If A and B are two events, prove that \( P(A \cap B) \geq 1 - P(A^c) - P(B^c) \)


We are asked to prove the inequality:

\[
P(A \cap B) \geq 1 - P(A^c) - P(B^c)
\]

### Proof:

1. **Recall the complement rules:**

   By definition, the complement of an event \(A\) (denoted \(A^c\)) is the event that \(A\) does not occur. Hence:
   \[
   P(A^c) = 1 - P(A)
   \]
   Similarly, for event \(B\):
   \[
   P(B^c) = 1 - P(B)
   \]

2. **Use the fact that probabilities are bounded by 1:**

   The probability of any event must satisfy \( 0 \leq P(E) \leq 1 \) for any event \( E \). Specifically:
   \[
   P(A^c) \leq 1 \quad \text{and} \quad P(B^c) \leq 1
   \]

3. **Use the union bound (Boole's inequality):**

   The probability of the union of two events \(A^c\) and \(B^c\) satisfies the following inequality:
   \[
   P(A^c \cup B^c) \leq P(A^c) + P(B^c)
   \]
   By De Morgan's law, we know that:
   \[
   A^c \cup B^c = (A \cap B)^c
   \]
   Therefore, the probability of the complement of \(A \cap B\) can be written as:
   \[
   P((A \cap B)^c) = P(A^c \cup B^c)
   \]
   Substituting this into the union bound inequality, we get:
   \[
   P((A \cap B)^c) \leq P(A^c) + P(B^c)
   \]

4. **Relating the complement to the original event:**

   The probability of the complement of \(A \cap B\) is related to the probability of \(A \cap B\) as follows:
   \[
   P((A \cap B)^c) = 1 - P(A \cap B)
   \]
   Substituting this into the inequality derived from the union bound, we get:
   \[
   1 - P(A \cap B) \leq P(A^c) + P(B^c)
   \]

5. **Rearranging the inequality:**

   To isolate \(P(A \cap B)\), subtract \(P(A^c) + P(B^c)\) from both sides:
   \[
   P(A \cap B) \geq 1 - P(A^c) - P(B^c)
   \]

This completes the proof.

### Conclusion:
We have shown that:
\[
P(A \cap B) \geq 1 - P(A^c) - P(B^c)
\]

---

Would you like further details or have any questions?

Here are 5 related questions to deepen your understanding:
1. How does the inequality change if \( A \) and \( B \) are independent events?
2. What is the interpretation of \( P(A^c \cup B^c) \) in a real-world example?
3. Can this inequality be strict, and under what conditions?
4. How does this inequality relate to conditional probabilities?
5. How would the inequality look for three events \( A \), \( B \), and \( C \)?

**Tip:** Understanding complement events is crucial for solving probability problems involving unions and intersections.

If A and B are two events, prove that P(A ∩ B) ≥ 1 - P(A^c) - P(B^c)

Discover how to prove the inequality P(A ∩ B) ≥ 1 - P(A^c) - P(B^c) using key concepts from probability theory, including complement rules, De Morgan's law, and the union bound. Suitable for advanced high school students in Grades 11-12.

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