A survey of 80 employers was conducted asking about which meals they regularly eat at the office. 20 employees eat breakfast at the office. 50 eat lunch, and 15 eat both breakfast and lunch. if an employee is randomly selected from this group, what is the probability of randomly selecting an employee who eats breakfast or lunch at the office? explain your answer below
To find the probability of randomly selecting an employee who eats breakfast or lunch at the office, we need to consider employees who eat breakfast, lunch, or both.
Let's define the events:
A = employees who eat breakfast at the office
B = employees who eat lunch at the office
We are interested in finding the probability of selecting an employee who eats breakfast (A) or lunch (B) or both (A ∩ B).
The total number of employees who eat breakfast or lunch at the office is:
|A ∪ B| = |A| + |B| - |A ∩ B|
|A ∪ B| = 20 + 50 - 15
|A ∪ B| = 55
The total number of employees surveyed was 80.
Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is:
P(A ∪ B) = |A ∪ B| / total number of employees
P(A ∪ B) = 55 / 80
P(A ∪ B) = 0.6875
So, the probability of randomly selecting an employee who eats breakfast or lunch at the office is 0.6875 or 68.75%.