I've recently learned about Exponential Distribution and when it's appropriate to be used. There was the following example given.

The duration of IC555 integrated circuits under normal operation conditions are well modeled by an exponential distribution with mean 2000 days. What is the probability such an IC lasts more than 10 years?

$$P(X> 10 \cdot 365.25) = e^{- \frac{3652.5}{2000}} \approx 0.1610 $$


You have a specific IC555 that has already lasted 3 years. What is the probability that it will last another 10 years?

$$P(X> 13 \cdot 365.25 | X> 3 \cdot 365.25)= P(X> 10 \cdot 365.25) = e^{- \frac{3652.5}{2000}} \approx 0.1610 $$

Which doesn't make sense to me because in the first scenario we know the probability it would last more than 10 years and in the second one we know it will last at least 13 years and they are the same. I tried thinking about it differently by filling in 13 years directly without knowing any time about how much it lasted like this:

$$P(X> 13 \cdot 365.25) = e^{- \frac{4748.25}{2000}} \approx 0.09310 $$

which is clearly less than the second case (i.e. when knowing it already lasted for 3 years). So it lasts 13 years in both cases but somehow knowing that it lasted 3 years doesn't provide any information and causes us to treat it as lasting 10 years. Which is confusing. Can someone explain to me why this is the way it is and what piece of information am I missing?

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    $\begingroup$ I think @whuber's answer here will help you understand this: stats.stackexchange.com/questions/369562/… The geometric distribution also shares this "memoryless" property. $\endgroup$ – StatsStudent May 11 at 19:17
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    $\begingroup$ And, if you want to introduce depreciation, you may look into Weibull distribution. $\endgroup$ – gunes May 11 at 20:32

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