats breakfast is 20/80 = 1/4, the probability of selecting an employee who eats lunch is 50/80 = 5/8, and the probability of selecting an employee who eats both breakfast and lunch is 10/80 = 1/8.

Therefore, the probability of selecting an employee who eats breakfast or lunch at the office is:

P(breakfast or lunch) = P(breakfast) + P(lunch) - P(breakfast and lunch)
= 1/4 + 5/8 - 1/8
= 2/8 + 5/8 - 1/8
= 6/8
= 3/4
= 5/8

Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 5/8.