Show the probability that the sum of these numbers is odd is 1/2 Setting
Let $S$ be a set of integers where at least one of the integers is odd. Suppose we pick a random subset $T$ of $S$ by including each element of $S$ independently with probability $1/2$, Show that
$$\text{Pr}\Big[\Big(\sum_{i \in T} i\Big) = \text{odd}\Big] = \frac{1}{2}$$
Solution given
Let $x$ be an odd integer in $T$. Then for every subset $S \subseteq T - \{x\}$, exactly one of $S$ and $S \cup \{x\}$ has an odd total, and both are subsets of $T$. Thus exactly half of the subsets have an odd total.
Problem
The solution makes very little sense to me. If someone has an alternate solution, or could pose the solution in a more explicable manner, it'd be really nice. 

This is a proof by pairing - it establishes a bijection between sets with even sums and sets with odd sums, to show that there are the same number of subsets of each kind.
Let $U$ be the set which contains all the elements of $S$ except for the. odd integer $x$.
Every subset $T$ of $S$ either contains $x$ or it doesn't. If it doesn't then $T$ is a subset of $U$. We can add the element $x$ to any set $T\subset U$ to get a subset $V$. We match $T$ with $V$. One of these sets has an even sum, and the other has an odd sum.
Do we get all the possible sets $V$? - Well if we have a $V$ containing $x$ we can strike out the element $x$ and get a set $T\subset U$. We have a perfect matching. We can't tell which of the two sets in each matching has an odd sum, but we know precisely one of them will.
We therefore know that half of the sets have odd sums and half have even sums.
