Explain the term independent events. Also show that, if A and B are independent events then i) ii) AC and B are also independent. AC and BC are also independent.

### Independent Events

Two events \( A \) and \( B \) are called **independent** if the occurrence of one event does not affect the probability of the other event occurring. Mathematically, events \( A \) and \( B \) are independent if:

\[
P(A \cap B) = P(A) \times P(B)
\]

This means that the probability of both events happening together is equal to the product of their individual probabilities.

### Proof of Independence of Complementary Events

#### 1. **\( A^C \) and \( B \) are Independent**

Let \( A^C \) denote the complement of event \( A \), meaning that \( A^C \) represents the event that \( A \) does not occur. We want to show that \( A^C \) and \( B \) are independent, given that \( A \) and \( B \) are independent.

We know from the definition of independent events that:

\[
P(A \cap B) = P(A) \times P(B)
\]

Now, the complement of \( A \), \( A^C \), is everything except \( A \), so:

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

Substitute \( P(A \cap B) = P(A) \times P(B) \) into the equation:

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

Factor out \( P(B) \):

\[
P(A^C \cap B) = P(B) \times (1 - P(A))
\]

Since \( 1 - P(A) = P(A^C) \), we have:

\[
P(A^C \cap B) = P(A^C) \times P(B)
\]

This shows that \( A^C \) and \( B \) are independent.

#### 2. **\( A^C \) and \( B^C \) are Independent**

Next, we show that \( A^C \) and \( B^C \), the complements of events \( A \) and \( B \), are also independent.

We know that:

\[
P(A^C \cap B^C) = 1 - P(A \cup B)
\]

Using the inclusion-exclusion principle, we can write:

\[
P(A \cup B) = P(A) + P(B) - P(A \cap B)
\]

Since \( P(A \cap B) = P(A) \times P(B) \) (due to the independence of \( A \) and \( B \)), we substitute this into the equation:

\[
P(A^C \cap B^C) = 1 - (P(A) + P(B) - P(A) \times P(B))
\]

Simplify:

\[
P(A^C \cap B^C) = 1 - P(A) - P(B) + P(A) \times P(B)
\]

This can be factored as:

\[
P(A^C \cap B^C) = (1 - P(A)) \times (1 - P(B)) = P(A^C) \times P(B^C)
\]

Thus, \( A^C \) and \( B^C \) are independent.

---

### Conclusion

We have shown that:

- If \( A \) and \( B \) are independent events, then:
  1. \( A^C \) and \( B \) are also independent.
  2. \( A^C \) and \( B^C \) are also independent.

Would you like further clarification on any part of this proof? 

Here are some related questions for deeper understanding:
1. What are dependent events, and how do they differ from independent events?
2. How does the inclusion-exclusion principle apply in probability?
3. Can two mutually exclusive events be independent? Why or why not?
4. How does conditional probability relate to independent events?
5. What are some real-life examples of independent events?

**Tip:** When solving probability problems, always check whether events are independent before applying the multiplication rule for combined probabilities.

Explain the term independent events. Also show that, if A and B are independent events then i) AC and B are also independent. AC and BC are also independent.

Explore the concept of independent events in probability and learn how to prove that complementary events (A^C and B, A^C and B^C) are independent if A and B are independent. This detailed explanation includes key probability formulas and theorems, ideal for Grades 10-12.

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