A survey of 80 employees was conducted asking about which meals they regularly eat the office. Twenty employees eat breakfast at the office, 50 eat lunch, and 15 eat 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?
To find the probability of randomly selecting an employee who eats breakfast or lunch at the office, we need to consider the 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
The total number of employees surveyed is 80.
Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is:
P(eats breakfast or lunch) = Number of employees who eat breakfast, lunch, or both / Total number of employees surveyed
P(eats breakfast or lunch) = 55 / 80
P(eats breakfast or lunch) = 0.6875
Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 0.6875 or 68.75%.