A colleague and I have tried two different approaches to this problem, both of which seem to make sense but are resulting in very different answers.

Suppose we have some units undergoing B hours of testing and are interested in the probability of failure in the next hour (or a generic next time t) given successful completion of B. We have some data showing that about 15% of the units failed in the B hours of testing at various times, and have tried fitting various models to estimate the answer.

A favored approach by both of us is to use a defective subpopulation distribution: We estimate the distribution parameters (call the distribution CDF) and the proportion defective p; no problem there. The problem is how to use the model to estimate the “next time t” probability of failure.

My approach is based on conditional reliability: R(t|survived B)=R(B+t)/R(B). I set R(t)=1-p*CDF(t), plug into the conditional reliability equation, and take one minus the answer to convert back to probability of failure.

My colleague takes a different approach: He says the probability of failure is equal to p*Prob(failure in next hour|member of defective subpopulation)=p(CDF(B+t)-CDF(B))/(1-CDF(B))

Both approaches seem to make sense but the answers are off by about an order of magnitude. I of course think mine is correct but can’t clearly explain why his isn’t. Any ideas?

  • $\begingroup$ Your approach makes sense to me. Do you know where your colleague is getting their idea from? Your colleague's answer seems strange to me, but it might be based on a method I'm not familiar with. It looks like they're fitting a Cure Model (some observations will never be defective), but they would be missing a term in their model. $\endgroup$
    – Eli
    Aug 4, 2021 at 13:30
  • $\begingroup$ If you add (1-p) to your colleague's answer, do you get a similar answer to your approach? $\endgroup$
    – Eli
    Aug 4, 2021 at 13:41
  • $\begingroup$ @Eli my colleague's approach follows from this line of reasoning: If we want to estimate the probability of failure, first assume we're dealing with a member of the defective subpopulation, then calculate the probability of failure given membership in that subpopulation. I.e. P(DS)*P(failure | DS) $\endgroup$ Aug 4, 2021 at 17:38
  • $\begingroup$ Adding (1-p) does not give a similar answer, but if you replace my colleague's denominator with (1-p*CDF(B)) the answer is the same. $\endgroup$ Aug 4, 2021 at 17:39

1 Answer 1


After thinking through this a bit more I was able to confirm that my approach is correct and identify where my colleague's approach went wrong: His use of 'p' (the probability of being in the defective subpopulation) in p*Prob(failure in next hour|member of defective subpopulation) should be replaced with Prob(member of defective subpopulation | survived B).

That probability can be computed with a fairly straightforward application of Bayes' rule and a bit of algebra (or a calculator) then shows that our answers are the same.


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