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 eat breakfast or lunch at the office? Explain how you got the answer...
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 or lunch at the office.
From the survey data, we know that 20 employees eat breakfast, 50 employees eat lunch, and 15 employees eat both breakfast and lunch.
To find the total number of employees who eat either breakfast or lunch, we can add the number of employees who eat breakfast and the number of employees who eat lunch, and then subtract the number of employees who eat both breakfast and lunch to avoid double counting them.
Total employees who eat either breakfast or lunch = (20 + 50) - 15 = 55
Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 55/80, which simplifies to 11/16. Therefore, the probability is 11/16 or approximately 0.6875.
wrong
I apologize for the error in my previous response. Let's reevaluate the probability calculation.
First, we know that there are 20 employees who eat breakfast, 50 employees who eat lunch, and 15 employees who eat both breakfast and lunch.
To find the total number of employees who eat either breakfast or lunch, we can add the number of employees who eat breakfast (20), the number of employees who eat lunch (50), and then subtract the number of employees who eat both breakfast and lunch (15) to avoid double counting.
Total employees who eat either breakfast or lunch = 20 + 50 - 15 = 55
Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 55/80, which simplifies to 11/16. Therefore, the probability is 11/16 or approximately 0.6875.
shorter
To find the probability of randomly selecting an employee who eats breakfast or lunch at the office, we add the number of employees who eat breakfast (20) to the number of employees who eat lunch (50), and then subtract the number of employees who eat both breakfast and lunch (15) to avoid double counting.
So, the total number of employees who eat either breakfast or lunch is 20 + 50 - 15 = 55.
Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 55/80, which simplifies to 11/16 or approximately 0.6875.
what is the completely right answer
To find the probability of randomly selecting an employee who eats breakfast or lunch at the office, we first determine the total number of employees who eat breakfast or lunch.
There are 20 employees who eat breakfast, 50 employees who eat lunch, and 15 employees who eat both breakfast and lunch.
To find the total number of employees who eat either breakfast or lunch, we sum the numbers of employees who eat breakfast and lunch separately. But, since the 15 employees who eat both breakfast and lunch have been counted twice, we need to subtract this overlap.
Total employees who eat either breakfast or lunch = 20 + 50 - 15 = 55
Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 55/80, which simplifies to 11/16 or approximately 0.6875. So, the correct answer is a probability of 11/16.