Comment (and solution of a simple special case.) This has been here for a while, apparently without helpful comments.
This appears to be a generalization of a 'multivariate hypergeometric'
distribution.
You might start with a simplified set of weights. Let an urn contain
balls labeled from 1 through 8. And suppose their respective weights are $w = (2, 2, 1, 1, 1, 1, 1, 1)/10.$ If you withdraw $k = 2$ balls from the urn without replacement, what is the probability
you get the ball labeled '$1$'?
Get 1 on the first draw: $P(\text{1 on 1st}) = (2/10)(8/8) = .2.$
Get 1 on the second draw: Either 21, or something other than 1 or 2
on the first, then 1 on the second.
$P(\text{2 then 1}) = (2/10)(2/8) = .05.$
$P(\text{3 then 1}) = (1/10)(2/9) = 2/90 \approx 0.0222.$
$P(\text{1 on 2nd}) = 0.05 + 6(2/90) \approx 0.05 + 0.1333 = 0.1833.$
Finally, $P(\text{1 in two draws}) \approx 0.2 + 0.1833 = 0.3833.$
Even this simple problem turned out to surprise me by its intricacy and lack of symmetry. But perhaps, you can find patterns to simplify more complicated outcomes.
R statistical software does weighted random sampling in a way that
would allow you to check some of your analytic solutions. As a prototype, here is a
simulation of the simple example just above. Results are mainly
accurate to three places.
m = 10^6; d1 = d2 = numeric(m)
n = 2; pop = 1:8; w = c(2,2,1,1,1,1,1,1)/10
for (i in 1:m) {
draw = sample(pop, n, prob=w)
d1[i] = draw[1]; d2[i] = draw[2] }
mean(d1 ==1 | d2 ==1) # '|' signifies union
## 0.383483
round(table(d1)/m,3)
## d1
## 1 2 3 4 5 6 7 8
## 0.200 0.199 0.100 0.100 0.100 0.100 0.100 0.100
round(table(d2)/m,3)
## d2
## 1 2 3 4 5 6 7 8
## 0.184 0.184 0.105 0.105 0.106 0.106 0.105 0.105
round(table(d1,d2)/m,3)
## d2
## d1 1 2 3 4 5 6 7 8
## 1 0.000 0.050 0.025 0.025 0.025 0.025 0.025 0.025
## 2 0.050 0.000 0.025 0.025 0.025 0.025 0.025 0.025
## 3 0.022 0.022 0.000 0.011 0.011 0.011 0.011 0.011
## 4 0.022 0.022 0.011 0.000 0.011 0.011 0.011 0.011
## 5 0.022 0.022 0.011 0.011 0.000 0.011 0.011 0.011
## 6 0.022 0.022 0.011 0.011 0.011 0.000 0.011 0.011
## 7 0.022 0.022 0.011 0.011 0.011 0.011 0.000 0.011
## 8 0.022 0.022 0.011 0.011 0.011 0.011 0.011 0.000
Best Answer
Ok. I'm posting two answers because one is very general, and the other (this one) only works nicely when the weights are positive - but I wrote this one first because I thought of it first.
Assume the weights are non-negative.
Notice that if I multiply all the weights by a fixed positive number $k$, then
$$ \frac{\sum_{i=1}^nkw_ix_i}{\sum_{i=1}^nkw_i} = \frac{k\sum_{i=1}^nw_ix_i}{k\sum_{i=1}^nw_i} = \frac{\sum_{i=1}^nw_ix_i}{\sum_{i=1}^nw_i},$$
which is to say that it doesn't change anything. It also doesn't affect the order. So let us assume that we have multiplied by the correct $k$ so that $\sum_i w_i = n$.
Notice also that if I add some constant $M$ to each $x_i$, then we have both
$$\frac{\sum_{i=1}^nw_i(x_i+M)}{\sum_{i=1}^nw_i} = \frac{\sum_{i=1}^nw_ix_i}{\sum_{i=1}^nw_i} + \frac{\sum_{i=1}^nw_iM}{\sum_{i=1}^nw_i} = \frac{\sum_{i=1}^nw_ix_i}{\sum_{i=1}^nw_i} + M,$$ and $$\frac{\sum_{i=1}^n(x_i+M)}{n} = \frac{\sum_{i=1}^nx_i}{n} + M,$$ so that the difference between the weighted mean and the arithmetic mean is the same. So we can assume that all the $x_i$ are positive.
Then
$$ \frac{\sum_{i=1}^nw_ix_i}{\sum_{i=1}^nw_i} = \frac{\sum_{i=1}^nw_ix_i}{n} < \frac{\sum_{i=1}^nw_1x_i}{\sum_{i=1}^nw_i} = \frac{w_1\sum_{i=1}^nx_i}{n},$$
and since the weights are positive, $w_1 < 1$. (If $w_1 = 1$ and the other weights are all $0$, then $x_1 + 0 + 0 + \ldots < x_1 + x_2 + \ldots$, so only the strictly positive case is interesting).
So we conclude that
$$ \frac{\sum_{i=1}^nw_ix_i}{\sum_{i=1}^nw_i} < \frac{\sum_{i=1}^nx_i}{n}.$$