Finding joint P.M.F A bin of 4 lightbulbs has 2 defective ones. Assume that you repeatedly take a (without replacement) lightbulb out of the bin and test them. Let $N_1$ be the rank (in [1..4]) of the first defective lightbulb found and let $N_2$ be the rank (in [1..4], with $N_2 > N_1$) of the second defective one found. Compute the joint p.m.f. of $X = N_1$ and $Y = N_2−N_1$, and the marginal p.m.f.’s for $X$ and $Y$ .

Just to clarify: What is  definition of rank in this question? Is it the # of tests made until the first defective lightbulb is found?
My attempt:
Well, there are only $4\choose2$ ways of choosing 2 defective lightbulbs from a group of 4.
Joint PMF: P(X,Y) = $\frac{1}{6} * \frac{1}{2} = \frac{1}{12}$ ?
But i'm confused because ..
bad, bad, good, good = $\frac{1}{4} *\frac{1}{3} *\frac{1}{2} * 1  = \frac{1}{24}$ 
bad, good, bad, good = $\frac{1}{4} *\frac{2}{3} *\frac{1}{2} * 1 = \frac{1}{12}$
For this kind of question, is there a specific kind of probability distribution that I could use? (e.g., hypergeometric distribution,negative binomial distribution.)
Would appreciate any tips/guide/help on how to tackle this question. Thank you. 
 A: Instead of looking for formulas, let us compute. After all, there are only $4$ lightbulbs in the game.
First we calculate $\Pr(X=1,Y=1)$. For this to happen, we must get a bad, then a bad. The probability the first is bad is $\frac{2}{4}$. Given that the first is bad, the probability the second is bad is $\frac{1}{3}$. Thus $\Pr(X=1,Y=1)=\frac{1}{6}$. 
Next we calculate $\Pr(X=1,Y=2)$. We must get bad, good, bad. This has probability $\frac{2}{4}\cdot \frac{2}{3}\cdot \frac{1}{2}$.
We can use the joint distribution to calculate the marginals. But it is in this case just as easy to start anew. For $X$, we want $\Pr(X=1)$, $\Pr(X=2)$, $\Pr(X=3)$, and similar things for $Y$.
For example, the event $Y=2$ can happen in two ways, bad, good, bad and good, bad, good, bad. Find the probability of each and add.
A: You are looking for: $\mathsf P(N_1=n, N_2-N_1=m)$ which is the probability that the draw $n$ is the first defective bulb, and draw $n+m$ is second defective bulb.
There are how many equally probable ways to arrange the bulbs and each such arrangement corresponds to a distinct pair of $(N_1, N_2)$ values.
$$\begin{align} \mathsf P(X=n, Y=m) ~=~ & 
\mathsf P(N_1=n, N_2-N_1=m) \\[1ex]~=~& \mathsf P(N_1=n, N_2=m+n)\\[1ex] ~=~& \boxed{?}\quad \Big[n\in\{1,2,3\}, m\in\{1,..,4-n\} \Big]\end{align}$$
