First of all, the total availability or uptime of a cluster depends on how large a part of the cluster is needed to be active for the whole cluster to be considered 'up'.

- Is one functioning machine enough? That would mean that any single machine can take the full load if needed.
- Do all of them need to be active at the same time? That is, there is no redundancy.
- Or perhaps two out of three online are sufficient? This would allow for a larger workload than the first case.

As you found out, the first two cases are quite simple to calculate.
Let the probability of a single server being online at any given time *p* = 0.95. Now, for three servers, the probability that they are all online at the same time is *p*^{3} = 0.857375.

For the opposite case, where at least one machine should be active at a given time, it's easier to calculate by inverting the problem and looking at the probabilities of the machines being *offline*. The probability that a single machine is offline is *q* = 1-*p* = 0.05, and hence the probability that they are all down at the same time is *q*^{3} = 0.000125, giving probability 1-*q*^{3} = 1-(1-*p*)^{3} = 0.999875 that at least one is up.

The 2 out of 3 case is slightly harder to calculate. There are four possible situations where at least two out of three servers are up. 1) ABC are up, 2) AB are up, 3) AC are up, 4) BC are up. The probabilities for all these are, respectively, *ppp*, *ppq*, *pqp* and *qpp*. Since the cases are disjoint, the probabilities can be added together, giving a total A = *p*^{3} + 3 *p*^{2}*q* = 0.992750.

(This can be expanded to more machines. The factors are the well known binomial coefficients, so counting the different cases by hand works mostly as an exercise.)

Of course, calculations like this are much easier to deal with by using a ready-made computer program... At least one online calculater can be found here:

http://stattrek.com/online-calculator/binomial.aspx

Entering the input values: probability of success = 0.95, number of trials = 3, number of successes = 2, we get the result "Cumulative Probability: P(X ≥ 2) = 0.99275". Some other related values are also given, and the online tool makes it easy to play with other numbers too.

And yes, all of the above assumes that the servers fail independently, that is a) I ignored any problems affecting the cluster as a whole, b) there isn't anything like component aging that would make it likely for the servers to fail at or nearly at the same time.