Fill-In-The-Blanks
Consider $1\_2\_3\_4\_5\_6\_7\_8\_9$, where each blank is filled with $+$ or $-$ independently with probability $\tfrac12$. Find the probability the result equals $0$.
Answer
The value is
\[
S=1\pm2\pm3\pm4\pm5\pm6\pm7\pm8\pm9.
\]
Even terms ($2,4,6,8$) never affect parity (evens don't make odds be even, or evens be odd). The odd part is
\[
1\pm3\pm5\pm7\pm9,
\]
a sum of $5$ odd numbers, which is always odd.
Hence $S$ is always odd, so it cannot be $0$ (even). Therefore,
\[
P(S=0)=0.
\]
Notes and comments
Comment 1: Note that $-3,-5,-7,-9$.