We assume **independence**. This may not be entirely realistic, but we need _some_ assumption or information about possible interactions between the components.
The system works if and only if $1$ and $4$ work and at least one of $2$ or $3$ works.
The probability that $2$ or $3$ **both** fail is $(0.1)(0.1)$. So the probability at least one works is $1-(0.1)^2=0.99$.
Thus our system reliability is $(0.9)(0.99)(0.9)$.