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I have a question about this specific problem:

A communications network has a built in safeguard system against failures.In this system if line I fails, it is bypassed and line II is used. If line II also fails,it is bypassed and line III is used. The probability of failure of any one of these three lines is .01, and the failures of these lines are independent events. What is the probability that this system of three lines does not completely fail?

My intuition, which is wrong for this problem, tells me to do (0.99)^3, since that is the probability of the network's safeguard not failing.

This is however wrong, and I am supposed to do 1-(0.01)^3.

I guess I am wondering what the difference is and if anybody can give me any intuition into understanding how these two solutions are different.

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    $0.99^3$ corresponds to the probability that all three lines work correctly, but this is not the only situation in which the system as a whole works correctly. You are allowed to have one or more of the individual lines failing, so long as at least one is working, while the system as a whole still operates correctly and your calculations do not account for this. – 2017-02-13

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The probability that all three lines fail is $0.01^3$. The complement of this is that at least one of the three lines works, which is $1$ minus this probability.

What you calculated was the probability that all three lines worked. This isn't quite what you needed. The system can work with three lines working, two lines, or just one:

$$1 - 0.01^3 = 0.99^3 + 3 \cdot 0.99^2 \cdot 0.01 + 3 \cdot 0.99 \cdot 0.01^2.$$

So in other words you got the first term on the right side, but not the second and third term (which are the probabilities that exactly two and exactly one line worked, respectively).