Let $X$ be an n element set, and let $A_1, ..., A_m$ be not necessarily distinct, nonempty subsets of $X$. Prove that there exists a subset from the list such that every element of this subset is contained in at least $m/n$ sets among $A_1, ..., A_m$.
I have spent quite a time to come up with something but I am stuck. My first idea was to create a random variable $O_x$ so it represents the number of occurences for a given number $x$. I was able to see $E[O_x] \geq m/n$ but the rest didn't go as I expected.
Now I think of another approach. It would be to calculate the probability of having 1 element among $m/n$ sets and raising that expression to power of $k$-- the cardinality of the wanted set, if I could show it is greater than zero it should be cleared, right?
There is also proof by contradiction. If you assume the result is false, then this means for all $A_i$, there exists some $x \in A_i$ such that x is contained in at most $m/n$ sets among $A_1,...,A_m$. I think with this maybe I can show $|\cup A_i| < m$, but this is a contradiction as $A_1,...,A_m$ is guaranteed to be at least $|m|$ since they are all non-empty.
Anyway, I think I am lost on this one and I would kindly ask for a hint or a directive.
The statement is false for $m=0$ so let's assume $m\gt0$, which I'm sure is what you intended.
Write $[m]=\{1,\dots,m\}$.
For $x\in X$ let $I_x=\{i\in[m]:x\in A_i\}$.
Let $B=\{x\in X:|I_x|\lt\frac mn\}$.
If $B=\emptyset$ any $A_i$ will work, so let's assume $B\ne\emptyset$. Then $$\sum_{x\in B}|I_x|\lt\sum_{x\in B}\frac mn=|B|\frac mn\le|X|\frac mn=m$$ and so $$\left|\bigcup_{x\in B}I_x\right|\le\sum_{x\in B}|I_x|\lt m.$$ Therefore we can choose an index $i\in[m]$ such that $i\notin\bigcup_{x\in B}I_x$, which means that $A_i\cap B=\emptyset$, which means that $|I_x|\ge\frac mn$ for every $x$ in $A_i$.