To solve this problem, we can use the principle of set theory and the concept of probability.
Part 1: Calculate the Probability
Let:
- \( B \) = the set of employees who eat breakfast
- \( L \) = the set of employees who eat lunch
From the survey:
- \( |B| = 20 \) (employees who eat breakfast)
- \( |L| = 50 \) (employees who eat lunch)
- \( |B \cap L| = 10 \) (employees who eat both breakfast and lunch)
To find the number of employees who only eat breakfast, we can use the formula: \[ |B \text{ only}| = |B| - |B \cap L| \] Substituting in the values: \[ |B \text{ only}| = 20 - 10 = 10 \]
To find the number of employees who only eat lunch: \[ |L \text{ only}| = |L| - |B \cap L| \] Substituting in the values: \[ |L \text{ only}| = 50 - 10 = 40 \]
Now, we can add the number of employees who only eat breakfast and those who only eat lunch: \[ \text{Total who eat only breakfast or only lunch} = |B \text{ only}| + |L \text{ only}| \] \[ = 10 + 40 = 50 \]
The probability of randomly selecting an employee who only eats breakfast or only eats lunch is given by the ratio of the favorable outcomes to the total number of employees: \[ P(\text{only breakfast or only lunch}) = \frac{\text{number of employees who only eat breakfast or lunch}}{\text{total number of employees}} \] \[ = \frac{50}{80} \]
Now, simplifying the fraction: \[ = \frac{5}{8} \]
Therefore, the answer to Part 1 is 5 eighths.
Part 2: Explanation of the Solution
To find the probability of selecting an employee who only eats breakfast or only eats lunch, I first identified the relevant sets based on the survey results.
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Set Definitions: I designated the set \( B \) for breakfast eaters and \( L \) for lunch eaters.
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Counts: I used the provided numbers to establish:
- Total breakfast eaters: 20
- Total lunch eaters: 50
- Employees eating both meals: 10.
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Exclusive Groups Calculation: By computing the exclusive categories:
- For employees who only eat breakfast, I subtracted those who eat both meals: \( 20 - 10 = 10 \).
- For employees who only eat lunch, I similarly performed: \( 50 - 10 = 40 \).
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Total Count of Exclusive Eaters: I combined these exclusive counts to get \( 10 + 40 = 50 \) employees who eat only one of the two meals.
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Calculating Probability: Finally, I computed the probability as the ratio of favorable outcomes (50) to total outcomes (80), which simplifies to \( \frac{5}{8} \).
Using set theory allowed for clear differentiation of overlapping groups, ensuring accuracy in determining the exclusive counts and probability calculation.