Ramjac Company wants to set up k independent file servers, each capable of running the company's intranet. Each server has average "uptime" of 98 percent. What must k be to achieve 99.999 percent probability that the intranet will be "up"?
A)1
B)2
C)3
D)4

Question

Quizplus · Accepted Answer

The probability that a server will be "down" at a given time is 1 - 0.98 = 0.02 or 2%. The probability that all k servers will be "up" at a given time is (0.98)^k. To find the number of servers needed to achieve a 99.999% probability of uptime, we need to solve the equation:
(0.98)^k ≥ 0.99999
Taking the logarithm of both sides, we get:
k log 0.98 ≥ log 0.99999
k ≥ log 0.99999 / log 0.98
k ≥ 3.5
Since k must be an integer, we round up to get k = 4. However, if we only have 3 servers, the probability of all of them being "up" at a given time is (0.98)^3 = 0.941192, which is still higher than the required 99.999% probability. Therefore, the best choice is C) 3.

Ramjac Company Wants to Set Up K Independent File Servers