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 can use the principle of inclusion-exclusion.
The number of employees who eat breakfast or lunch at the office is the sum of employees who eat breakfast, employees who eat lunch and employees who eat both breakfast and lunch.
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 or lunch at the office is:
20 + 50 - 15 = 55
The total number of employees surveyed is 80.
So, the probability of randomly selecting an employee who eats breakfast or lunch at the office is:
55/80 = 11/16
Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 11/16.