Part 1: 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? (2 points)
To find the probability of randomly selecting an employee who eats breakfast or lunch at the office, we need to calculate the total number of employees who eat either breakfast, lunch, or both.
Total employees who eat breakfast: 20
Total employees who eat lunch: 50
Total employees who eat both: 10
Total employees who eat either breakfast or lunch = Total employees who eat breakfast + Total employees who eat lunch - Total employees who eat both
= 20 + 50 - 10
= 60
So, out of the 80 employees surveyed, 60 eat either breakfast or lunch at the office.
Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is:
Probability = Number of employees who eat either breakfast or lunch / Total number of employees
= 60/80
= 3/4
= 0.75
So, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 0.75 or 75%.
is it 5/8?
Let's re-evaluate the probability calculation:
Total employees who eat breakfast: 20
Total employees who eat lunch: 50
Total employees who eat both: 10
Total employees who eat either breakfast or lunch = Total employees who eat breakfast + Total employees who eat lunch - Total employees who eat both
= 20 + 50 - 10
= 60
So, out of the 80 employees surveyed, 60 eat either breakfast or lunch at the office.
Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is:
Probability = Number of employees who eat either breakfast or lunch / Total number of employees
= 60/80
= 3/4
= 0.75
So, the correct probability of randomly selecting an employee who eats breakfast or lunch at the office is indeed 0.75 or 75%.
is it 9/8?
The probability of an event occurring cannot exceed 1, which is equivalent to 100%. Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office cannot be 9/8.
The correct probability is 3/4 or 75%. This is because out of the 80 employees surveyed, 60 eat either breakfast or lunch at the office, resulting in a probability of 60/80 = 3/4 = 0.75.
well my test doesnt have a 3/4 option and now im confused
I apologize for the confusion. Let's reevaluate the calculation to ensure accuracy.
Total employees who eat breakfast: 20
Total employees who eat lunch: 50
Total employees who eat both: 10
Total employees who eat either breakfast or lunch = Total employees who eat breakfast + Total employees who eat lunch - Total employees who eat both
= 20 + 50 - 10
= 60
Therefore, out of the 80 employees, 60 eat either breakfast or lunch at the office.
To find the probability of randomly selecting an employee who eats breakfast or lunch, divide the number of employees who eat either breakfast or lunch by the total number of employees:
Probability = Number of employees who eat either breakfast or lunch / Total number of employees
= 60/80
= 3/4
= 0.75
If the test does not have these options available, you may want to check if there is an equivalent representation such as 75% or try to calculate the probability using a different approach. Let me know if you would like to proceed with a different method.
ok well pick one from below
9/8
5/8
7/8
2/8
To find the probability of randomly selecting an employee who eats breakfast or lunch at the office, we calculated that out of the 80 employees surveyed, 60 eat either breakfast or lunch.
Therefore, the probability can be represented as a fraction by simplifying 60/80:
60/80 = 3/4
Since 3/4 is not listed as an option, let's convert it to a fraction with a denominator of 8:
3/4 = 6/8
Among the options provided, 6/8 simplifies to 3/4. Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office equivalent to 6/8.