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? Explain as simplified fraction.
To find the probability of randomly selecting an employee who eats breakfast or lunch at the office, we need to find the total number of employees who eat breakfast, the total number of employees who eat lunch, and the employees who eat both.
Total employees who eat breakfast = 20
Total employees who eat lunch = 50
Employees who eat both = 15
To avoid double counting employees who eat both breakfast and lunch, we need to subtract those who eat both from the total count.
Total employees who eat breakfast or lunch = Total employees who eat breakfast + Total employees who eat lunch - Employees who eat both
= 20 + 50 - 15
= 55
Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 55/80, which is the simplified fraction for this probability.