Final answer:
During the binary search for the target value of 15, the mid variable progresses through the indices of 5, 2, and finally 3 where the target value is found. The correct choice is B. 5, 2, 3.
Step-by-step explanation:
The question asks about the progression of the mid variable value during a binary search operation in an ordered array to locate the target value of 15.
We begin by setting the initial left (low) pointer at 0 and the right (high) pointer at the length of the array minus 1, which is 11 in this case. The first middle (mid) pointer would be calculated as (low + high) / 2, which is (0 + 11) / 2 = 5.5. Since we're dealing with indices, we take the integer part, so mid = 5. The value at index 5 is 24, which is greater than 15, so we move our right pointer to the mid-1 index, which is 4.
The next calculation would then be (0 + 4) / 2 = 2, so mid = 2. The value at index 2 is 9, which is less than 15, so we now move our low pointer to mid+1, which is 3. The next mid calculation is thus (3 + 4) / 2 = 3.5, so mid = 3. The value at index 3 is 15, which is the target value we were searching for.
Therefore, the progression of the mid variable would be 5, 2, 3, indicating that the correct choice is B. 5, 2, 3.