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 eats breakfast or lunch at the office? Put in fraction form.
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.
There are a total of 80 employees surveyed.
- 20 employees eat breakfast
- 50 employees eat lunch
- 10 employees eat both breakfast and lunch
To find the total number of employees who eat breakfast or lunch (or both), we can add the number of employees who eat breakfast, the number of employees who eat lunch, and then subtract the employees who eat both breakfast and lunch to avoid double counting:
20 (breakfast) + 50 (lunch) - 10 (both) = 60
So, there are 60 employees who eat breakfast, lunch, or both out of the total 80 employees surveyed.
Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is:
Number of employees who eat breakfast, lunch, or both / Total number of employees surveyed
= 60/80
= 3/4
Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 3/4.