How to Prove that P (Exactly one of A, B, C occurs) = P(A) + P(B) + P(C)-2P(A∩B)-2P (B∩C) -2P(A∩C)+3P(A∩B∩C)?

To prove that the probability of exactly one of the events \( A \), \( B \), or \( C \) occurring is equal to the expression

\[
P(A) + P(B) + P(C) - 2P(A \cap B) - 2P(B \cap C) - 2P(A \cap C) + 3P(A \cap B \cap C),
\]

we need to break down the problem using principles from probability theory, particularly the inclusion-exclusion principle.

### Step 1: Definition of the Events
Let's denote the event "exactly one of \( A \), \( B \), or \( C \) occurs" by \( E \). The event \( E \) can be written as:

\[
E = (A \cap B^c \cap C^c) \cup (A^c \cap B \cap C^c) \cup (A^c \cap B^c \cap C)
\]

Here:
- \( A^c \) is the complement of \( A \), meaning event \( A \) does not occur.
- \( A \cap B^c \cap C^c \) represents the event where \( A \) occurs but neither \( B \) nor \( C \) occur.
- Similarly for the other terms.

### Step 2: Probability of \( E \)
The probability of \( E \) (exactly one event occurring) is:

\[
P(E) = P(A \cap B^c \cap C^c) + P(A^c \cap B \cap C^c) + P(A^c \cap B^c \cap C)
\]

### Step 3: Calculate Each Probability Using Inclusion-Exclusion
We calculate each of the probabilities above using the principle of inclusion-exclusion.

1. **\( P(A \cap B^c \cap C^c) \)**:
   \[
   P(A \cap B^c \cap C^c) = P(A) - P(A \cap B) - P(A \cap C) + P(A \cap B \cap C)
   \]

2. **\( P(A^c \cap B \cap C^c) \)**:
   \[
   P(A^c \cap B \cap C^c) = P(B) - P(A \cap B) - P(B \cap C) + P(A \cap B \cap C)
   \]

3. **\( P(A^c \cap B^c \cap C) \)**:
   \[
   P(A^c \cap B^c \cap C) = P(C) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)
   \]

### Step 4: Sum the Probabilities
Now, sum the probabilities:

\[
P(E) = \left[P(A) - P(A \cap B) - P(A \cap C) + P(A \cap B \cap C)\right] + \left[P(B) - P(A \cap B) - P(B \cap C) + P(A \cap B \cap C)\right] + \left[P(C) - P(A \cap C) - P(B \cap C) + P(A \cap B \cap C)\right]
\]

### Step 5: Simplify the Expression
Combine all the terms:

\[
P(E) = P(A) + P(B) + P(C) - 2P(A \cap B) - 2P(A \cap C) - 2P(B \cap C) + 3P(A \cap B \cap C)
\]

### Conclusion
This expression matches exactly with the one given in the problem. Therefore, we have proved that:

\[
P(\text{Exactly one of } A, B, C \text{ occurs}) = P(A) + P(B) + P(C) - 2P(A \cap B) - 2P(B \cap C) - 2P(A \cap C) + 3P(A \cap B \cap C)
\]

This completes the proof.

Would you like more details or have any other questions? 

Here are 5 related questions:
1. How would this formula change if we wanted the probability of exactly two events occurring?
2. What happens to the formula if the events \( A \), \( B \), and \( C \) are mutually exclusive?
3. Can you derive the formula for the probability that at least one of \( A \), \( B \), or \( C \) occurs?
4. How can this proof be extended to four events \( A \), \( B \), \( C \), and \( D \)?
5. What is the geometric interpretation of the inclusion-exclusion principle?

**Tip**: When working with probabilities, drawing a Venn diagram can be very helpful in visualizing the relationships between different events.

Explore how to prove the probability of exactly one event occurring using the inclusion-exclusion principle in probability theory. Learn the formula step-by-step with detailed explanations and examples.

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