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 would your answer be 98 or 5/8 or 7/8 or 2/8
To find the probability of randomly selecting an employee who eats breakfast or lunch at the office, we need to consider employees who eat breakfast, lunch, or both.
Number of employees who eat breakfast = 20
Number of employees who eat lunch = 50
Number of employees who eat both breakfast and lunch = 10
Total number of employees who eat breakfast, lunch, or both = 20 + 50 - 10 = 60
Total number of employees surveyed = 80
Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is:
Probability = Number of employees who eat breakfast, lunch, or both / Total number of employees surveyed
Probability = 60 / 80
Probability = 3/4
So, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 3/4 or 0.75.