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Form the family $\mathcal{F}=\{A\subseteq [n]:{\sum }_{i\in A}{c}_i>1/2\}$. The nonnegative weights make it an up-set. It is an intersecting family, since two disjoint members would have combined weight exceeding one. The no-tie hypothesis makes it a self-dual set family: exactly one of $A,A^c$ belongs.

Let $a_j=|\mathcal{F}\cap (\genfrac{}{}{0ex}{}{[n]}{j})|/(\genfrac{}{}{0ex}{}{n}{j})$. Self-duality gives $a_{n-j}=1-a_j$. The Erdős-Ko-Rado theorem gives $a_j\le j/n$ for $j<n/2$; at $j=0$ this is immediate. If $n$ is even, $a_{n/2}=1/2$.

The biased measure of a set family is ${\mu }_p(\mathcal{F})={\sum }_j\left(\genfrac{}{}{0ex}{}{n}{j}\right)a_j{p}^j(1-p{)}^{n-j}$. By independence of the Bernoulli random variables, it equals $\mathbb{P}(Z\ge 1/2)$, since equality is excluded. Subtract $p={\sum }_j\left(\genfrac{}{}{0ex}{}{n}{j}\right)(j/n)p^j(1-p{)}^{n-j}$ and pair complementary levels. The complementary-layer bound for biased measure givesBoth factors are nonnegative for $p\ge 1/2$. The endpoints $p=1/2,1$ also follow directly, or by continuity. Thus the weighted Bernoulli majority bound is