Suppose that we have a computer server. The probability that server works normally is p. Also we use n copies of the server in order to increase its reliability. What is the chance the whole system works normally ?

There are 5 choices:

1) (p-1)^n 
2) 1 - (1-p)^n  
3) 1 - p^n 
4) p^n - 1 
5) (1-p)^n - 1.

So,i think that 4 & 5 are stupid choices because there can come up a negative probability result. Also 1 just looks wrong because if we increase n the probability goes down, and it should be the other way around. My intuition tells me that 2 is the correct , but at the same time i cant reject 3.


Your intuition is correct, but better to get there by logic.

The probability that a particular server is down is $(1 - p)$, the probability that all servers are down is $(1-p)^n$, so the probability that at least one is up is $1 - (1-p)^n$, i.e. option (2).

(This assumes from your statement about increasing reliability that if at least 1 server is up the system is "working normally". You could otherwise interpret "working normally" to mean that all the servers are up - the answer to this is $p^n$)

  • $\begingroup$ Thanks for your answer! "Working Normally" means that at least one server is up. $\endgroup$ – Andr. Ludw. Jun 2 '14 at 12:13
  • $\begingroup$ Glad it helps. I've previously heard "system is available" as perhaps less ambiguous than "working normally" $\endgroup$ – Tom Collinge Jun 2 '14 at 12:17
  • $\begingroup$ yeah indeed, wrong terminology i guess :D $\endgroup$ – Andr. Ludw. Jun 2 '14 at 12:17

Your intuition on 4) and 5) is good. It can also be used to reject 1) on the basis that that formula also gives negative numbers is some cases (namely when $n$ is odd).

Finally, you can reject 3) on the basis that if each server works perfectly ($p=1$), then so should the system -- or, if you prefer, on the basis that if none of the servers works at all ($p=0$), then having $n$ of them shouldn't suddenly make the system work perfectly (especially since you could just let $n=1$).


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