If `P(A|B) gt P(A)`, then which of the following is correct:
(A) `P(B | A) lt; P(B)` (B) `P(AnnB) lt P(A)* P(B)` (c)` P(B|A) gt P(B)` (D) `P(B | A) = P(B)`

Explanation

To solve the problem, we need to analyze the condition given and derive the implications step by step.

### Step 1: Understand the Given Condition
We are given that:
\[ P(A|B) > P(A) \]
This means that the probability of event A occurring given that event B has occurred is greater than the probability of event A occurring independently.

### Step 2: Use the Definition of Conditional Probability
By the definition of conditional probability, we have:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]
Substituting this into our condition gives:
\[ \frac{P(A \cap B)}{P(B)} > P(A) \]

### Step 3: Rearranging the Inequality
Multiplying both sides by \( P(B) \) (assuming \( P(B) > 0 \)):
\[ P(A \cap B) > P(A) \cdot P(B) \]
This shows that the probability of both A and B occurring together is greater than the product of their individual probabilities.

### Step 4: Analyze the Options
Now, we need to analyze the options provided:

- **Option (A)**: \( P(B|A) < P(B) \)
- **Option (B)**: \( P(A \cap B) < P(A) \cdot P(B) \)
- **Option (C)**: \( P(B|A) > P(B) \)
- **Option (D)**: \( P(B|A) = P(B) \)

From our derived condition \( P(A \cap B) > P(A) \cdot P(B) \), we can conclude that option (B) is incorrect.

### Step 5: Find \( P(B|A) \)
Now, let's find \( P(B|A) \):
Using the definition of conditional probability:
\[ P(B|A) = \frac{P(A \cap B)}{P(A)} \]
Since we know \( P(A \cap B) > P(A) \cdot P(B) \), we can substitute:
\[ P(B|A) = \frac{P(A \cap B)}{P(A)} > \frac{P(A) \cdot P(B)}{P(A)} = P(B) \]
This shows that:
\[ P(B|A) > P(B) \]
Thus, option (C) is correct.

### Conclusion
The correct answer is:
**(C) \( P(B|A) > P(B) \)**