6. Consider two events A and B. Prove that P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

To prove that \( P(A \cup B) = P(A) + P(B) - P(A \cap B) \), we will use basic principles of set theory and probability.

### Definitions:
- \( A \cup B \) is the event that either \( A \), \( B \), or both occur (union of events).
- \( A \cap B \) is the event that both \( A \) and \( B \) occur (intersection of events).
- \( P(A) \) is the probability of event \( A \).
- \( P(B) \) is the probability of event \( B \).
- \( P(A \cup B) \) is the probability that either \( A \), \( B \), or both occur.
- \( P(A \cap B) \) is the probability that both events occur simultaneously.

### Proof:
1. **Start with the interpretation of \( P(A \cup B) \)**:
   The probability of \( A \cup B \) represents the total probability of either event \( A \), \( B \), or both happening.

2. **Break \( A \cup B \) into parts**:
   We can divide the region representing \( A \cup B \) into three mutually exclusive parts:
   - The part where only \( A \) occurs.
   - The part where only \( B \) occurs.
   - The part where both \( A \) and \( B \) occur.

   Mathematically:
   \[
   P(A \cup B) = P(\text{only } A) + P(\text{only } B) + P(A \cap B)
   \]

3. **Rewrite \( P(A) \) and \( P(B) \)**:
   The probability of \( A \) includes both the part where only \( A \) occurs and where both \( A \) and \( B \) occur:
   \[
   P(A) = P(\text{only } A) + P(A \cap B)
   \]
   Similarly, the probability of \( B \) includes both the part where only \( B \) occurs and where both \( A \) and \( B \) occur:
   \[
   P(B) = P(\text{only } B) + P(A \cap B)
   \]

4. **Add \( P(A) \) and \( P(B) \)**:
   Now, let's add the expressions for \( P(A) \) and \( P(B) \):
   \[
   P(A) + P(B) = \left( P(\text{only } A) + P(A \cap B) \right) + \left( P(\text{only } B) + P(A \cap B) \right)
   \]
   Simplifying:
   \[
   P(A) + P(B) = P(\text{only } A) + P(\text{only } B) + 2P(A \cap B)
   \]

5. **Account for overcounting \( P(A \cap B) \)**:
   Notice that \( P(A \cap B) \) has been added twice (once in \( P(A) \) and once in \( P(B) \)). To correct this, we subtract \( P(A \cap B) \) once:
   \[
   P(A) + P(B) - P(A \cap B) = P(\text{only } A) + P(\text{only } B) + P(A \cap B)
   \]

6. **Conclusion**:
   The right-hand side is exactly the expression for \( P(A \cup B) \). Thus, we have shown:
   \[
   P(A \cup B) = P(A) + P(B) - P(A \cap B)
   \]

This completes the proof.

---

Would you like more details or have any questions?

Here are 5 related questions for further exploration:
1. How does the formula change if \( A \) and \( B \) are mutually exclusive events?
2. How can you extend this formula to three events \( A \), \( B \), and \( C \)?
3. What is the interpretation of \( P(A \cup B \cup C) \) for three events?
4. How does the inclusion-exclusion principle apply in probability for multiple events?
5. How does the formula for \( P(A \cup B) \) apply to real-world scenarios?

**Tip:** When events are independent, \( P(A \cap B) = P(A) \times P(B) \), simplifying the expression for \( P(A \cup B) \).

Consider two events A and B. Prove that P(A ∪ B) = P(A) + P(B) − P(A ∩ B).

Discover how to prove the probability formula P(A ∪ B) = P(A) + P(B) − P(A ∩ B) using principles of set theory and probability. This proof involves understanding the union and intersection of events, and the Inclusion-Exclusion Principle. Suitable for students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation