If A and B are two events, then `P(A) + P(B) = 2P(AnnB)` if and only if

A. `P(A)+P(B)=1`
B. `P(A)=P(B)`
C. `P(A)=P(B)`
D. None of these

Options

A. `P(A)+P(B)=1`
B. `P(A)=P(B)`
C. `P(A)+P(B)gt1`
D. None of these

Answer

To solve the problem, we need to analyze the given equation and the conditions provided.

### Given:
We have two events A and B. The equation we need to analyze is:

\[ P(A) + P(B) = 2P(A \cap B) \]

### Step 1: Rearranging the Equation
We can rearrange the equation to express it in a different form:

\[ P(A) + P(B) - 2P(A \cap B) = 0 \]

### Step 2: Using the Formula for Probability
We know from probability theory that:

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

Using this, we can express \( P(A) + P(B) \) in terms of \( P(A \cup B) \):

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

### Step 3: Substituting Back into the Equation
Now, substituting \( P(A) + P(B) \) back into our rearranged equation gives us:

\[ P(A \cup B) + P(A \cap B) - 2P(A \cap B) = 0 \]

This simplifies to:

\[ P(A \cup B) - P(A \cap B) = 0 \]

### Step 4: Conclusion from the Simplified Equation
From the simplified equation, we can conclude that:

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

This means that the probability of the union of A and B is equal to the probability of their intersection. This situation occurs only when A and B are the same event or one event is a subset of the other.

### Step 5: Analyzing the Options
Now, let’s analyze the given options:

- **A. \( P(A) + P(B) = 1 \)**: This is not necessarily true, as we have shown that the sum can be equal to twice the intersection.

- **B. \( P(A) = P(B) \)**: This can be true, as we have seen that if both events have the same probability, it can satisfy the equation.

- **C. \( P(A) = P(B) \)**: This is identical to option B and is also true.

- **D. None of these**: This is incorrect since we have valid options.

### Final Answer:
The correct answer is **B and C** (both are valid).