Find the odd one out among the four digit sequences by using a simple parity (even/odd) count test: “5 8 7 8”, “6 4 8 2”, “5 7 8 8”, “9 7 4 8”.

Introduction / Context:
With short numeric strings, a fast way to detect an outlier is to count evens and odds. Parity patterns are objective, quick to compute mentally, and often deliberately used by test setters. Here, each option has four digits. The odd one should break a parity pattern held by the other three.

Given Data / Assumptions:

• A: 5 8 7 8
• B: 6 4 8 2
• C: 5 7 8 8
• D: 9 7 4 8

Concept / Approach:
Count evens in each string. Option A has 8 and 8 as evens (2 evens). Option B has 6, 4, 8, 2 — all four are even (4 evens). Option C has 8 and 8 (2 evens). Option D has 4 and 8 (2 evens). So, three strings have exactly two even digits; one string is composed entirely of even digits. That all-even string cleanly stands out as the outlier.

Step-by-Step Solution:
1) Mark evens in each option.2) Tally even counts: A=2, B=4, C=2, D=2.3) Select the unique case with 4/4 evens: 6 4 8 2.

Verification / Alternative check:
Sum or product checks are unnecessary; parity already yields a unique, unambiguous choice. Any other simple property (e.g., presence of repeated digits) does not separate exactly one item here.

Why Other Options Are Wrong:
Options A, C, and D share the property “exactly two evens,” so they form the majority group.

Common Pitfalls:
Overcomplicating with divisibility rules; when a crisp parity split exists, use it.

Final Answer:
6 4 8 2