Thanks very much for explaining it so patiently to me, A Bear of Little
Brain [reference to Pooh Bear]
Darryl Gove wrote on 29/08/12 2:53 PM:
> On 28 August 2012 21:05, steve jenkin <stev...@gmail.com> wrote:
>> Darryl Gove wrote on 29/08/12 11:26 AM:
>> The probability of experiencing no failures over 50 years is (99/100)
>> ^ 50. So the probability of experiencing at least one is 1-(99/100)^50
>> = 0.39.
>> Thinking a little more on this.
>> If I have a server and run it 4-5 years and there's an averaged change of
>> failure of 6%/year of drives,
>> then over the 5 year life of a single drive:
>> - Prob. No Failure/yr = 1 - prob(failure/yr) = 1 - .06 = .94
>> - prob No failure in 5 yr = 0.94 ^ 5 = 0.734
>> Now if I have 3 drives, is the Probability of No drives failing in a single
>> year the product of 1 - sum(prob failure)??
> Not the sum - you can only add the probabilities of mutually exclusive events.
>> ie. 0.94 * 0.94 * 0.94 = 0.831
> Yes, this is the probability of no drive failing during one year.
>> or 1 - (0.06 + 0.06 +.06) = 1 - 0.18) = .82
> No, this is not right (imagine that the probability of a drive failing is 0.5 :)
>> I'm guessing having working out the yearly rate of "No Fails this year" for
>> a group of drives, then the probability for getting through the entire 5
>> years with no failures is the product:
>> ie. p1 * p2 * p3 * p4 * p5
> 0.831^5 = 0.396
> So the total probability is (prob working for one year)^( #drives * #years )
>> Steve Jenkin, Info Tech, Systems and Design Specialist.
>> 0412 786 915 (+61 412 786 915)
>> PO Box 48, Kippax ACT 2615, AUSTRALIA
>> stev...@gmail.comhttp://members.tip.net.au/~sjenkin >>
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Steve Jenkin, Info Tech, Systems and Design Specialist.
0412 786 915 (+61 412 786 915)
PO Box 48, Kippax ACT 2615, AUSTRALIA