Step-by-step Solution:

\[Step 1:\] Analyze Statement I Statement I: If \( A \subset B \), then \( B \) can be expressed as \( B = A \cup (\overline{A} \cap B) \) and \( P(A) > P(B) \). 1. Expression for \( B \): - If \( A \subset B \), then \( B = A \cup (\overline{A} \cap B) \) is true. This is because \( \overline{A} \cap B \) represents the part of \( B \) that is not in \( A \), and the union of \( A \) and \( \overline{A} \cap B \) gives \( B \). 2. Probability Inequality \( P(A) > P(B) \): - If \( A \subset B \), then \( P(A) \leq P(B) \), not \( P(A) > P(B) \). This part of the statement is false. Conclusion: Statement I is partially true and partially false. \[Step 2:\] Analyze Statement II Statement II: If \( A \) and \( B \) are independent events, then \( (A \) and \( \overline{B}) \), \( (\overline{A} \) and \( B) \), and \( (\overline{A} \) and \( \overline{B}) \) are also independent. 1. Independence of \( A \) and \( \overline{B} \): - If \( A \) and \( B \) are independent, then \( A \) and \( \overline{B} \) are also independent. This is because: \[ P(A \cap \overline{B}) = P(A) - P(A \cap B) = P(A) - P(A)P(B) = P(A)(1 - P(B)) = P(A)P(\overline{B}). \] 2. Independence of \( \overline{A} \) and \( B \): - Similarly, \( \overline{A} \) and \( B \) are independent because: \[ P(\overline{A} \cap B) = P(B) - P(A \cap B) = P(B) - P(A)P(B) = P(B)(1 - P(A)) = P(\overline{A})P(B). \] 3. Independence of \( \overline{A} \) and \( \overline{B} \): - \( \overline{A} \) and \( \overline{B} \) are also independent because: \[ P(\overline{A} \cap \overline{B}) = 1 - P(A \cup B) = 1 - [P(A) + P(B) - P(A \cap B)] = 1 - P(A) - P(B) + P(A)P(B) = (1 - P(A))(1 - P(B)) = P(\overline{A})P(\overline{B}). \] Conclusion: Statement II is true. \[Step 3:\] Final Conclusion - Statement I: Partially true and partially false. - Statement II:True. Thus, the most appropriate answer is: Correct Answer: \(\boxed{D}\)