Consider the array arr = [1, 2, 3, 4, 5] and the

Solution

Correct Answer: B (Step 1: [1, 2, 9, 3, 4, 5]. Step 2: pop(3) removes 3, resulting in [1, 2, 9, 4, 5]. Oh, wait. arr.pop(3) removes the element at index 3. After arr.insert(2, 9), arr is [1, 2, 9, 3, 4, 5]. Index 3 is 3. So arr.pop(3) removes 3. The array becomes [1, 2, 9, 4, 5]. Let me re-evaluate the options. My initial thought was [1, 2, 9, 5]. Let's re-trace. Initial: [1, 2, 3, 4, 5] 1. arr.insert(2, 9): [1, 2, 9, 3, 4, 5] (index 0=1, 1=2, 2=9, 3=3, 4=4, 5=5) 2. arr.pop(3): removes element at index 3, which is 3.    Result: [1, 2, 9, 4, 5] So, option A is correct. My previous dry run was wrong. Let's ensure the options are correct. A) [1, 2, 9, 4, 5] - This is the correct result. B) [1, 2, 9, 5] - Incorrect. C) [1, 9, 2, 4, 5] - Incorrect. D) [1, 9, 2, 5] - Incorrect. E) [1, 2, 3, 9, 5] - Incorrect. Correct Answer: A