Serial Number in a Geometric Distribution I won't bother posting the whole exercise.Basically, we've got 2000 pc's and 12 of them are malfunctioning. 
At some point, the exercise writes:
We choose the pc's until we find a malfunctioning one, and let Y be the serial number of the first malfunctioning one. What is the distribution of Y,expected value and variance? 
I reckon that the distribution is geometric.
My question is what does "let Y be the serial number of the first malfunctioning one" mean in mathematical terms? Does its meaning change if we use replacement?
 A: Let the serial numbers be 1, 2, ..., 2000; sample then in serial-number order.
There are two formulations of the geometric distribution. Most commonly in beginning courses the random variable $Y$ is the number of the trial (serial number) on which the first defective PC is found. 
Using that formulation, and denoting by $p$ the probability of finding a defective PC on any one trial, and $q = 1 - p,$ we have the
geometric distribution
$$P(Y = k) = q^{k-1}p,\;\;\mathrm{for}\;\;k = 1, 2, \dots.$$
As mentioned in the comment this is only an approximate model because the draws of PCs are not exactly independent, and because there is a largest value $k = 1989.$
Assuming the exact geometric distribution, $E(Y) = 1/p$ and
$V(Y) = q/p^2.$ It is not clear from your question, but I assume you have covered this distribution in your course and discussed how to find the mean and variance, so that those derivations are not part of what you need to show.
In your case $p = 12/2000 = 0.006,$ so that $q = 0.994,$ and you can compute the mean and variance from there. On average, you will have to look at about 167 PCs before you find the first defective one.
In order to illustrate the consequences of using an approximate
model instead of an exact one, consider $P(Y = 5).$
For a geometric random variable $$P(Y = 5) = 0.006(0.994)^4 = 0.005857291.$$
An exact computation for your situation would be
$$P(Y = 5) = \frac{1988}{2000}\frac{1987}{1999}\frac{1986}{1998}
\frac{1985}{1997}\frac{12}{1996} = 0.005868922.$$
This difference in the third significant digit is small for
any one probability. However, the cumulative effect of these
differences makes a relatively small, yet noticeable, difference in $E(Y).$
Simulation of exact model. The simulation below randomly arranges the 12 bad PCs among
the 1988 good ones, and then looks for the location of the
first bad one in sequence. Averaging a million such experiments
we approximate $E(Y) \approx 154.$ (Also, $SD(Y) \approx 142.$) On theoretical grounds, the margin of simulation
error for the mean is very likely less than 2. (And a second simulation also rounded to 154.) So it is clear that the
the true situation with a finite sample gives a little smaller mean than
does the geometric approximation.
 PC = c(rep(1,12), rep(0,1988))
 m = 10^6;  y = numeric(m)
 for(i in 1:m) {
   perm= sample(PC, 2000);  y[i]=match(1, perm) }
 mean(y)
 ## 154.1165
 sd(y)
 ## 142.0756

Intuitive argument. An intuitive justification for a mean of 154 is as follows:
Suppose you had one defective PC in 2000; 'on average' you would
expect the defective one to be about half-way through the sequence
for a waiting time of about 1000. Suppose you had two defectives,
then your average waiting time would be $2000/3 \approx 667.$
With 12 defective PC's you would guess your average wait to be
$2000/13 \approx 154.$
Addendum: The alternative formulation of the geometric
distribution counts the number of good PCs encountered
waiting for the first bad one. See the Wikipedia article on
'geometric distribution'.  B.T.
