Darryl,

Thanks very much for explaining it so patiently to me, A Bear of Little

Brain [reference to Pooh Bear]

cheers

steve

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.

>>

>> Darryl,

>>

>> 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

> Yes.

>

> 0.831^5 = 0.396

>

> So the total probability is (prob working for one year)^( #drives * #years )

>

> Darryl.

>

>> cheers

>> steve

>>

>>

>> --

>> Steve Jenkin, Info Tech, Systems and Design Specialist.

>> 0412 786 915 (+61 412 786 915)

>> PO Box 48, Kippax ACT 2615, AUSTRALIA

>>

>>

stev...@gmail.com http://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

stev...@gmail.com http://members.tip.net.au/~sjenkin