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I have a problem where I'm trying to derive an expression for the availability of a server cluster.

Suppose I have 100 servers, where each one of them individually has a probability of failure 0.05. Now, I'm trying to find an expression for the probability that at most 3 of the servers fail.

Currently I have reasoned that:

$Pr$(at most 3 of the servers fail) = $Pr$(0, 1, 2, or 3 of the servers fail) = $Pr$(0 fail) + $Pr$(1 fails) + $Pr$(2 fail) + $Pr$(3 fail)

$Pr$(0 servers fail) = $0.95^{100} = 0.00592$

$Pr$(1 server fails) = $0.95^{99} \times 0.05^1 = 0.05623$

$Pr$(2 server fails) = $0.95^{98} \times 0.05^2 = 0.00906$

$Pr$(3 server fails) = $0.95^{97} \times 0.05^3 = 0.00702$

Thus, $Pr$(at most 3 of the servers fail) = $0.00592 + 0.05623 + 0.00906 + 0.00702 = 0.07823$.

Is that correct? That probability of 7.8% seems like a really small number for having at most 3 servers fail.

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  • $\begingroup$ You're forgetting some binomial coefficients. $\endgroup$
    – paw88789
    Oct 14, 2014 at 23:07
  • $\begingroup$ Yes, you are very close., however you need to ask the question for each of $\textit{which}$ server(s) failed. $\endgroup$
    – JMoravitz
    Oct 14, 2014 at 23:09

1 Answer 1

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When calculating the probabilities you should also take into account which of the servers will fail. Thus, the correct probabilities are

$\Pr$(0 servers fail) = $0.95^{100} = 0.00592$, (correct since $\binom{100}{0}=1$)

$\Pr$(1 server fails) = $\binom{100}{1}\times 0.95^{99} \times 0.05^1 = 0.03116$

$\Pr$(2 server fails) = $\binom{100}{2}\times 0.95^{98} \times 0.05^2 = 0.08118$

$\Pr$(3 server fails) = $\binom{100}{3}\times 0.95^{97} \times 0.05^3 = 0.13958$

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  • $\begingroup$ Thanks. So when I was earlier computing that $Pr$(2 server fails) = $0.95^{98}×0.05^2$, does that imply that I was computing the probability that the first 98 servers were good and the last 2 servers failed? $\endgroup$ Oct 14, 2014 at 23:21
  • $\begingroup$ Yes, exactly. Actually you were computing the probability that two specific servers failed (f.e. as you say the last 2, or f.e. the servers in places 54 and 87 etc.) But you need to calculate the probability for any two. $\endgroup$
    – Jimmy R.
    Oct 14, 2014 at 23:24
  • $\begingroup$ Thank you. Now that I look further, I find it kind of extraordinary that $Pr$(0 servers fail) = $0.00592$. That seems really low when each of the servers seems pretty stable with each having probability $0.95$ of failing. But the math proves it, I guess. $\endgroup$ Oct 14, 2014 at 23:30
  • $\begingroup$ 0.95 of not failing you mean. Yes, the point you mention is exactly the problem in inferential statistics, in case you want to make many simultaneous inferences. If you are "only" 95 percent sure that each of 100 statements is correct then there is 1-0.00592=0.99408 probability that at least one statement is wrong! $\endgroup$
    – Jimmy R.
    Oct 14, 2014 at 23:34

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