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Failure rate of a system

If a system has a 10% independent chance of failing in any given hour, what are the chances of it failing in a given 2 hour period or n-hours period? Note: 10% failure probability in 1 hour has nothing to do with 10% of the time . It's just that a system has a 10% independent chance of failing in any given hour

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sammy123
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    This question is probably a better fit for http://math.stackexchange.com/ – cegfault Dec 01 '12 at 06:12
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    This might help - http://stackoverflow.com/a/6819375/1791606 – qooplmao Dec 01 '12 at 06:17
  • @Qoop- Ya, I checked that but in that question, failing probability is 10% of the time but in this question, failing probability of 10% is not 10% of the time, it's just related to the system. I hope you understand what I mean. Thanks for your help. I appreciate that. – sammy123 Dec 01 '12 at 06:22

1 Answers1

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  • Let Pfail be the probability that the system fails in any given hour.
  • Then Pnofail, the probability that the system does not fail in any given hour, is 1 - Pfail.
  • The chance of it not failing in 2 hours is (Pnofail)2, since it must independently not-fail in each of those hours, and the joint probability of two independent events is the product of the probability of each event (that is, P(A ∩ B) = P(A)*P(B)).
  • More generally, then, the chance of it not failing in n hours is (Pnofail)n .
  • The chance of it failing in n hours is 1 - (chance of not failing in n hours).

You should be able to work it out from there.

Matt Ball
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