Even-parity check: For the 9-bit data word 100111100, determine whether an even-parity system with parity bit = 0 is a correct combination (i.e., total number of 1s must be even).

Difficulty: Easy

Correct Answer: Incorrect

Explanation:


Introduction / Context:
Parity bits provide simple error detection. In an even-parity system, the total count of 1s across the data bits plus the parity bit must be even. Verifying parity requires counting ones accurately and checking whether the parity setting produces an even total.

Given Data / Assumptions:

  • Data word: 100111100 (9 bits).
  • Parity scheme: even parity.
  • Proposed parity bit value: 0.


Concept / Approach:
Count the number of 1s in the data. If the count is odd, the parity bit must be 1 to make the total even; if the count is even, the parity bit should be 0. This is independent of byte ordering or endianness since parity is a scalar property of the bit set.

Step-by-Step Solution:

Write data: 1 0 0 1 1 1 1 0 0.Count ones: positions 1, 4, 5, 6, 7 → total = 5.Even parity requires total ones (data + parity) to be even.With parity bit = 0, total remains 5 (odd) → parity condition not met.


Verification / Alternative check:

Set parity bit = 1 instead: total would be 6 (even), which would satisfy even parity.


Why Other Options Are Wrong:

Correct: Fails because 5 ones + 0 parity is odd.Odd parity reference / Depends on endianness: The scheme is explicitly even parity; endianness does not affect the count of ones.


Common Pitfalls:

Miscounting ones in longer bit strings.Confusing odd vs. even parity conventions.


Final Answer:

Incorrect

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