Solve the question: In a class of 25 students, 19 study at most one of Art and Music and 9 study exactly one of Art and Music. Two students are chosen at random from the class.
(a) Draw a Venn diagram to represent the above information showing the number studying both Art and Music and the number studying neither Art nor Music.
(b) Find in simplest fraction form the probability that both the students chosen study neither Art nor Music.

Let's solve the problem step by step.

### Part (a) - Drawing a Venn Diagram

Given:
- There are 25 students in total.
- 19 students study at most one of Art and Music.
- 9 students study exactly one of Art and Music.

Let:
- \( A \) represent the set of students studying Art.
- \( M \) represent the set of students studying Music.
- \( x \) be the number of students studying both Art and Music.

We know:
1. The total number of students is \( |A \cup M \cup (A \cap M)^c | = 25 \).
2. The number of students who study neither Art nor Music is \( n_0 \).
3. The number of students studying exactly one of Art or Music is \( |A \setminus M| + |M \setminus A| = 9 \).
4. The number of students studying at most one of Art and Music is \( |A \setminus M| + |M \setminus A| + n_0 = 19 \).

Let's break it down:

- Let \( |A \setminus M| = a \) (only Art) and \( |M \setminus A| = b \) (only Music).
- The number of students studying both Art and Music is \( x \).
- The number of students studying neither is \( n_0 \).

We have:
\[ a + b + x + n_0 = 25 \]
\[ a + b + n_0 = 19 \]
\[ a + b = 9 \]

From the second equation, \( n_0 = 19 - 9 = 10 \).

Substituting \( n_0 = 10 \) in the first equation:
\[ 9 + x + 10 = 25 \]
\[ x = 6 \]

So:
- \( x = 6 \) (both Art and Music)
- \( n_0 = 10 \) (neither Art nor Music)

Thus:
- \( a = 9 - b \) (only Art)
- \( b = 9 - a \) (only Music)

To maintain consistency:
- \( |A| = a + x \) (total in Art)
- \( |M| = b + x \) (total in Music)

### Venn Diagram Representation:
- 6 students study both Art and Music.
- 10 students study neither Art nor Music.
- 3 students study only Art (since \( a = 3 \)).
- 6 students study only Music (since \( b = 6 \)).

### Part (b) - Probability Calculation

We need to find the probability that both randomly chosen students study neither Art nor Music.

Number of students studying neither Art nor Music: \( n_0 = 10 \).

The probability of selecting one student who studies neither Art nor Music:
\[ \text{P(first student)} = \frac{10}{25} = \frac{2}{5} \]

The probability of selecting another student who also studies neither Art nor Music, given the first student was already chosen:
\[ \text{P(second student | first)} = \frac{9}{24} = \frac{3}{8} \]

So, the combined probability is:
\[ \text{P(both)} = \frac{2}{5} \times \frac{3}{8} = \frac{6}{40} = \frac{3}{20} \]

### Final Answer
The probability that both students chosen study neither Art nor Music is \( \frac{3}{20} \).

Would you like any further details or clarifications?

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Here are 8 related questions:
1. How would the Venn diagram change if the total number of students increased to 30?
2. What is the probability that both students study exactly one of Art and Music?
3. What is the probability that one student studies Art only and the other Music only?
4. If a third student is chosen, what is the probability all three study neither Art nor Music?
5. How does the probability change if 3 students are chosen instead of 2?
6. What is the probability that at least one of the chosen students studies both Art and Music?
7. How many students would be required for the probability of both chosen students studying neither to be exactly 0.25?
8. What if there were three subjects instead of two, how would you model that situation?

**Tip:** Understanding the relationships in a Venn diagram is key to solving probability problems involving overlapping sets.

Learn how to calculate the probability that two randomly chosen students from a class of 25 study neither Art nor Music. This problem involves set theory, Venn diagrams, and probability calculations. Suitable for high school students.

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