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. 

We can disregard all even elements from $S$, as they don't affect the parity of the sum. So let $N$ be the number of elements from $S$, all of which are odd. Say those elements are $x_1, x_2, \ldots, x_N$.
Let $S_k = \{x_1, \ldots, x_k\}$ (partial set of the first $k$ elements of $S$), and let $t_k$ be the sum of the elements from $S_k$ that get picked.
Element $x_1$ is picked with probability $1/2$. So the $t_1$ is odd with probability $1/2$.
Element $x_2$ is also picked with probability $1/2$. So the parity of $t_2$ differs from that of $t_1$ with probability $1/2$. This implies that $t_2$ is again odd with probability $1/2$.
You can proceed similarly for all $x_3, \ldots, x_k$. (Of course this could be made more formal with induction).
