Summing $N \sim Pois(\lambda)$ independent random variables Let the number of complaints $X$ have a $Poisson$ distribution with mean, $λ=2$ and let $U$ be a random variable with $P(U=u) = 0.1u$, for $u=1,2,3,4$. (Calculate the probabilities $P(U = x),$ for $x=0,1,2,3,4.$)
For $N=0$, we have 
$P(N=0)=\frac{e^{-2}2^x}{x!}=\frac{e^{-2}}{1}≈0.1353352832.$
I just don't understand why they give me the random variable $U.$ Am I missing something?
Edit:
The previous problem states:
For a r.v. $N$ with $Poisson$ distribution and $U_i$ with $P(U=u)=0.1u$, for $u=1,2,3,4.$ All random variables are independent. 
Find the expected value of $S,$ defined as: $$S=\sum_{i=1}^N U_i.$$

Note: Confusion seems to have emerged about the statement of the problem. For clarity, here is the problem addressed in this Answer:

Let $N \sim Pois(2)$ be the number of complaints received by a firm in a week. The cost of servicing each complaint is a random variable $U$ with
  distribution given by $P(U = 1) = .1,\;P(U = 2) = .2,\;P(U = 3) = .3,\;P(U = 4) = .4.$ Thus the weekly cost of servicing complaints is the sum $S = \sum_{i=1}^N U_i,$ where $U_i$ are distributed as $U$ and random variables $N, U_1, U_2, \dots$ are independent.

It seems that each complaint has consequences and the "cost" (in dollars
time, or whatever) of a complaint is expressed by $U.$
For a similar Question, please see this page and also
these lecture notes. This kind of random variable $S$ is known
as a "random sum of random variables."
A result, often derived by a conditional argument, is that
$E(S) = E(N)E(U).$ I will leave it to you to verify that
$E(N) = 2$ and $E(U) = 3.$
There is also a result for $V(S),$ not
required here, that is not so 'obvious'.
It is easy to get a good idea of some characteristics of
the distribution of $S$ by simulation in R. The random
function rpois determines the number $N$ of complaints, 
and the $U_i$ can be simulated by sampling from the
integers 1 through 4 in the appropriate proportions.
Results are for 100,000 repetitions of your experiment.
 B = 10^5;  lam = 2;  s = numeric(m)
 for (i in 1:m) {
   n = rpois(1, lam)
   s[i] = sum(sample(1:4, n, repl=T, prob=(1:4)/10)) }
 mean(s);  sd(s)
 ## 5.985     # approx E(N)E(U) = 2(3) = 6
 ## 4.471518  # approx V(S) = E(N)V(U) + V(N)[E(U)]^2

A histogram of the simulated values of $S$ approximates
the distribution of $S.$ Notice that $S = 0$ whenever
$N = 0,$ and you have already noted that $P(N = 0) = .135.$

