Suppose that a bowl contains 100 chips: 30 are labelled 1, 20 are labelled 2, and 50 are labelled 3. The chips are thoroughly mixed, a chip is drawn, and let $X$ be the number on the chip.
1) Compute $P(X = x)$ for every real number $x$.
2) Suppose the first chip is replaced, a second chip is drawn, and let $Y$ be the number on the second chip. Compute $P(Y = y)$ for every real number $y$.
3) Let $W = X + Y$ and compute $P(W = w)$ for every real number $w$.

I have the solution for this:
I understand part 1 and 2, but not 3.
We define $W = X+Y$, and so this means that $W(w)=X(w)+Y(w), \forall w\in\mathbb{R^1}$
First my question is, why do we not consider $W(1)$? We have defined $X(1)$ and $Y(1)$, so why not $W(1)$?
Secondly, and probably the most important question is, how do they get $W(2)=0.12$? From the equation $W(2)=X(2)+Y(2)=0.2+0.2 = 0.4$. But they don't follow the equation. Why?
There's two questions I'm asking here.

 A: There are five possibilities for the sum of the numbers on the two chips: $$ X+Y \in \{2,3,4,5,6\}$$



*

*If $X+Y=2$, then $(X,Y)$ must be $(1,1)$. 

*If $X+Y=3$, then $(X,Y)$ can be $(1,2)$ or $(2,1)$. 

*If $X+Y=4$, then $(X,Y)$ can be $(1,3)$, $(2,2)$, or $(3,1)$. 

*If $X+Y=5$, then $(X,Y)$ can be $(2,3)$ or $(3,2)$. 

*If $X+Y=6$, then $(X,Y)$ must be $(3,3)$. 



The probability that $X+Y=2$ is the same as the probability that both $X=1$ and $Y=1$. Since the chip is replaced, $X=1$ and $Y=1$ are independent events, so we simply multiply their probabilities to find the probability that both occur:
$$P(X+Y=2) = P(X=1)P(Y=1) = 0.3^2 = 0.09$$
The probability that $X+Y=3$ is the same as the probability that both $X=1$ and $Y=2$, plus the probability that both $X=2$ and $Y=1$:
$$P(X+Y=3) = P(X=1)P(Y=2) + P(X=2)P(Y=1)= 0.3\cdot 0.2 \cdot 2= 0.12$$

The same reasoning applies for the other three cases, and we get the final answers of $0.09, 0.12, 0.34, 0.2$, and $0.25$. 
As an extra check, when you finish a problem like this, make sure that all of the probabilities add up to $1$ (since there are no other possibilities for the value of $X+Y$, this should be the case:
$$0.09 + 0.12 + 0.34 + 0.2 + 0.25 = 1$$
A: $W = X + Y$ is a random variable that is computed as
$P(W=w) = \Sigma_{x=1}^3 P(X=x)\times P(Y=w-x) $
Hence, it should now be clear that $P(W=1) = P(X=1)\times P(Y=0)$. Since $P(Y=0)=0, P(W=1)=0$
Similarly other values can be calculated.
eg $$P(W=3) = P(X=1)\times P(Y=2) + P(X=2)\times P(X=1)$$ which gives the required answer.
