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Odd-parity validation example: For an ODD-parity system, consider data = 011011100 (9 bits). With parity bit = 0, does the word pass an odd-parity check? Evaluate.

Difficulty: Easy

Correct
Incorrect
Valid only if start/stop bits are ignored
Cannot be determined without clock polarity

Correct Answer: Correct

Explanation:

Introduction / Context:
Parity schemes append one extra bit to help detect transmission errors. In an odd-parity system, the total number of 1s in the complete word (data + parity) must be odd. We check whether a given data word paired with a stated parity bit satisfies this rule.

Given Data / Assumptions:

Data bits: 011011100.
Parity type: odd parity.
Parity bit provided: 0.

Concept / Approach:
Count the number of 1s in the data. If the count is already odd, then an odd-parity bit of 0 keeps the total odd. If the count is even, the parity bit must be 1 to make the total odd.

Step-by-Step Solution:

Count ones in 011011100 → ones at positions 2,3,5,6,7 = 5 ones.Five is odd, so to maintain odd parity, parity bit should be 0.Total ones including parity = 5 + 0 = 5 (still odd).Therefore, the provided combination passes an odd-parity check.

Verification / Alternative check:
If one data bit flips (e.g., to 011011101), ones become 6; with parity 0 total is 6 (even) and the checker flags an error.

Why Other Options Are Wrong:

Incorrect: Miscounts ones or misapplies odd-parity rule.Valid only if start/stop bits are ignored: Framing bits are not part of the payload parity in typical UART schemes.Cannot be determined: Sufficient information is provided (data and parity).

Common Pitfalls:
Off-by-one in counting ones; confusing even vs odd parity; mixing parity with checksums/CRCs.

Odd-parity validation example: For an ODD-parity system, consider data = 011011100 (9 bits). With parity bit = 0, does the word pass an odd-parity check? Evaluate.

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