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.

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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
Difficulty rating: 1310
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 .

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