Question
A conservator installs two lamps at the same time. Lamp A has an exponentially distributed lifetime with failure rate 0.002 per hour, while lamp B's lifetime is exponential with rate 0.003 per hour. Their lifetimes are independent.
The conservator replaces whichever lamp fails first. What is the probability that the first replacement is for lamp A, and what is the probability it is for lamp B? Derive the probabilities from the lifetime distributions rather than simply comparing the lamps' mean lifetimes.
Get hint
Hint 1
To be the first failure at time t, lamp A must fail near t and lamp B must still be operating at t.
Check your answer
Solution
Let \(T_A\sim\mathrm{Exp}(\lambda_A)\) and \(T_B\sim\mathrm{Exp}(\lambda_B)\) be independent, with rates \(\lambda_A=0.002\) and \(\lambda_B=0.003\) per hour. For lamp A to fail first at time \(t\), A must fail then while B survives beyond \(t\). Independence gives
\[\Pr(T_A<T_B)=\int_0^{\infty}f_A(t)\Pr(T_B>t)\,dt.\]
Using \(f_A(t)=\lambda_Ae^{-\lambda_A t}\) and \(\Pr(T_B>t)=e^{-\lambda_B t}\),
\[\Pr(T_A<T_B)=\int_0^{\infty}\lambda_Ae^{-(\lambda_A+\lambda_B)t}\,dt=\frac{\lambda_A}{\lambda_A+\lambda_B}=\frac{2}{5}.\]
The same calculation with A and B interchanged gives \(\Pr(T_B<T_A)=\lambda_B/(\lambda_A+\lambda_B)=3/5\). The two probabilities sum to one; the equality event has probability zero because these are independent continuous lifetimes.
Lamp B does have a shorter mean lifetime (\(1/0.003\) versus \(1/0.002\) hours), but comparing those means alone does not determine a race probability for arbitrary lifetime distributions. The exact 2/5–3/5 split follows from the exponential densities, their constant hazard rates and independence.
Assumptions
Lamp lifetimes are independent exponential random variables with constant failure rates in the same per-hour units.
Both lamps start operating at the same time and are observed continuously until the first failure.
There is no shared shock that could cause simultaneous failure.
Final Answer
\(\Pr(\text{A fails first})=\frac{0.002}{0.002+0.003}=\frac{2}{5}\); \(\Pr(\text{B fails first})=\frac{3}{5}\).