← Bytedance Interview Insights
I'd seen this problem before so I knew the answer was 1/4, but actually deriving it cleanly under pressure is another thing.
Model the two break points as independent uniform random variables on [0,1]. The three pieces form a triangle if and only if each piece is less than 1/2. Compute the probability by integrating over the region in the unit square where this condition holds.
Pro tip: Emphasize that the triangle condition is equivalent to the longest piece being less than the sum of the other two, which simplifies to each piece < 1/2. This geometric insight avoids messy algebra and shows strong probabilistic intuition.
Let X and Y be the two break points, independently and uniformly distributed on [0,1]. Assume X < Y without loss of generality (by symmetry).
The three pieces have lengths X, Y - X, and 1 - Y. The triangle condition is that each length is less than 1/2.
The conditions X < 1/2, Y - X < 1/2, and 1 - Y < 1/2 simplify to X < 1/2, Y < X + 1/2, and Y > 1/2.
In the unit square (X,Y) with X < Y, the valid region is a triangle with vertices (0, 1/2), (1/2, 1/2), and (1/2, 1). Its area is 1/8. Since the total area of the region X < Y is 1/2, the conditional probability is (1/8) / (1/2) = 1/4.
The probability is 1/4. Optionally, verify by simulation or mention that the result is a classic problem.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.