Competition problem — Combinatorics

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Combinatorics Difficulty 4.7 AIME Find the answer

5. As shown in Figure 3, in a $4 \times 4$ dot array, the probability of forming a triangle by randomly selecting three points is:

Pick one

Solution

5. A.

The number of ways to choose any three points (disregarding order) is $\mathrm{C}_{16}^{3}=560$ , and the number of ways that do not form a triangle (i.e., the three points are collinear) is 44. Therefore,
$P=\frac{560-44}{560}=\frac{129}{140} .$

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.