`P(A cup B)=P(A cap B)` if and only if the relation between P(A) and P(B) is …………

Options

A. P(A)+P(B) = 2P(A)P`(B/A)`
B. P(A)+P(B) = 2P(A)P`(A/B)`
C. P(A) +P(B) = 2P(A)P(B)
D. P(A) + P(B) = P(A)P(B)

Answer

To solve the problem, we need to analyze the relationship between the probabilities of events A and B given the condition that \( P(A \cup B) = P(A \cap B) \).

### Step-by-Step Solution:

1. **Understanding the Given Condition**:
We start with the equation given in the problem:
\[ P(A \cup B) = P(A \cap B) \]

2. **Using the Formula for Union of Two Events**:
We know from probability theory that:
\[ P(A \cup B) = P(A) + P(B) - P(A \cap B) \]
Substituting this into our condition gives us:
\[ P(A) + P(B) - P(A \cap B) = P(A \cap B) \]

3. **Rearranging the Equation**:
Rearranging the equation leads to:
\[ P(A) + P(B) = 2P(A \cap B) \]

4. **Expressing \( P(A \cap B) \)**:
We can express \( P(A \cap B) \) in terms of conditional probabilities:
\[ P(A \cap B) = P(A | B) \cdot P(B) = P(B | A) \cdot P(A) \]
Substituting this into our equation, we can write:
\[ P(A) + P(B) = 2 \cdot P(A | B) \cdot P(B) \]
or
\[ P(A) + P(B) = 2 \cdot P(B | A) \cdot P(A) \]

5. **Finding the Relationship**:
From the equation \( P(A) + P(B) = 2P(A \cap B) \), we can derive that:
- If \( P(A) + P(B) \) is equal to twice the probability of their intersection, it indicates a specific relationship between the probabilities of A and B.
- This implies that the events A and B are dependent on each other in a specific way.

6. **Conclusion**:
The relationship between \( P(A) \) and \( P(B) \) can be summarized as:
\[ P(A) + P(B) = 2P(A \cap B) \]
This indicates that both events A and B must occur together, and hence, they are not independent. The correct answer to the question is that the relation between \( P(A) \) and \( P(B) \) is such that:
\[ P(A) + P(B) = 2P(A \cap B) \]