A survey of 80 employees was conducted asking about which meals they regularly eat at the office. 20 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?(2 points)
Responses
98
9 eighths
58
5 eighths
78
7 eighths
28
To find the probability of randomly selecting an employee who eats breakfast or lunch at the office, we need to add the number of employees who eat breakfast, the number of employees who eat lunch, and then subtract the number of employees who eat both breakfast and lunch (to avoid double counting).
Number of employees who eat breakfast = 20
Number of employees who eat lunch = 50
Number of employees who eat both breakfast and lunch = 10
Adding the employees who eat breakfast and lunch separately:
20 + 50 - 10 = 60
So, there are 60 employees who eat breakfast or lunch at the office.
Total number of employees surveyed = 80
Probability = Number of employees who eat breakfast or lunch / Total number of employees surveyed
Probability = 60 / 80
Probability = 3 / 4
Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 3/4 or 75%.