Which lightbulb burns out first?

Question

Three lightbulbs are switched on together. Their lifetimes are independent and exponentially distributed with failure rates $\lambda_1 = 1$ , $\lambda_2 = 2$ , and $\lambda_3 = 3$ (per year).

(a) How long, on average, until the first bulb fails? (b) What is the probability that bulb $3$ (the highest-rate one) is the first to fail?

Accepted Answer

Exponential lifetimes "compete," and the minimum of independent exponentials is the key object. (a) Time of first failure. The first failure happens after time T = \min(T1, T2, T3). For independent exponentials, the minimum is itself exponential with rate equal to the sum of the rates: T \sim 	ext{Exponential}(\lambda1 + \lambda2 + \lambda3) = 	ext{Exponential}(6). Why: the chance all three survive past time t is e^{-\lambda1 t}e^{-\lambda2 t}e^{-\lambda3 t} = e^{-(\lambda1+\lambda2+\lambda3)t}. So E[T] = \frac{1}{\lambda1 + \lambda2 + \lambda3} = \frac{1}{6}	ext{ year} \approx 2 	ext{ mo…

Which lightbulb burns out first?

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