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 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? Explain how you got the answer.
To find the probability of randomly selecting an employee who eat breakfast or lunch at the office, we use the principle of inclusion-exclusion.
Let A be the event that an employee eats breakfast at the office, and let B be the event that an employee eats lunch at the office.
From the information given:
- P(A) = 20/80 = 1/4
- P(B) = 50/80 = 5/8
- P(A and B) = 10/80 = 1/8
We want to find P(A or B), which is the probability of an employee eating breakfast or lunch at the office. This can be calculated using the formula:
P(A or B) = P(A) + P(B) - P(A and B)
= 1/4 + 5/8 - 1/8
= 2/8 + 5/8 - 1/8
= 6/8
= 3/4
Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 3/4.
Select from the answers:
1. 9/8
2. 5/8
3. 7/8
4. 2/8
The correct answer from the options provided is:
3. 7/8
Explanation: The correct probability of randomly selecting an employee who eats breakfast or lunch at the office is 3/4, which is equivalent to 6/8. So, the closest option provided is 7/8, which is the correct answer.