A pump running two years so far

Question

A pump's time to failure is modeled as exponential with a mean of 4 years. It has been running for 2 years without failing.

Given the 2 years already in service, what is the probability it keeps running at least 1 more year?

Accepted Answer

The exponential is memoryless, so a pump that has not yet failed is treated as good as new. The 2 years already logged do not make failure any more imminent: P(T s + t \mid T s) = P(T t). So the 2 years drop out and we only need the chance a fresh exponential wait exceeds 1 year. With mean 4 years, the rate is \lambda = 	frac{1}{4} per year, so for t = 1, P(	ext{at least 1 more year}) = e^{-\frac{1}{4}\cdot 1} = e^{-0.25} \approx 0.779. One year is only a quarter of the average lifetime, so surviving that extra stretch is quite likely, about 78%, and this figure would be identical for a bra…

A pump running two years so far

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