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 given by:

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

we can start by breaking down what it means for "exactly one" of the events to occur. Let's denote the probability that exactly one of the events \(A\), \(B\), or \(C\) occurs as \(P(\text{exactly one})\).

### Step 1: Define the probability of exactly one event occurring.
The event "exactly one of \(A\), \(B\), \(C\) occurs" can be divided into three mutually exclusive events:
1. \(A\) occurs and \(B\) and \(C\) do not occur.
2. \(B\) occurs and \(A\) and \(C\) do not occur.
3. \(C\) occurs and \(A\) and \(B\) do not occur.

Thus,

\[
P(\text{exactly one}) = P(A \cap B^c \cap C^c) + P(B \cap A^c \cap C^c) + P(C \cap A^c \cap B^c)
\]

### Step 2: Express each of these probabilities using the inclusion-exclusion principle.
Each of these probabilities can be expressed as follows:

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

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

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

### Step 3: Add up the probabilities.
Now, sum up the three expressions:

\[
P(\text{exactly one}) = \left[P(A) - P(A \cap B) - P(A \cap C) + P(A \cap B \cap C)\right] + \left[P(B) - P(B \cap A) - P(B \cap C) + P(A \cap B \cap C)\right] + \left[P(C) - P(C \cap A) - P(C \cap B) + P(A \cap B \cap C)\right]
\]

### Step 4: Simplify the expression.
Combine all the terms:

\[
P(\text{exactly one}) = P(A) + P(B) + P(C) - [P(A \cap B) + P(A \cap C) + P(B \cap C)] - [P(A \cap B) + P(A \cap C) + P(B \cap C)] + 3P(A \cap B \cap C)
\]

This simplifies to:

\[
P(\text{exactly one}) = P(A) + P(B) + P(C) - 2[P(A \cap B) + P(A \cap C) + P(B \cap C)] + 3P(A \cap B \cap C)
\]

### Conclusion
Thus, the probability of exactly one of the events \(A\), \(B\), or \(C\) occurring is indeed:

\[
P(\text{exactly one}) = P(A) + P(B) + P(C) - 2P(A \cap B) - 2P(A \cap C) - 2P(B \cap C) + 3P(A \cap B \cap C)
\]

Would you like to explore this proof further, or do you have any questions?

Here are some related questions you might find interesting:
1. How would this proof change if we considered "exactly two" of the events occurring?
2. What if the events \(A\), \(B\), and \(C\) are independent? How does the formula simplify?
3. Can we generalize this result for more than three events?
4. What is the probability that "at least one" of the events \(A\), \(B\), or \(C\) occurs?
5. How can this formula be applied in real-world scenarios?

**Tip:** When dealing with probabilities involving multiple events, breaking down complex events into simpler, mutually exclusive events can make the calculations easier.

Explore a detailed proof of the probability that exactly one of the events A, B, or C occurs using the inclusion-exclusion principle. Understand how to calculate probabilities of mutually exclusive events and the intersections of sets. Suitable for advanced high school students and beyond.

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