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? Explain how I got 7/8 step by step in 3 sentences.
The total number of employees who eat either breakfast or lunch at the office is the sum of those who eat breakfast (20), lunch (50), and both breakfast and lunch (10), which equals 80. Thus, the probability of randomly selecting an employee who eats either breakfast or lunch is 80/80 = 1. To get the probability of selecting an employee who eats breakfast or lunch, we subtract the probability of selecting an employee who eats neither from 1, which is 1 - 1/8 = 7/8.
