K independent file server. Each server has average "uptime" of 98%. What must k be to achieve 99.999% probability that it will be "up"?

1 Answer
MattyMatty
Feb 17, 2018

#K=3#

Explanation:

#P["1 server is up"] = 0.98#
#=> P["at least 1 server out of K servers is up"] =#
#1 - P["0 servers out of K servers are up"] = 0.99999#
#=> P["0 servers out of K servers are up"] = 0.00001#
#=> (1-0.98)^K = 0.00001#
#=> 0.02^K = 0.00001#
#=> K log(0.02) = log(0.00001)#
#=> K = log(0.00001)/log(0.02) = 2.94#
#=> " We must take at least 3 servers, so K=3."#

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