Odd parity generation: An 8-bit binary word is applied to an odd-parity generator. The parity bit will be 1 only if ________.

Introduction / Context:
Parity bits provide a simple error-detection mechanism. With odd parity, the total count of 1s across data bits plus the parity bit must be odd. Understanding when the parity bit becomes 1 is essential for designing and checking parity logic.

Given Data / Assumptions:

• 8 data bits (any pattern of 1s and 0s).
• Odd-parity convention: total ones (data + parity) is odd.
• Single parity bit added to the word.

Concept / Approach:
Let k be the number of 1s in the 8-bit word. Odd parity sets parity_bit so that k + parity_bit is odd. Therefore parity_bit = 1 when k is even, and parity_bit = 0 when k is odd. The numeric value of the word is irrelevant; only the count of 1s matters.

Step-by-Step Solution:

Verification / Alternative check:
Example: data = 10110000 has k = 3 (odd) → parity_bit = 0; total ones = 3 (odd). Another: data = 11000000 has k = 2 (even) → parity_bit = 1; total ones = 3 (odd).

Why Other Options Are Wrong:

• “number is odd/even”: The arithmetic value of the byte does not determine parity; the Hamming weight does.
• “number of 1s is odd”: For odd k, parity bit is 0, not 1.
• “MSB is 1”: Parity depends on all bits, not only the MSB.

Common Pitfalls:
Confusing odd vs even parity definitions; focusing on numeric value rather than bit count; omitting the parity bit in the final odd-count requirement.

Final Answer:
the number of 1s in the word is even