Find the distribution of $Y+X$ Consider the following two experiments: the first has outcome $X$ taking on
the values $0, 1$ and $2$ with equal probabilities; the second results in an (in-
dependent) outcome $Y$ taking on the value $3$ with probability $1/4$ and $4$ with
probability $3/4$. Find the distribution of $Y+X$
Attempt: I assigned the probability $1/3$ to $0,1$ and $2$ and the I set $P(S=3)=m(3)m(0)+m(0)m(3)+m(1)m(2)=7/18$, but I felt like I was getting the wrong answer. How would you solve this?
 A: In this case you can enumerate all the possibilities and use independence to find their probabilities. Then you can divide up the possibilities according to what the sum $X+Y$ is.
We have:
$$
P(X=0,Y=3) = \frac{1}{3}\frac{1}{4} = \frac{1}{12}\\
P(X=1,Y=3) = \frac{1}{3}\frac{1}{4} = \frac{1}{12}\\
P(X=2,Y=3) = \frac{1}{3}\frac{1}{4} = \frac{1}{12}\\
P(X=0,Y=4) = \frac{1}{3}\frac{3}{4} = \frac{1}{4}\\
P(X=1,Y=4) = \frac{1}{3}\frac{3}{4} = \frac{1}{4}\\
P(X=2,Y=4) = \frac{1}{3}\frac{3}{4} = \frac{1}{4}
$$
The values of $X+Y$ in each of these cases is $3,4,5,4,5,6$ respectively, so 
$$ P(X+Y=3) =\frac{1}{12}\\
P(X+Y=4) =\frac{1}{12}+\frac{1}{4} = \frac{1}{3}\\
P(X+Y=5) =\frac{1}{12}+\frac{1}{4} = \frac{1}{3}\\
P(X+Y=6) =\frac{1}{4}\\
$$
A: 
Attempt: I assigned the probability $1/3$ to $0,1$ and $2$ and the I set $P(S=3)=m(3)m(0)+m(0)m(3)+m(1)m(2)=7/18$, but I felt like I was getting the wrong answer. How would you solve this?

You have different probability measures for the two random variables. You will only confuse yourself (and others) by using the same symbol. Do not do that!
$\mathsf P(X+Y=3) ~{= m(0)n(3)+m(1)n(2)+m(2)n(1)\\= \mathsf P(X=0)~\mathsf P(Y=3)+\mathsf P(X=1)~\mathsf P(Y=2)+\mathsf P(X=2)~\mathsf P(Y=1) \\ = \tfrac 13 \tfrac 14 +\tfrac 13\tfrac 0~+\tfrac 13\tfrac 0~ \\ = \tfrac{1}{12}}$
