# Re: backup archive format saved to disk

```On Thu, Dec 07, 2006 at 08:41:23AM -0600, Mike McCarty wrote:
> Andrew Sackville-West wrote:
> >On Wed, Dec 06, 2006 at 01:11:06PM -0600, Mike McCarty wrote:
> >
> >>Andrew Sackville-West wrote:
>
> [Mike wrote]
>
> >>Let p be the probability of failure of each disc, independently of the
> >>other. There are four mutually independent events which comprise the
> >>space. Both discs may fail [Pr = p^2]. The first disc may fail, while
> >>the second does not [Pr = p(1-p)]. The second disc may fail, while the
> >>first does not [Pr = (1-p)p]. Both discs may survive [Pr = (1-p)(1-p)].
> >>
> >>So, the probability that at least one disc fails is 1-(1-p)(1-p).
> >>For p = 0.01, that is 0.0199.
> >
> >      --------------------^^^^
> >
> >you aren't using an old pentium are you? ;-)
>
> If p = 0.01, then (1-p) = 0.99, and (1-p)(1-p) = 0.9801, so that
> 1 - (1-p)(1-p) = 0.0199 exactly.

that was a joke...

>
> >>I'll grant you this is not markedly different from 2%, but it is also
> >>not simply 2p.
> >
> >
> >huh. been a long time since math class for me. I was under the
> >impression that the probability was 2p. chance of one failing is added
> >to the chance of the other failing give the chance of both
> >failing. But I'll take your word for it :)
>
> If the probability of the first one failing is 60% and the probability
> of the second one failing is 60%, then I suppose you believe that the
> probability of one or the other or both failing is 120%.
>

no I don't. consider me thinking aloud.

> The probability in this case works out to 1-(0.4)(0.4) = 0.84.
>
> The way to think of it is that it is 1 - Pr(none fails).
> This extends to any number of independent events.
> The problem with trying to add those probabilities is that
> the events are not disjoint, so the probabilities do not add.
>

its starting to come back to me.

> You can however decompose it into events A, B, and C which are
> disjoint, as A = {disc one fails, but disc two does not},
> B = {disc two fails, but disc A one does not} and
> C = {both discs fail}. This is all the ways one or more discs
> fail, and the events are disjoint.
>
> Pr(A) = p(1-p)
> Pr(B) = (1-p)p
> Pr(C) = p^2
>
> So, Pr(at least one disc fails) = p - p^2 + p - p^2 + p^2
> = 2p - p^2
>
> This technique gets more complicated as the number of events
> increases.
>
> NB: 1 - (1-p)^2 = 1 - (1 - 2p + p^2) = 2p - p^2, as computed
> above.

doesn't this involve pascal's triangle somewhere? my brain is too
fuzzy for this stuff. I need to get a good book and retrain my brain.

thanks for the refresher!

A
```

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