Let `A` and `B` be two non-null events such that `AsubeB`. Then, which of the following statements is always correct? (1) (P(A/B)=P(B)−P(A) (2) P(A/B)>P(A) (3) P(A/B)≤P(A) (4) P(A/B)=1

Options

A. `p(A//B)=p(B)-P(A)`
B. `P(A//B)gtP(A)`
C. `P(A//B)leP(A)`
D. `P(A//B)=1`

Answer

To solve the problem, we need to analyze the relationship between two events \( A \) and \( B \) where \( A \subseteq B \). We will evaluate each of the given options to determine which statement is always correct.

### Step-by-step Solution:

1. **Understanding the Relationship**:
Since \( A \subseteq B \), every outcome in event \( A \) is also in event \( B \). This means that the probability of \( A \) occurring cannot exceed the probability of \( B \) occurring.

2. **Using the Conditional Probability Formula**:
The conditional probability \( P(A|B) \) is defined as:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]
Since \( A \subseteq B \), we have \( A \cap B = A \). Therefore, we can rewrite the formula as:
\[ P(A|B) = \frac{P(A)}{P(B)} \]

3. **Evaluating Each Option**:
- **Option (1)**: \( P(A|B) = P(B) - P(A) \)
- This is incorrect because \( P(A|B) = \frac{P(A)}{P(B)} \) does not equal \( P(B) - P(A) \).

- **Option (2)**: \( P(A|B) > P(A) \)
- This is incorrect. Since \( P(A|B) = \frac{P(A)}{P(B)} \) and \( P(B) \) is greater than \( P(A) \) (because \( A \subseteq B \)), it follows that \( P(A|B) < P(A) \).

- **Option (3)**: \( P(A|B) \leq P(A) \)
- This is correct. Since \( P(A|B) = \frac{P(A)}{P(B)} \) and \( P(B) \) is greater than or equal to \( P(A) \), it follows that \( P(A|B) \leq P(A) \).

- **Option (4)**: \( P(A|B) = 1 \)
- This is incorrect. \( P(A|B) = 1 \) only if \( A = B \). Since \( A \) is a subset of \( B \), this is not always true.

4. **Conclusion**:
The only statement that is always correct is option (3): \( P(A|B) \leq P(A) \).

### Final Answer:
The correct statement is (3) \( P(A|B) \leq P(A) \).