Three uniform lengths forming a triangle

Question

Three lengths $X,Y,Z$ are drawn independently from $\text{Unif}(0,1)$ . What is the probability that a triangle can be built with these three side lengths?

Short answer

$\tfrac12$ . Failure means one length exceeds the sum of the other two; these three events are disjoint and equally likely. The region $Z>X+Y$ in the unit cube is a tetrahedron of volume $\tfrac16$ , so the failure probability is $\tfrac12$ and success is $\tfrac12$ .

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Lessons behind this question

Continuous Random Variables and Probability Density Functions
Cumulative Distribution Functions
Sample Spaces, Events, and the Probability Axioms

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