2 Bulgarian Winter Tournament

Let $0<p<q<1$. Notice that $p^{\ast }=\frac{q-p}{1-p}\in (0,1)$. We construct the sets $X$ and $Y$ as follows: For each element $i\in \{1,2,\dots ,n\}$ we put $i$ in $X$ with probability $p$ (independently of each other), and for each element $j\in \{1,2,\dots ,n\}\setminus X$ we put $j$ in $Y$ with probability $p^{\ast }$. Then $P(x\in X\cup Y)=p+(1-p)p^{\ast }=q$. It is easy to see that $f(p)=P(X\in \mathcal{A})$, and $f(q)=P(X\cup Y\in \mathcal{A})$, but since $\mathcal{A}$ has the property from the condition we have that $X\subseteq X\cup Y\Rightarrow f(p)<f(q)$. $\square$