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 given condition and the options provided.

**Step 1: Understand the given condition.**
We are given that \( P(A|B) > P(A) \). By the definition of conditional probability, we can express \( P(A|B) \) as:
\[ P(A|B) = \frac{P(A \cap B)}{P(B)} \]
Thus, the condition can be rewritten as:
\[ \frac{P(A \cap B)}{P(B)} > P(A) \]

**Step 2: Rearranging the inequality.**
Multiplying both sides by \( P(B) \) (assuming \( P(B) > 0 \)):
\[ P(A \cap B) > P(A) \cdot P(B) \]
This means that the probability of both events A and B occurring together is greater than the product of their individual probabilities.

**Step 3: Analyze the options.**
Now, let’s analyze each option given in the question:

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

From our earlier result \( P(A \cap B) > P(A) \cdot P(B) \), we can conclude that option (B) is incorrect because it contradicts our finding.

**Step 4: Find \( P(B|A) \).**
Now, we can find \( P(B|A) \):
\[ P(B|A) = \frac{P(A \cap B)}{P(A)} \]
Since we know \( P(A \cap B) > P(A) \cdot P(B) \), we can substitute:
\[ P(B|A) = \frac{P(A \cap B)}{P(A)} > \frac{P(A) \cdot P(B)}{P(A)} = P(B) \]
Thus, \( P(B|A) > P(B) \), which means option (C) is correct.

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

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