Q:

Do distributions weights always sum up to 1

I have been given the following problem: Let $Z=(Z_1,Z_2,Z_3)$ be a triple coin toss with $p=1/4$. Let $Y:=\frac{1}{2}Z_1 + \left(\frac{1}{2}\right)^2Z_2+\left(\frac{1}{2}\right)^3Z_3$ be a random variable $\in [0,1]$. What are the probability weights of $Y$. 

I have been given this problem before, but I was just asked to compute the probability for $Z$, which is fairly straightforward looking at 4 different outcomes (0 hits = $\frac{1}{8}$, 1 hit = $\frac{3}{8}$, 2 hits = $\frac{3}{8}$, 3 hits = $\frac{3}{8}$).
What I'm confused about:
1) Why am I looking at a continuous $[0,1]$ variable?
2) I computed the weights by just plugging in all possible combinations, but the weights don't sum up to 1. They sum up to 3.5. Why is that? And can I now just take weights that describe the same outcome (e.g. (hit, no hit, no hit), (no hit, hit, no hit), (no hit, not hit, hit)) sum them, divide by 3.5 and that's it? I have a feeling this is not how to do it.
Thanks!
/edit: There are a couple of follow up questions on my exercise sheet. I don't care too much about the exact answer (right away at least. eventually, of course I care about it :D). But really, what I want is get an understanding of the random variable I'm looking at. I'm missing something!

A:

You may be over thinking this: there are eight possible values for the triplet $Z$ and each corresponds to a different $Y$.  These are:
   Z         Y       probability
(0,0,0)      0          27/64
(0,0,1)     1/8          9/64
(0,1,0)     1/4          9/64
(0,1,1)     1/4          3/64
(1,0,0)     1/2          9/64 
(1,0,1)     5/8          3/64 
(1,1,0)     3/4          3/64 
(1,1,1)     7/8          1/64 

