If A and B are independent events, then A' and B' are also

Options

A. not independent
B. also independent
C. mutually exclusive
D. none of these

Answer

To prove that if A and B are independent events, then A' (the complement of A) and B' (the complement of B) are also independent, we can follow these steps:

### Step-by-step Solution:

1. **Understanding Independence**:
- Two events A and B are independent if:
\[ P(A \cap B) = P(A) \cdot P(B) \]

2. **Complement of Events**:
- The complement of an event A is denoted as A' (or A complement), and it represents the event that A does not occur. Similarly, for event B, its complement is denoted as B'.

3. **Finding the Probability of Complements**:
- The probabilities of the complements are given by:
\[ P(A') = 1 - P(A) \]
\[ P(B') = 1 - P(B) \]

4. **Finding the Probability of the Intersection of Complements**:
- We need to find \( P(A' \cap B') \). By using De Morgan's law, we know:
\[ A' \cap B' = (A \cup B)' \]
- Therefore,
\[ P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B) \]

5. **Using the Formula for the Union of Two Events**:
- The probability of the union of two events can be expressed as:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
- Since A and B are independent, we substitute:
\[ P(A \cap B) = P(A) \cdot P(B) \]
- Thus:
\[ P(A \cup B) = P(A) + P(B) - P(A) \cdot P(B) \]

6. **Substituting Back to Find \( P(A' \cap B') \)**:
- Now substituting this back into our equation for \( P(A' \cap B') \):
\[ P(A' \cap B') = 1 - (P(A) + P(B) - P(A) \cdot P(B)) \]
- Simplifying this gives:
\[ P(A' \cap B') = 1 - P(A) - P(B) + P(A) \cdot P(B) \]

7. **Rearranging**:
- Rearranging the terms:
\[ P(A' \cap B') = (1 - P(A))(1 - P(B)) = P(A') \cdot P(B') \]

8. **Conclusion**:
- Since we have shown that:
\[ P(A' \cap B') = P(A') \cdot P(B') \]
- This means that A' and B' are independent events.

### Final Statement:
Thus, if A and B are independent events, then A' and B' are also independent.

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