# Probability question involving paths through computer network

I am working the following probability problem which I came across online:

A computer network has 5 paths to its destination. Each path has a 20% chance of working and each path is independent. Am I guaranteed that at least one path will work?

I first answered it like:

P(1st path working OR 2nd path working OR 3rd path working OR 4th path working OR 5th path working) = 0.2 + 0.2 + 0.2 + 0.2 + 0.2 = 1

but apparently that is not correct. Can someone explain why?

Thanks!

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So if the probability that each path works were $30\%$ instead of $20\%$, would you have said that the probability of at least one path working is $0.3+0.3+0.3+0.3+0.3=1.5$? –  Dilip Sarwate Apr 21 '12 at 22:40

## 2 Answers

What you're doing has no right to work, because if there had been 6 paths each with 20% reliability, the same method would yield a probability of $1.2$ -- but probabilities cannot be more than $1$.

Your fundamental error is that you're adding independent probabilities. Doing so is never correct, because it amounts to ignoring the possibility that both independent events occur at the same time (which they can do because they're independent).

The smart way to do questions like this is to compute the probability that all paths will fail at the same time (which involves multiplication which is allowed for independent probabilities), and then invert that result.

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Thanks. So under what conditions do you add probabilities? –  jasonbogd Apr 21 '12 at 16:46
Probabilities of events that are known never to happen at the same time can be added freely. (And addition is used internally in the correct rule for independent events: $P(A\cup B)=P(A)+P(B)-P(A)P(B)$ when $A$ and $B$ are independent). –  Henning Makholm Apr 21 '12 at 16:52

It is clear that the probability isn't $1$, just imagine the scenario where all paths fail. Surely this scenario can happen since each path has a positive probability of failing and they are independent.

The probability that at least one path will work is $1 - P(\text{all paths fail})$. Since the paths are independent, the probability that all paths fail simultaneously is exactly the product of their probabilities to fail.

Thus, $P(\text{at least one path works}) = 1 - P(\text{all paths fail}) = 1 - (0.80)^5 = 0.67232$.

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@Peter: Yeah, read it too quickly. Fixed now. –  Brandon Carter Apr 21 '12 at 21:28