A survey of 80 employees was conducted asking about which means they regulary eat @ da office. Twenty employees eat breakfast at da office, 50 eat lunch, n 15 ear both breakfast n lunch. If an an employee who eats breakfreast or lunch @ da office? HINT: think mutually exclusive n mutually inclusive formulas explain as simplified fraction

1 answer

To find out how many employees eat breakfast, lunch, or both at the office, we can use the principle of inclusion-exclusion. Let's break down the information:

  • Let \( A \) be the set of employees who eat breakfast at the office.
  • Let \( B \) be the set of employees who eat lunch at the office.

From the survey:

  • Number of employees in \( A \) (breakfast) = 20
  • Number of employees in \( B \) (lunch) = 50
  • Number of employees in both \( A \) and \( B \) (both breakfast and lunch) = 15

We want to find the number of employees who eat either breakfast or lunch or both, which is represented by the union of sets \( A \) and \( B \) (denoted as \( A \cup B \)).

The formula for the union of two sets using inclusion-exclusion is:

\[ |A \cup B| = |A| + |B| - |A \cap B| \]

Where:

  • \( |A| \) is the number of employees who eat breakfast (20)
  • \( |B| \) is the number of employees who eat lunch (50)
  • \( |A \cap B| \) is the number of employees who eat both (15)

Substituting the values into the formula:

\[ |A \cup B| = 20 + 50 - 15 = 55 \]

This means that 55 employees eat breakfast or lunch or both at the office.

Now, to express this as a simplified fraction of the total number of employees (80):

\[ \text{Fraction} = \frac{|A \cup B|}{\text{Total Employees}} = \frac{55}{80} \]

To simplify the fraction \( \frac{55}{80} \):

  1. Find the greatest common divisor (GCD) of 55 and 80. The GCD is 5.
  2. Divide the numerator and the denominator by their GCD:

\[ \frac{55 \div 5}{80 \div 5} = \frac{11}{16} \]

So, the final answer is that \(\frac{11}{16}\) of the employees eat breakfast or lunch at the office.