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.

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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 $P(A \cup B)$.