= Solution
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 $a_j=|\mathcal F\cap\binom{[n]}j|/\binom nj$. 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 $\mu_p(\mathcal F)=\sum_j\binom nj a_jp^j(1-p)^{n-j}$. By independence of the <Bernoulli random variables>, it equals $\mathbb P(Z\ge1/2)$, since equality is excluded. Subtract $p=\sum_j\binom nj(j/n)p^j(1-p)^{n-j}$ and pair complementary levels. The <complementary-layer bound for biased measure> gives
$$
\mu_p(\mathcal F)-p=\sum_{j<n/2}\binom nj\left(\frac jn-a_j\right)\left[p^{n-j}(1-p)^j-p^j(1-p)^{n-j}\right]\ge0.
$$
Both factors are nonnegative for $p\ge1/2$. The endpoints $p=1/2,1$ also follow directly, or by continuity. Thus the <weighted Bernoulli majority bound> is
$$
\boxed{\mathbb P(Z\ge1/2)\ge p}.
$$
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