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 employees randomly selected from this group what is the probability of randomly selecting an employee who eats breakfast or lunch at the office? what is the answer please tell me the real answer i need it so bad you guys keep giving me the wrong answer so give me the right answer please and its not 3/4 or 75% give me the right answer now
The correct answer for the probability of randomly selecting an employee who eats breakfast or lunch at the office can be found by adding the number of employees who eat breakfast at the office (20) to the number of employees who eat lunch at the office (50), and then subtracting the number of employees who eat both breakfast and lunch (10) to avoid double-counting:
Probability = (20 + 50 - 10) / 80
Probability = 60 / 80
Probability = 3/4 or 75%
theres no answer that says 3/4 give me the right answer please
I apologize for the confusion. Let's calculate the probability of randomly selecting an employee who eats breakfast or lunch at the office with the correct steps:
Number of employees who eat breakfast = 20
Number of employees who eat lunch = 50
Number of employees who eat both breakfast and lunch = 10
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 either breakfast or lunch (or both) without double-counting:
Total = Employees who eat breakfast + Employees who eat lunch - Employees who eat both
Total = 20 + 50 - 10 = 60
Probability = Number of employees who eat breakfast or lunch / Total employees
Probability = 60 / 80
Probability = 3/4 or 75%
Therefore, the correct answer is 3/4 or 75%. Let me know if you would like further assistance.
no its not give me the right answer please
I apologize for the confusion earlier. Let's reevaluate the calculation with the provided data:
Number of employees who eat breakfast = 20
Number of employees who eat lunch = 50
Number of employees who eat both breakfast and lunch = 10
To find the probability of randomly selecting an employee who eats breakfast or lunch at the office without double-counting, we need to use the Principle of Inclusion-Exclusion:
Probability = (Number of employees who eat breakfast + Number of employees who eat lunch - Number of employees who eat both) / Total number of employees
Probability = (20 + 50 - 10) / 80
Probability = 60 / 80
Probability = 3/4 or 0.75
Therefore, the correct answer to the probability of randomly selecting an employee who eats breakfast or lunch at the office is 3/4 or 0.75 (75%). If you have further questions or need clarification, please let me know.
im going to unlive my self if you do not give me the right answer!
I apologize for the confusion earlier. Let's calculate the probability of randomly selecting an employee who eats breakfast or lunch at the office with the correct computation:
Given data:
Number of employees who eat breakfast = 20
Number of employees who eat lunch = 50
Number of employees who eat both breakfast and lunch = 10
To find the probability of selecting an employee who eats breakfast or lunch at the office, we need to consider the total unique count of employees who eat either breakfast or lunch:
Total unique count = Employees who eat breakfast + Employees who eat lunch - Employees who eat both
Total unique count = 20 + 50 - 10 = 60
Probability = Total unique count / Total number of employees
Probability = 60 / 80
Probability = 0.75 or 75%
Therefore, the correct answer is 0.75 or 75%. Thank you for your patience, and I apologize for any confusion caused. If you have any more questions or need further assistance, please let me know.