Finding Joint PMF of bivariate disttibution Problem: Let $W$ equal the weight of laundry soap in a 1-kilogram box that is distributed in Southeast Asia. Suppose $P(W<1)=0.02$ and $P(W>1.072)=0.08$. Call a box of soap light, good, or heavy depending on whether $W<1$, $1 \leq W \leq 1.072$, or $W>1.072$. In $n=50$ independent observations of these boxes, let $X$ equal the number of light boxes and $Y$ the number of good boxes.

Question 1: What is the joint PMF of $X$ and $Y$? I was told to consider that as a trinomial distribution, but why? 
   Can someone explain the problem solving process?
Question 2: Given $X=3,$ how is $Y$ distributed conditionally? I am not sure what kind of distribution can this problem relate to.

 A: First of all we have the propabilities to select a single boxes which are light, good, or heavy: 
$p_1=0.02, p_2=0.9, p_3=0.08$. 
The probabilities of selecting a specific box are mutually independent. Now we pick 50 boxes randomly, where $X=x,Y=y, Z=z$. $x,y,z$ are the numbers of selected light, good and heavy boxes, respectively. 
We are selecting  $n=50$ boxes. And we know that $x+y+z=50$. Therefore $z=50-x-y$. That means that the number of selected heavy boxes is determined by the selected number of the other two kinds of boxes. Therefore the joint pmf is
$$P(X=x,Y=y)=\binom{50}{x,y,50-x-y}\cdot 0.02^x\cdot 0.9^x\cdot 0.08^{50-x-y}$$
A: Consider a sequence of 50 boxes, $x$ of them are light, $y$ good, and $50 - x - y$ are heavy. Obviously, $x+y \leq 50$. Assuming this condition holds, you have ${50 \choose x}$ ways to select positions for light boxes, and ${50 - x \choose y}$ ways to select positions for good boxes, giving the total number of favorable sequences as ${50 \choose x}{50 - x \choose y}$. Finally, the probability of a sequence with these parameters is $0.02^x0.9^y0.08^{50 - x - y}$, and your final answer is $$\mathbb{P}[X = x, Y = y] = {50 \choose x}{50 - x \choose y}0.02^x0.9^y0.08^{50 - x - y}$$
For conditional distribution, you have $$\mathbb{P}[Y = y|X = 3] = \frac{\mathbb{P}[Y = y, X = 3]}{\mathbb{P}[X = 3]}$$ The numerator was derived in the first question, and the denominator you can calculate using the law of total probability: $$\mathbb{P}[X = 3] = \sum_{y=1}^{47}\mathbb{P}[X = 3,Y = y]$$
