# Probability of a machine failing when components fail independently

Our machine is made out of 3634 units of component A which has failure rate of 10%, 1656 units of component B with failure rate of 35% and 3368 units of component C with failure rate of 55%. Assuming the failure of each typo of component are independent of each other, and the machine fails if only one component fails. What is the failure rate of the given machine?

Here is my approach,

Calculate the probability of failure of one component in each type using binomial distribution.

Here is the probability of 1 component failure out of 3634 of type A, $$P(X=1) = {1 \choose 3634} 0.1^{3634}(1- 0.1)^{1 - 3634} = 2.105822^{-164}$$

Similarity I calculated 1 component failure of B and C.

Since their failure are independent therefore I multiplied failure rate of the components to get failure rate of the machine,

dbinom(1, 3634, .1) * dbinom(1, 1656, .35) * dbinom(1, 3368, 0.55)


The output of the R command is zero, which means the probability of the machine failing is close to zero. (The first two term returns very small number and the final term returns zero)

I have the 2 following questions.

1. Am I solving this problem correctly?
2. Is it correct to interpret that the probability of the machine failing is very close to zero.

Your machine is composed of 8658 components, and if any of them breaks down the machine breaks down. And, we seem to be implicitly talking about a lifetime behavior or behavior in a fixed time interval of a component, i.e. the failure rate of a component being $$10\%$$ means that the component will either fail with probability $$0.1$$ or not fail at all. In this setup, intuitively, it seems quite possible for the machine to break down as soon as possible it starts working (given the probabilities); so the probability of machine failing will be close to $$1$$, not $$0$$.
We'll find the probability that the machine does not fail, by calculating the probability that each of the components doesn't fail at all, and finally subtract this number from $$1$$. We can't focus on failure of only one component, e.g. $$X=1$$ in your case, because two or more components could also fail. The easiest way is to calculate no fail at all, and complement it. For each type, we have (let $$X_m$$ be the number of failing components of type $$m$$): $$P(X_A=0)=0.9^{3634},P(X_B)=0.65^{1656},P(X_C)=0.45^{3368}$$ and we'll multiply these numbers to get the probability of machine not failing, which probably gives a number close to $$0$$ and complement of it will be close to $$1$$. Its probable that you'll have overflow issues with dbinom method. Instead we could calculate the probability using logarithms:
log10(0.1) * 3634 + log10(0.65) * 1656 + log10(0.45) * 3368

which gives -5111.796, i.e. $$\approx P(\text{Failure}) = 1-10^{-5111.796}$$