Solution
Exponential Sticks
Show the problem again
Three rods have lengths that are independent exponential random variables with the same mean. What is the probability that the three rods can be assembled into a (non-degenerate) triangle? Give a reduced fraction.
Worked solution
The answer is 1/4. The rods fail to form a triangle exactly when one rod is at least as long as the other two combined, and at most one rod can do that. So P(no triangle) = 3·P(X₃ ≥ X₁ + X₂). To compute P(X₃ > X₁ + X₂): first X₃ must exceed X₁, which has probability 1/2. Given that, by memorylessness the excess X₃ − X₁ is again exponential with the same mean, independent of X₂, so it exceeds X₂ with probability 1/2. Thus P(X₃ > X₁ + X₂) = 1/4, P(no triangle) = 3/4, and P(triangle) = 1/4. For comparison, three independent uniform lengths on [0, 1] form a triangle with probability 1/2; the heavy right tail of the exponential makes one dominant rod much more likely.
Source: Original problem written for AxiomIQ.