<s>[INST] 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?
 [/INST] 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 
$$ P(X+Y=3) =\frac{1}{12}\\
P(X+Y=4) =\frac{1}{12}+\frac{1}{4} = \frac{1}{3}\\
P(X+Y=5) =\frac{1}{12}+\frac{1}{4} = \frac{1}{3}\\
P(X+Y=6) =\frac{1}{4}\\
$$
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