Let's solve each part of the problem step by step using the information provided and set theory.

Given Data:

• Total number of students: $100$
• Students in Spanish ($S$): $28$
• Students in French ($F$): $26$
• Students in German ($G$): $16$
• Students in both Spanish and French ($S \cap F$): $12$
• Students in both Spanish and German ($S \cap G$): $4$
• Students in both French and German ($F \cap G$): $6$
• Students in all three classes ($S \cap F \cap G$): $2$

(a) Probability that a student is not in any of the language classes

We use the principle of inclusion-exclusion to find the total number of students enrolled in at least one class:

$|S \cup F \cup G| = |S| + |F| + |G| - |S \cap F| - |S \cap G| - |F \cap G| + |S \cap F \cap G|$

Substituting the values:

$|S \cup F \cup G| = 28 + 26 + 16 - 12 - 4 - 6 + 2 = 50$

So, $50$ students are taking at least one class. The number of students not taking any language class is:

$100 - 50 = 50$

Therefore, the probability that a randomly chosen student is not in any of the language classes is:

$P(\text{not in any class}) = \frac{50}{100} = 0.5$

(b) Probability that a student is taking exactly one language class

To find the number of students taking exactly one language class, calculate:

• Students only in Spanish: $|S| - (|S \cap F| + |S \cap G| - |S \cap F \cap G|) = 28 - (12 + 4 - 2) = 14$
• Students only in French: $|F| - (|F \cap S| + |F \cap G| - |S \cap F \cap G|) = 26 - (12 + 6 - 2) = 10$
• Students only in German: $|G| - (|G \cap S| + |G \cap F| - |S \cap F \cap G|) = 16 - (4 + 6 - 2) = 8$

Total number of students taking exactly one language class is:

$14 + 10 + 8 = 32$

So, the probability that a randomly chosen student is taking exactly one language class is:

$P(\text{exactly one class}) = \frac{32}{100} = 0.32$

(c) Probability that at least 1 of the 2 chosen students is taking a language class

First, find the probability that a randomly chosen student is not taking any language class, which is $0.5$.

For 2 students chosen randomly, the probability that both are not taking any class is:

$P(\text{both not in any class}) = 0.5 \times 0.5 = 0.25$

The probability that at least one of the two chosen students is taking a language class is:

$P(\text{at least 1 in a class}) = 1 - P(\text{both not in any class}) = 1 - 0.25 = 0.75$

Summary of Results:

(a) Probability that a student is not in any class: $0.5$

(b) Probability that a student is taking exactly one language class: $0.32$

(c) Probability that at least 1 of the 2 students is taking a language class: $0.75$

---

Would you like any more details on these calculations or have further questions?

Related Questions:

1. How would the probabilities change if the number of students in each class were different?
2. What is the probability that a student is enrolled in at least two classes?
3. How would you calculate the probability if three students were chosen instead of two?
4. What is the probability that a randomly chosen student is enrolled in both Spanish and German but not French?
5. How would adding a fourth language class affect these probabilities?

Tip: The Principle of Inclusion-Exclusion is a powerful tool for finding the count of elements in unions of sets by accounting for overlaps.