For any two events A and B, if `P (A uu B) = P ( A nn B, ` then

Options

A. `P (A) = P (B)`
B. `P (A) gt P (B)`
C. `P (A) lt P (B)`
D. None of these

Answer

To solve the problem, we need to analyze the given condition:

**Given:**
\[ P(A \cup B) = P(A \cap B) \]

We need to derive what this implies about the events A and B.

### Step 1: Use the formula for the probability of the union of two events.

The formula for the probability of the union of two events is:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]

### Step 2: Substitute the given condition into the formula.

From the given condition, we can substitute \( P(A \cup B) \) with \( P(A \cap B) \):
\[ P(A \cap B) = P(A) + P(B) - P(A \cap B) \]

### Step 3: Rearrange the equation.

Now, we can rearrange the equation:
\[ P(A \cap B) + P(A \cap B) = P(A) + P(B) \]
\[ 2P(A \cap B) = P(A) + P(B) \]

### Step 4: Divide both sides by 2.

Dividing both sides by 2 gives:
\[ P(A \cap B) = \frac{P(A) + P(B)}{2} \]

### Step 5: Analyze the implications.

This indicates that the probability of the intersection of A and B is equal to the average of the probabilities of A and B.

### Step 6: Consider the case when A and B are independent.

If A and B are independent events, then:
\[ P(A \cap B) = P(A) \cdot P(B) \]

### Step 7: Set up the equation for independent events.

From our previous result:
\[ P(A) \cdot P(B) = \frac{P(A) + P(B)}{2} \]

### Step 8: Rearrange this equation.

Rearranging gives us:
\[ 2P(A) \cdot P(B) = P(A) + P(B) \]

### Step 9: Set up a quadratic equation.

Rearranging further leads to:
\[ 2P(A)P(B) - P(A) - P(B) = 0 \]

### Step 10: Factor the quadratic equation.

This can be factored as:
\[ (2P(A) - 1)(2P(B) - 1) = 1 \]

### Conclusion:

The only way for \( P(A \cap B) \) to equal \( P(A \cup B) \) is if both probabilities are zero. Therefore, we conclude that:
\[ P(A) = 0 \quad \text{and} \quad P(B) = 0 \]

### Final Result:

Thus, the correct conclusion is:
- Both events A and B must have probabilities equal to zero.

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