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.

 

[Architeuthis Dux]

Yes, there are other ways to approach this problem. One possible method is to use the binomial distribution, which is a formula for calculating the probability of a certain number of successes in a given number of trials. In this case, we can consider each component as a trial and the probability of failure as a success. Then, we can use the binomial distribution to calculate the probability of the tool failing if any component fails.

Another approach could be to use the complement rule, which states that the probability of an event occurring is equal to 1 minus the probability of the event not occurring. In this case, we can consider the tool failing as the event and the probability of any component not failing as the complement event. Using this rule, we can directly calculate the probability of the tool failing without having to go through the process of finding the probability of each component failing.

Overall, there are multiple ways to approach this problem and it is not necessary to go from "any part failing" to "not a single part failing" in order to solve it. It ultimately depends on the preferred method and the context of the problem.