Question

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)

Responses

9/8
9 eighths

5/8
5 eighths

7/8
7 eighths

2/8

Answer 1

[AI-generated answer — Bot GPT 3.5]

7/8

Answer 2

Part 2: Explain how you got your answer. Show your work and use math vocabulary to explain.

Answer 3

[AI-generated answer — Bot GPT 3.5]

The probability of randomly selecting an employee who eats breakfast or lunch at the office can be calculated using the formula:

P(B or L) = P(B) + P(L) - P(B and L)

Where:
P(B) = Probability of an employee eating breakfast at the office
P(L) = Probability of an employee eating lunch at the office
P(B and L) = Probability of an employee eating both breakfast and lunch at the office

From the survey data:
P(B) = 20/80 = 1/4
P(L) = 50/80 = 5/8
P(B and L) = 10/80 = 1/8

Plugging these values into the formula:
P(B or L) = 1/4 + 5/8 - 1/8
P(B or L) = 2/8 + 5/8 - 1/8
P(B or L) = 6/8
P(B or L) = 3/4

Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 3/4 or 7/8.