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.
Best Answer
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) = $\binom{100}{3}\times 0.95^{97} \times 0.05^3 = 0.13958$