To find the probability of randomly selecting an employee who eats breakfast or lunch at the office, we need to use the formula for the probability of the union of two events:
P(A or B) = P(A) + P(B) - P(A and B)
In this case, let A = eating breakfast at the office and B = eating lunch at the office.
P(A) = 20/80 = 1/4 (since 20 out of 80 employees eat breakfast at the office)
P(B) = 50/80 = 5/8 (since 50 out of 80 employees eat lunch at the office)
P(A and B) = 10/80 = 1/8 (since 10 out of 80 employees eat both breakfast and lunch at the office)
Now, plug these values into the formula:
P(A or B) = 1/4 + 5/8 - 1/8
P(A or B) = 2/8 + 5/8 - 1/8
P(A or B) = 6/8
P(A or B) = 3/4
Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 3/4.
Therefore, the answer is option C - 7/8.
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 A- 9/8 B- 5/8 C- 7/8 D- 2/8 Question 2
Part 2: Explain how you got your answer. Show your work and use math vocabulary to explain.
1 answer