To find the probability of randomly selecting an employee who eats either breakfast or lunch at the office, we can apply the principle of inclusion-exclusion.
Let:
- \( B \) = the number of employees who eat breakfast
- \( L \) = the number of employees who eat lunch
- \( B \cap L \) = the number of employees who eat both breakfast and lunch
From the survey, we have:
- \( B = 20 \) (employees who eat breakfast)
- \( L = 50 \) (employees who eat lunch)
- \( B \cap L = 15 \) (employees who eat both)
Using the inclusion-exclusion principle, the number of employees who eat breakfast or lunch is given by:
\[ B \cup L = B + L - B \cap L \]
Substituting the values we have:
\[ B \cup L = 20 + 50 - 15 = 55 \]
So, 55 employees eat either breakfast or lunch.
Now, to find the probability \( P \) of randomly selecting an employee who eats breakfast or lunch, we use the formula:
\[ P(B \cup L) = \frac{B \cup L}{\text{Total number of employees}} = \frac{55}{80} \]
Next, we can simplify this fraction:
\[ \frac{55}{80} = \frac{11}{16} \]
Therefore, the probability of randomly selecting an employee who eats breakfast or lunch at the office is:
\[ \boxed{\frac{11}{16}} \]