1. A certain electronic device contains three components, with probabilities of failure being $0.1$, $0.2$, and $0.3$, respectively. If one, two, or three components fail, the probabilities of the device malfunctioning are $0.25$, $0.6$, and $0.9$, respectively. Find the probability that the device malfunctions.
Let the generating function be
$$
\varphi(x)=(0.1 x+0.9)(0.2 x+0.8)(0.3 x+0.7) \text {, }
$$

where the coefficient of the
$x$ term corresponds to the “probability of one component failing”,
$x^{2}$ term corresponds to the “probability of two components failing”,
$x^{3}$ term corresponds to the “probability of three components failing”.
And $\varphi(x)=0.006 x^{3}+0.092 x^{2}+0.398 x+0.504$,

Thus, the required probability is
$$
\begin{array}{l}
p=0.006 \times 0.9+0.092 \times 0.6+0.398 \times 0.25 \\
=0.1601
\end{array}
$$