Hi!
> >If 1 disk has a 1/1000 chance of failure, then
> >2 disks have a (1/1000)^2 chance of double failure, and
> >3 disks have a (1/1000)^2 * 3 chance of double failure
> >4 disks have a (1/1000)^2 * 7 chance of double failure
>
> After the first drive fails you have no redundancy, the
> chance of an additional failure is linear to the number
> of remaining drives.
>
> Assume:
> p - probability of a drive failing in unit time
> n - number of drives
> F - probability of double failure
>
> The chance of a single drive failure is n*p. After that
<pedantic>
Actually it is not. Imagine 100 drives with 10% failure rate each. You
can't have probability of 1000%...
</>
Pavel
--
Thanks, Sharp!
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