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Start the planConsider three straight lines \(A, B\) and \(C\) in a plane. \(4\) unique points are marked on line \(A,\) \(3\) unique points are marked on line \(B\) and \(6\) unique points are marked on line \(C.\) If we randomly select three marked points, find the probability that the three points form a triangle. Note that no other points, except the points on the same line, are collinear
A. \(\dfrac{25}{286}\)
B. \(\dfrac{72}{286}\)
C. \(\dfrac{90}{286}\)
D. \(\dfrac{189}{286}\)
E. \(\dfrac{261}{286}\)
[spoiler]OA=E[/spoiler]
Source: e-GMAT
A. \(\dfrac{25}{286}\)
B. \(\dfrac{72}{286}\)
C. \(\dfrac{90}{286}\)
D. \(\dfrac{189}{286}\)
E. \(\dfrac{261}{286}\)
[spoiler]OA=E[/spoiler]
Source: e-GMAT
