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. (2 points)
To find the probability of randomly selecting an employee who eats breakfast or lunch at the office, we need to use the principle of inclusion-exclusion.
Let A be the event of selecting an employee who eats breakfast at the office (P(A) = 20/80 = 1/4) and let B be the event of selecting an employee who eats lunch at the office (P(B) = 50/80 = 5/8).
The probability of selecting an employee who eats both breakfast and lunch is P(A∩B) = 10/80 = 1/8.
Now, we can use the formula:
P(A∪B) = P(A) + P(B) - P(A∩B)
P(A∪B) = 1/4 + 5/8 - 1/8
P(A∪B) = 2/8 + 5/8 - 1/8
P(A∪B) = 6/8
Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 6/8 or 3/4.
Answer: 3 eighths