Under the given assumptions, the number $X$ of components that fail has binomial distribution, with $n=400$ and $p=0.005$. The probability that $X\le 2$ can be computed using $\binom{400}{0}p^0(1-p)^{400}+\binom{400}{1}p^1(1-p)^{399}+\binom{400}{2}p^2(1-p)^{398}.$
However, here $n$ is large and $np=2$ is moderate. So the binomial is well approximated by the probability that a Poisson with $\lambda=2$ is $\le 2$. This probability is $e^{-2}+e^{-2}\frac{2^1}{1!}+e^{-2}\frac{2^2}{2!}.$
Remark: The answer simplifies to $5e^{-2}$, which is about $0.6766764$.
I computed the "exact" answer using the binomial distribution, without the Poisson approximation. The calculator gives $0.676677$. So in this case, the Poisson approximation is very very good.
In real work, one can get away with much less precise approximations. The assumptions under which we make our calculations, such as independence, are seldom absolutely true. And the number $0.005$ is undoubtedly a rough approximation. So the numbers we obtain from a probability calculation, even if that calculation is "exact," can usually only be treated as a rough approximation to the truth.