Solution
Twice as Long
Show the problem again
Two batteries have independent lifetimes X and Y, each exponentially distributed with the same mean. What is the probability that X > 2Y, that is, the first battery lasts more than twice as long as the second? Give a reduced fraction.
Worked solution
The answer is 1/3. Take the mean to be 1 (the answer does not depend on the scale). Conditioning on Y = y, the chance that X exceeds 2y is e^(−2y). Averaging over the density of Y, P(X > 2Y) = ∫₀^∞ e^(−2y) e^(−y) dy = 1/3. A cleaner view: 2Y is exponential with rate 1/2, so P(X > 2Y) is the probability that a rate-1/2 clock rings before a rate-1 clock, which is (1/2)/(1/2 + 1) = 1/3.
Source: Classic exponential-distribution exercise in wide circulation; no traceable original source. Statement written for AxiomIQ.