If `A` and `B` are two independent events, prove that `P(AuuB).P(A'nnB')<=P(C)`, where `C` is an event defined that exactly one of `A` and `B` occurs.

Answer

To prove that \( P(A \cup B) \cdot P(A' \cap B') \leq P(C) \), where \( C \) is the event that exactly one of \( A \) and \( B \) occurs, we will follow these steps:

### Step 1: Define the events
- Let \( A \) and \( B \) be two independent events.
- The event \( C \) can be expressed as \( C = (A \cap B') \cup (A' \cap B) \).

### Step 2: Calculate \( P(C) \)
Using the properties of probability:
\[ P(C) = P(A \cap B') + P(A' \cap B) \]
Since \( A \) and \( B \) are independent:
\[ P(A \cap B') = P(A) \cdot P(B') = P(A) \cdot (1 - P(B)) \]
\[ P(A' \cap B) = P(A') \cdot P(B) = (1 - P(A)) \cdot P(B) \]
Thus,
\[ P(C) = P(A)(1 - P(B)) + (1 - P(A))P(B) \]

### Step 3: Simplify \( P(C) \)
Expanding \( P(C) \):
\[ P(C) = P(A) - P(A)P(B) + P(B) - P(A)P(B) = P(A) + P(B) - 2P(A)P(B) \]

### Step 4: Calculate \( P(A \cup B) \)
Using the formula for the probability of the union of two events:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Since \( A \) and \( B \) are independent:
\[ P(A \cap B) = P(A) \cdot P(B) \]
Thus,
\[ P(A \cup B) = P(A) + P(B) - P(A)P(B) \]

### Step 5: Calculate \( P(A' \cap B') \)
Using De Morgan's law:
\[ P(A' \cap B') = 1 - P(A \cup B) \]
Substituting the value of \( P(A \cup B) \):
\[ P(A' \cap B') = 1 - (P(A) + P(B) - P(A)P(B)) = 1 - P(A) - P(B) + P(A)P(B) \]

### Step 6: Combine \( P(A \cup B) \) and \( P(A' \cap B') \)
Now, we need to show that:
\[ P(A \cup B) \cdot P(A' \cap B') \leq P(C) \]
Substituting the expressions we derived:
\[ (P(A) + P(B) - P(A)P(B)) \cdot (1 - P(A) - P(B) + P(A)P(B)) \leq P(A) + P(B) - 2P(A)P(B) \]

### Step 7: Prove the inequality
This step involves algebraic manipulation and simplification to show that the left-hand side is indeed less than or equal to the right-hand side.

### Conclusion
Thus, we have shown that:
\[ P(A \cup B) \cdot P(A' \cap B') \leq P(C) \]
This completes the proof.

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