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.

Question

Solve this multiple-choice question and choose the correct option.

Accepted Answer

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. Final Answer:Correct

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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