Prove that we can divide a group of 90 people into 3 groups each with 30 people so that each has a friend in group if each one has more than 30 friends. Friendships is a symmetric relation.
I tryed to think probabilistic.
Divide arbitrary group in 3 equal parts, so we can do that on $${1\over 6}{90\choose 30}{60\choose 30}$$ Let $D_i$ be an event $i$ has no friend in group. The number of such events is $$m_i = {1\over 2}{89-d_i\choose 29}{60\choose 30}$$ where $d_i$ is number of friends for $i$, so $$m_i\leq {1\over 2}{59\choose 29}{60\choose 30}$$ Now we have $$P(\bigcup D_i)\leq \sum _{i=1}^{90}P(D_i) \leq 90\cdot {3{59\choose 29}\over {90\choose 60}} <1$$ So the probability that $\bigcup D_i$ doesn't happend is more than $0$ and we are done.
Yes, this seems to me to be a correct application of the union bound. In particular, you have shown that the probability of the event $(\lnot D_1)\lor (\lnot D_2)\lor\cdots\lor (\lnot D_{90})$ is strictly less than $1$. It follows that the probability of the event $\lnot\left((\lnot D_1)\lor (\lnot D_2)\lor\cdots\lor (\lnot D_{90})\right)\iff D_1\land D_2\land\cdots\land D_{90}$ is strictly greater than $0$. So among the sample space, i.e. the set of all partitions of the $90$ individuals into three equally-sized groups, there is at least one for which $D_1\land D_2\land\cdots\land D_{90}$, as desired.