If a point is arbitrarily selected on a line segment, thus breaking it

Check the diagram below:



The larger segment will be at least twice as long as the smaller one if the breaking point is located either in the first third or the last third of the segment. Hence, the probability of this occurring is \(\frac{2}{3}\).

This problem can also be approached algebraically. If we consider the length of the original segment as 1 and the length of the first sub-segment as \(x\), then we are looking for values of \(x\) that satisfy \(\frac{x}{1 - x} \ge 2\) or \(\frac{1 - x}{x} \ge 2\). Solving the first inequality, we get \(x \ge \frac{2}{3}\), and for the second, we find \(x \le \frac{1}{3}\). Therefore, the probability that either \(x \ge \frac{2}{3}\) or \(x \le \frac{1}{3}\) holds is \(\frac{2}{3}\).


Answer: D