Parity check with odd parity: For the 9-bit data pattern 100011010, would this word pass an odd-parity check at the receiver (i.e., total count of 1s should be odd)? Choose the best evaluation.

Introduction / Context:
Parity is one of the simplest error-detection methods used in digital communications and storage. With odd parity, the total number of 1 bits in the transmitted unit (data plus any parity bit) must be odd. The receiver recomputes the number of 1s and checks whether the rule holds. This question asks you to determine whether the 9-bit word 100011010 would pass an odd-parity check when evaluated as received data (no extra parity bit shown).

Given Data / Assumptions:

• Word under test: 100011010 (9 bits shown explicitly).
• Odd parity rule: total count of 1s must be odd.
• Assume these 9 bits constitute the unit being checked (no hidden parity bit beyond what is shown).
• Noise may flip bits; parity aims only to detect, not correct, such flips.

Concept / Approach:
To test odd parity, count the number of 1s. If the count is odd, the unit passes; if the count is even, it fails. Parity is a lightweight integrity check and primarily detects any odd number of bit errors; it cannot detect all multi-bit error patterns (for example, two-bit flips can go undetected).

Step-by-Step Solution:

Verification / Alternative check:
To make this unit satisfy odd parity, you would append a parity bit of 1 (increasing the total number of 1s from 4 to 5). The placement (MSB or LSB) depends on protocol, but the value required is 1 to achieve an odd count.

Why Other Options Are Wrong:

Common Pitfalls:
Confusing odd with even parity, miscounting 1s, or assuming an implicit parity bit when none is shown. Always count exactly the bits under test and apply the specified parity rule.

Final Answer:
Incorrect — it fails the odd-parity check