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 A- 9/8 B- 5/8 C- 7/8 D- 2/8 Question 2 (the answer to this is 7/8, so explain why 7/8 is the answer using math vocabulary, and show your work)
To calculate the probability of randomly selecting an employee who eats breakfast or lunch at the office, we need to use the principle of inclusion-exclusion.
The formula for the probability of the union of two events A and B is given by:
P(A or B) = P(A) + P(B) - P(A and B)
In this case, P(Breakfast) = 20/80 = 1/4, P(Lunch) = 50/80 = 5/8, and P(Breakfast and Lunch) = 10/80 = 1/8.
Therefore,
P(Eat Breakfast or Lunch) = P(Breakfast) + P(Lunch) - P(Breakfast and Lunch)
= 1/4 + 5/8 - 1/8
= 2/8 + 5/8 - 1/8
= 6/8
= 3/4
= 0.875
So, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 7/8.