A = $\begin{bmatrix}a_{1}\\ a_{2} \\ . \\ . \\ .\\a_{n} \end{bmatrix}$ be a non zero vector.
X = $\begin{bmatrix}x_{1}\\ x_{2} \\ . \\ . \\ .\\x_{n} \end{bmatrix}$ be a randomly chosen vector
Since A is non zero, 'k' entries in A are '1's, where 0 $<$ k $\leq$ n.
n - k entries are '0's.
Moreover these n - k entries multiplied with corresponding entries in X vector is 0. So need not bother about these n - k entries
Let $a_{j_{1}}$, $a_{j_{2}}$, $a_{j_{3}}$, .... , $a_{j_{k}}$ be the entries in A which are '1's, where $j_{1}$ $<$ $j_{2}$ $<$ $j_{3}$ $<$ .... $<$ $j_{k}$
( Note : $j_{1}$ , $j_{2}$ , $j_{3}$ , .... , $j_{k}$ necessarily need not be a continuous sequence as well )
Now consider corresponding $x_{j_{1}}$ , $x_{j_{2}}$ , $x_{j_{3}}$ , ... , $x_{j_{k}}$ entries in X vector.
In order for $\sum_{i=1}^{n}$ $a_{i}$ $x_{i}$ to be odd we need odd number of entries to be '1's in $x_{j_{1}}$ , $x_{j_{2}}$ , $x_{j_{3}}$ , ... , $x_{j_{k}}$ and the rest '0's
So total number of vectors having odd number of '1's in $x_{j_{1}}$ , $x_{j_{2}}$ , $x_{j_{3}}$ , ... , $x_{j_{k}}$ so that $\sum_{i=1}^{n}$ $a_{i}$ $x_{i}$ being odd is
$\binom{k}{1}$ + $\binom{k}{3}$ + $\binom{k}{5}$ + ... + $\binom{k}{k}$ if k is odd = $2^{k-1}$
$\binom{k}{1}$ + $\binom{k}{3}$ + $\binom{k}{5}$ + ... + $\binom{k}{k-1}$ if k is even = $2^{k-1}$
Note : Following is not part of the answer
/*
Similarly,
In order for $\sum_{i=1}^{n}$ $a_{i}$ $x_{i}$ to be even we need even number of entries to be '1's in $x_{j_{1}}$ , $x_{j_{2}}$ , $x_{j_{3}}$ , ... , $x_{j_{k}}$ and the rest '0's
So total number of vectors having even number of '1's in $x_{j_{1}}$ , $x_{j_{2}}$ , $x_{j_{3}}$ , ... , $x_{j_{k}}$ so that $\sum_{i=1}^{n}$ $a_{i}$ $x_{i}$ being even is
$\binom{k}{0}$ + $\binom{k}{2}$ + $\binom{k}{4}$ + ... + $\binom{k}{k-1}$ if k is odd = $2^{k-1}$
$\binom{k}{0}$ + $\binom{k}{2}$ + $\binom{k}{4}$ + ... + $\binom{k}{k}$ if k is even = $2^{k-1}$
*/
The total number of combinations in $x_{j_{1}}$ , $x_{j_{2}}$ , $x_{j_{3}}$ , ... , $x_{j_{k}}$ is $2^{k}$
that is $\binom{k}{0}$ + $\binom{k}{1}$ + $\binom{k}{2}$ + ... + $\binom{k}{k}$ = $2^{k}$ or in other words $2_{j_{1}}$ $\times$ $2_{j_{2}}$ $\times$ $2_{j_{3}}$ $\times$ ... $\times$ $2_{j_{k}}$ = $2^{k}$
Now Probability = $\frac{Odd\,Combinations}{Total\,Combinations}$ = $\frac{2^{k-1}}{2^{k}}$ = 0.5
