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 15 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
To find the probability of randomly selecting an employee who eats breakfast or lunch at the office, we need to consider employees who eat breakfast, lunch, or both.
Number of employees who eat breakfast = 20
Number of employees who eat lunch = 50
Number of employees who eat both breakfast and lunch = 15
Therefore, the total number of employees who eat breakfast, lunch, or both = 20 + 50 - 15 = 55
So, the probability of randomly selecting an employee who eats breakfast or lunch at the office = Number of employees who eat breakfast or lunch / Total number of employees
= 55/80
= 11/16
≈ 0.6875
Therefore, the correct answer is not among the given options. It should be 11/16.