Two computers exchange ASCII messages using a simple parity check (single parity bit per block). Which error pattern will always remain undetected by this parity method?

Introduction / Context:
A simple parity check appends one parity bit so that the total number of 1s in a protected block is always even (even parity) or always odd (odd parity). The receiver recomputes the parity and flags an error if the parity does not match. While parity is inexpensive, it cannot detect all error patterns.

Given Data / Assumptions:

• ASCII transmission with a single parity bit per protected block (byte or larger block as defined).
• Parity decision rule: detect a mismatch in overall parity.
• We consider arbitrary bit inversions during transmission.

Concept / Approach:
Parity detects any error pattern that flips an odd number of bits, because an odd-bit flip toggles overall parity. Conversely, flipping an even number of bits leaves the parity unchanged, so the receiver sees a “valid” parity and accepts a corrupted block—an undetected error. This limitation motivates stronger error-detection codes such as checksums and CRCs.

Step-by-Step Solution:
Assume even parity for concreteness; the argument is symmetrical for odd parity.If 1, 3, 5, ... bits flip → total count of 1s changes by an odd number → parity toggles → error detected.If 2, 4, 6, ... bits flip → total count changes by an even number → parity unchanged → error undetected.

Verification / Alternative check:
Consider a simple 7-bit ASCII character plus parity. Flipping any two bits within the protected unit results in the same overall parity state. Generalizing to a whole block protected by one parity bit yields the same conclusion.

Why Other Options Are Wrong:
Odd-bit inversions (options a and c) are always detected because parity flips.

Even-bit inversions limited to a single byte (option d) are a subset of the more general undetectable pattern across the block; the question asks for what “will always” be undetected, and the correct, general statement is the even-bit case across the protected block.

None of the above: Incorrect because even-bit errors are the known blind spot.

Common Pitfalls:
Thinking parity can locate the bit in error—it cannot; it only signals presence/absence of an error and completely misses even-bit flips.

Final Answer:
Two bits or any even number of bits inverted in a block