Students at a performing arts college can choose to study one or more courses of acting, dancing or singing. The probability that a student is studying acting is 0.5, dancing 0.65, singing 0.55, acting or dancing 0.8, acting and singing 0.25, dancing and singing 0.25. Find the probability that a student studies all three courses.

We are given the following probabilities:

- \( P(A) = 0.5 \) (Probability of studying acting)
- \( P(D) = 0.65 \) (Probability of studying dancing)
- \( P(S) = 0.55 \) (Probability of studying singing)
- \( P(A \cup D) = 0.8 \) (Probability of studying acting or dancing)
- \( P(A \cap S) = 0.25 \) (Probability of studying acting and singing)
- \( P(D \cap S) = 0.25 \) (Probability of studying dancing and singing)

We are tasked with finding the probability that a student studies all three courses: \( P(A \cap D \cap S) \).

### Step 1: Use the inclusion-exclusion principle

We will use the inclusion-exclusion principle to relate the probabilities of individual events, pairs of events, and all three events.

The formula for the union of two events is:

\[
P(A \cup D) = P(A) + P(D) - P(A \cap D)
\]

However, to include the third event (singing), we apply the inclusion-exclusion principle for three sets:

\[
P(A \cup D \cup S) = P(A) + P(D) + P(S) - P(A \cap D) - P(A \cap S) - P(D \cap S) + P(A \cap D \cap S)
\]

### Step 2: Solve for \( P(A \cap D \cap S) \)

From the problem, we know \( P(A \cup D) = 0.8 \), but we don’t know \( P(A \cap D) \) yet. Let's calculate it first:

Using the formula for \( P(A \cup D) \):

\[
P(A \cup D) = P(A) + P(D) - P(A \cap D)
\]
\[
0.8 = 0.5 + 0.65 - P(A \cap D)
\]
\[
P(A \cap D) = 0.5 + 0.65 - 0.8 = 0.35
\]

Now we substitute into the inclusion-exclusion formula:

\[
P(A \cup D \cup S) = P(A) + P(D) + P(S) - P(A \cap D) - P(A \cap S) - P(D \cap S) + P(A \cap D \cap S)
\]

Since the problem does not explicitly give \( P(A \cup D \cup S) \), we can assume that all students study at least one of the three courses, meaning \( P(A \cup D \cup S) = 1 \).

Thus, we have:

\[
1 = 0.5 + 0.65 + 0.55 - 0.35 - 0.25 - 0.25 + P(A \cap D \cap S)
\]

Simplifying:

\[
1 = 1.7 - 0.85 + P(A \cap D \cap S)
\]
\[
1 = 0.85 + P(A \cap D \cap S)
\]
\[
P(A \cap D \cap S) = 1 - 0.85 = 0.15
\]

### Final Answer:
The probability that a student studies all three courses is \( \boxed{0.15} \).

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Would you like more details or have any questions about this problem?

Here are five related questions to deepen your understanding:
1. How would the solution change if the probability of studying acting or dancing was different?
2. Can you explain the inclusion-exclusion principle and its significance in probability theory?
3. What is the difference between dependent and independent events in probability?
4. How would you calculate the probability that a student studies only two courses?
5. How does Venn diagram visualization help with problems involving three events?

**Tip**: Always list known probabilities and use them in structured formulas like inclusion-exclusion to avoid confusion in multivariable probability problems.

Solve a probability problem involving the courses acting, dancing, and singing. Using the inclusion-exclusion principle, learn how to calculate the probability that a student studies all three courses, given specific probabilities. Suitable for students in Grades 10-12.

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