A system uses three servers with an availability of 0.95 each and has the required capacity as long as at least two of them are running. The servers fail independently, and failures of devices other than the servers are ignored. Which is the availability of this system?

Answer: B. 0.99275

Add the probabilities that at least two of the three servers are running. The probability that exactly two are running is 3 × 0.95² × 0.05 = 0.135375, and the probability that all three are running is 0.95³ = 0.857375, so the availability is 0.135375 + 0.857375 = 0.99275. 0.9025 is the value with the two required servers in series and leaves out the effect of the spare. 0.999875 is the value for a parallel system in which any one of the three is enough.

• A. This is the value for two servers in series (0.95²) and ignores the effect of the third, spare server.
• B. Correct. 3 × 0.95² × 0.05 + 0.95³ = 0.99275.
• C. This is the value for a parallel system in which one server is enough (1 − 0.05³).