Trying to understand a stat question and its answer

[FocusedWolf]

Homework Statement

A robotic insertion tool contains 21 primary components. The probability that any component fails during the warranty period is 0.01. Assume that the components fail independently and that the tool fails if any component fails. What is the probability that the tool fails during the warranty period? Round the answer to 3 significant digits.

correct answer .19027

p(fails) = 1 - p(works) = 1 - (1-.01)^21 = .19027

2. My question

This is the right answer, but what i want to know is... is their any way to avoid moving from p(failing) to 1- p(works)

I mean, is their someway to work with the .01 directly like .01 ^ (something) * something etc to get p(fails)? or is this the only way to do this problem... like is it a requirement to go from "any part failing" to "not a single part failing" in order to like "combine" the probabilities so we're not conscerned with individual parts but the whole thing?

 

[Dick]

Sure, you add up the probability of exactly n components failing for n=1 to 21. Use the binomial theorem. It's a lot of work. Using 1-p(failing)=p(works) is considered the clever way to do it. p(failing) is more about the sum of the whole than counting individual parts.