Proportion multiple components failing when knowing one component failing Let's consider a system having 3 similar components. The probability of having only one component failing is 10%.
If we don't take into account the sampling error, can I say that, the probability to have 2 components failing is $(10\%)^2 = 1\%$, and the probability to have 3 components failiing is $(10\%)^3 = 0.1\%$?
If I have probability to have 2 components failing = 10% and probability to have 3 components failing is 16%, can I conclude that there is another event which makes also the 3 components failing together?
And, last but not least, I didn't found any formula for that because I don't know the naming of this kind of problem in the statistics world. What should I have searched for?
Thanks a lot
P.S.: For context, I have a real case showing those probability of failure, for which we want to establish the root cause. Having seen that chain, and knowing that those 3 components are at some point managed together, and at another point independently,  those proba tends to tell me that there are two issues: one at common part, and one at independant part.
 A: If the failure probabilities are independent, and each of the $N$ components has an individual failure probability of $p$, then the number of failures $n$ will have a Binomial distribution
$$\Pr[n=k]=\begin{pmatrix}N\\ k\end{pmatrix}p^k(1-p)^{N-k} \text{ , } k=0,1,\ldots,N$$
Here the factor in parentheses is a binomial coefficient, and is read as "$N$ choose $k$", representing the number of distinct combinations of $k$ elements that can be chosen from a set of $N$.
In your case $N=3$ and $p=0.1$. In particular, the probability that $n=2$ is not given by $p^2=0.01$, because you must account for 1) the different combinations of two components that could fail, and 2) the probability that the third component doesn't fail, i.e.
\begin{align}
\Pr[n=2] &= \begin{pmatrix}3\\ 2\end{pmatrix}\times{0.1^2}\times{0.9} \\
&= 3\times{0.01}\times{0.9} \\
&= 2.7 \%
\end{align}
The binomial distribution only applies if the component failure probabilities are i.i.d.. From your description, this may not be the case.
