probability distribution $X, Y$ and $X+Y$ A box contains $5$ ticket, $\{ 0 , 0 , 0 , 4 , 4\}$. 
Drawing two tickets at random w/o replacement. 
$X$ be the sum of the first two draws and Y be the outcome of the first draw. 
Question: Find distribution of $X + Y$? 


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*What I did --
I found the distribution of $X$. There are only three possible sums: $0 , 4, 8$. And there are $10$ ways to get these sums:  $(0,0), (0,0), (0,4),(0,4),(0,0),(0,4),(0,4),(0,4),(0,4) (4,4)$. Therefore $P(0) = 3/10 , P(4) = 6/10, P(8) = 1/10$.


My other question:  Is there another way (different from what I did) to get the distribution of $X$?  And I have no idea how to get distribution of $Y$.  

For the distribution of $X$, I do not think there is a substantially better way than yours. 
For the distribution of $X+Y$, we do something similar. If it helps to keep track of things, draw a tree diagram. Let $W=X+Y$. 
Case $1$: Maybe the first pick was $0$. This has probability $\frac{3}{5}$. Given that this happened, the probability of picking a $0$ next is $\frac{2}{4}$. Then $W=0$. The probability of piking a $4$ is  $\frac{2}{4}$. Then $W=4$.  
Case $2$: Maybe the first pick was a $4$. This has probability $\frac{2}{5}$. In that case, the probability of next getting $0$ is $\frac{3}{4}$, and we get $W=8$. The probability of getting a $4$ is $\frac{1}{4}$. In that case $W=12$.
So $W$ takes on values $0$, $4$, $8$, $12$.
For the probability that $W=0$, locate all the ways $W$ can be $0$. This only happens in one way, $0$ then $0$. We can see that the probability is $\frac{3}{5}\frac{2}{4}$.
Similarly, the probability that $W=4$ is $\frac{3}{5}\frac{2}{4}$.
Similarly, $\Pr(W=8)=\frac{2}{5}\frac{3}{4}$ and the probability that $W=12$ is $\frac{2}{5}\frac{1}{4}$. 
As a partial correctness check, add up our four numbers. We should get $1$, and we do. 
