Odd parity assignment for data words For each 7- or 8-bit data word, determine the required parity bit P to achieve odd parity: • 1011101 • 11110111 • 1001101

Introduction / Context:
Parity bits provide simple error detection over a set of data bits. For odd parity, the total number of 1s—including the parity bit—must be odd. This question asks you to compute the parity bit P for three separate data words.

Given Data / Assumptions:

• Words: 1011101, 11110111, 1001101.
• Odd parity required: total count of 1s (data + P) must be odd.
• We append one bit P to each word independently.

Concept / Approach:
Count the number of 1s in each data word. If the count is already odd, set P = 0. If the count is even, set P = 1 to make the total odd. This follows directly from parity definition and requires only a popcount of each word.

Step-by-Step Solution:

Verification / Alternative check:
Add P and recount: Word 1 total ones 5, Word 2 total ones 7, Word 3 total ones 5 → all odd, confirming correctness.

Why Other Options Are Wrong:

• P = 1, 1, 0 and P = 1, 1, 1: Do not satisfy odd parity for the words with already-odd one counts.
• P = 0, 0, 0: Fails for the third word which has an even number of ones and requires P = 1.

Common Pitfalls:
Miscounting ones, assuming the same parity bit for all words, or forgetting that each word’s parity is computed independently.

Final Answer:
P = 0, P = 0, P = 1