To find the probability of randomly selecting an employee who eats breakfast or lunch at the office, we need to add the number of employees who eat breakfast and lunch separately and then subtract the number who eat both because they have been counted twice.
Number of employees who eat breakfast = 20
Number of employees who eat lunch = 50
Number of employees who eat both breakfast and lunch = 10
Total number of employees who eat either breakfast or lunch = 20 + 50 - 10 = 60
Total number of employees surveyed = 80
Probability of randomly selecting an employee who eat breakfast or lunch at the office = 60/80 = 3/4 = 0.75
Therefore, the probability is 7/8.
A survey of 80 employees was conducted asking about which meals they regularly eat at the office. Twenty 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?
9/8
5/8
7/8
2/8
1 answer
1 answer