To find the probability of 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 = 15
Total number of employees who eat breakfast or lunch = 20 + 50 - 15 = 55
Total number of employees surveyed = 80
Probability of selecting an employee who eats breakfast or lunch = Number of employees who eat breakfast or lunch / Total number of employees = 55 / 80 = 11 / 16 = 0.6875
Therefore, the answer is 7/8 (or 0.875) because out of the 80 employees surveyed, 7 out of 8 employees eat either breakfast or lunch at the office.
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 selecting an employee who eat breakfast or lunch at the office? THE NSWER IS 7/8 EXPLAIN WHY IT IS
1 answer