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 employees randomly selected from this group what is the probability of randomly selecting an employee who eats breakfast or lunch at the office? Is the answer nine eighths?
To find the probability of selecting an employee who eats breakfast or lunch at the office, we need to use the principle of inclusion-exclusion.
Let A be the event that an employee eats breakfast at the office, B be the event that an employee eats lunch at the office, and C be the event that an employee eats both breakfast and lunch at the office.
P(A) = 20/80 = 1/4
P(B) = 50/80 = 5/8
P(C) = 10/80 = 1/8
Now, we can find the probability of selecting an employee who eats breakfast or lunch at the office:
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
P(A ∪ B) = P(A) + P(B) - P(C)
P(A ∪ B) = 1/4 + 5/8 - 1/8
P(A ∪ B) = 2/8 + 5/8 - 1/8
P(A ∪ B) = 6/8 = 3/4
So, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 3/4, not nine eighths.