Question 3 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 eats breakfast or lunch at the office?
To find the probability of randomly selecting an employee who eats breakfast or lunch at the office, we first need to find the total number of employees who eat breakfast or lunch.
From the survey:
- 20 employees eat breakfast
- 50 employees eat lunch
- 15 employees eat both breakfast and lunch
To find the total number of employees who eat breakfast or lunch, we add the number of employees who eat breakfast (20) to the number of employees who eat lunch (50), and then subtract the employees who eat both (15) to avoid double-counting:
Total number of employees who eat breakfast or lunch = 20 + 50 - 15 = 55
Therefore, there are 55 employees who eat breakfast or lunch at the office.
The probability of randomly selecting an employee who eats breakfast or lunch at the office is:
Number of employees who eat breakfast or lunch / Total number of employees = 55 / 80 = 0.6875
So, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 0.6875 or 68.75%.