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 only eats breakfast or only eats lunch at the office?(2 points)

Question

Responses

98
9 eighths

58
5 eighths

78
7 eighths

2/8

Answer 1

To find the probability of selecting an employee who only eats breakfast or only eats lunch, we can start by calculating the number of employees who only eat breakfast and those who only eat lunch.

1. Employees who eat breakfast only:

   • Total breakfast eaters = 20
   • Employees who eat both breakfast and lunch = 10
   • Therefore, employees who eat breakfast only = 20 - 10 = 10

2. Employees who eat lunch only:

   • Total lunch eaters = 50
   • Employees who eat both breakfast and lunch = 10
   • Therefore, employees who eat lunch only = 50 - 10 = 40

3. Total employees who eat either breakfast only or lunch only:

   • Breakfast only + Lunch only = 10 + 40 = 50

4. Probability of selecting an employee who only eats breakfast or only eats lunch:

   • Total employees surveyed = 80
   • Probability = (Number of employees who eat breakfast only or lunch only) / Total employees
   • Probability = 50 / 80 = 5 / 8

So the probability of randomly selecting an employee who only eats breakfast or only eats lunch at the office is 5/8.

The correct response is: 5 eighths.