Solution
Finishing in Order
Show the problem again
Three machines have independent exponential lifetimes. Machine A has mean lifetime 12 months, machine B has mean 6 months, and machine C has mean 4 months. What is the probability that A fails first, then B, then C? Give a reduced fraction.
Worked solution
The answer is 1/15. In failures per year the rates are 1 for A, 2 for B and 3 for C. Among competing exponential clocks, the one that rings first is clock i with probability (rate i)/(sum of rates). So A fails first with probability 1/(1 + 2 + 3) = 1/6. After A fails, B and C are, by memorylessness, as good as new, so B fails before C with probability 2/(2 + 3) = 2/5. Multiplying, P = (1/6)(2/5) = 1/15.
Source: Classic exponential-distribution exercise in wide circulation; no traceable original source. Statement written for AxiomIQ.