2008 AMC 8 Problem 19

Attempt Problem 19 of the 2008 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

19.

Eight points are spaced at intervals of one unit around a 2×22\times2 square, as shown. Two of the 88 points are chosen at random. What is the probability that the points are one unit apart?
 14 \ \dfrac{1}{4}
 27 \ \dfrac{2}{7}
 411 \ \dfrac{4}{11}
 12 \ \dfrac{1}{2}
 47 \ \dfrac{4}{7}
Answer: B
Concepts:basic probabilitycounting pairs
Small hint:
Count the total number of unordered pairs of points.
Big hint:
There are exactly 88 adjacent unit segments around the square.
Solution:
Each dot has 22 dots that are one unit away from it.
Therefore, regardless of the choice of the first dot, 22 of the other 77 dots would be within one unit, so the probability is 27.\dfrac 27.
Thus, the answer is B .
LIVE by Po-Shen Loh
Programs

LIVE Classes

Self-Paced Classes

Stars

Scholarship

About Us

Who Is Po-Shen Loh?

Email Contact

Newsletter

Jobs

Become a Star

For a more Thought Full world.

© 2026 Expii, Inc.