If a piece of computer equipment will fail once a year and be unavailable for 1.5 hours, how do I calculate the probability of the failure occurring each hour of every day

  • $\begingroup$ Could you tell us the source of your question? Homework? Work assignment? Idle curiosity? As stated, there seems to be a lot of ambiguity as to what is meant. $\endgroup$ – Dilip Sarwate Dec 21 '11 at 16:21
  • $\begingroup$ How many hours are in a year? Do you count all of them, or only "working" hours of some kind? $\endgroup$ – Ross Millikan Dec 21 '11 at 16:30
  • $\begingroup$ Silly remark, but since it will only fail once a year, it is certain that it won't fail each hour of every day. $\endgroup$ – Marc van Leeuwen Dec 21 '11 at 18:20

If your question is:

"What is the probability of a failure occurring in a given hour?"

We will count the failure event itself, which happend once per year, giving:
Number of hours in a year = 365d * 24h = 8,760h
Number of failures per hour = 1 failure per year / 8,760h/y = 0.0001142 failures per hour, or:

0.01142% chance of experiencing a failure in a given hour.

If the question was:

"What is the probability of having the computer be unavailable within a given hour?"

(as is implied by the duration given), then:

The probability of the single failure overlapping with a given hour is:

Number of days in a year = 365d/y
Number of days of unavailability = 1.5h / 24h/d = 0.0625d of unavailability per year

0.01712% probability of having some unavailability within a given hour.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.