$\frac{\alpha+k-1}{k}$ is the minumum probability that verifies a condition for finite covers Let $k\in\mathbb{N}-\{0,1\}$ and $\alpha\in(0,1)$.
Prove that $\frac{\alpha+k-1}{k}$ is the minumum $p\in(0,1]$ such that for every probability space $(\Omega,\Sigma,\mathbb{P})$ and $\forall\{U_{1},...,U_{k}\}$ $\subseteq$ $\Sigma-\{\emptyset\}$ $∣$ $\Omega=\bigcup\limits_{i=1}^{k}U_{i}$ $ $ $\land$ $ $ $p\le\mathbb{P}(U_{i})$ $\forall$$i\in\{1,...,k\}$, you have $\alpha\le\mathbb{P}(\bigcap\limits_{i=1}^{k}U_{i})$.
EDIT: $ $ @AlbertParadek proves in his answer that the statement as it was written (see edit history) was false. The new one is true: the bound can be found by De Morgan's Law and union bound, and that it is minimum can be proven using the counting measure in a finite probability space.
 A: I dont think that $\frac{\alpha+k-1}{k}$ is the correct result though. Denote $U=\cap_{i=1}^k U_i$. Take (if possible) $P(U_i)=p$ such that $U_i\cap U_j=U$, which means that all of them contain $U$ but otherwise are disjoint. Then, by inclusion-exclusion principle, we have
$$
1=P(\cup U_i) = \sum_{i=1}^k P(U_i) - \sum_{i\neq j} P(U_i\cap U_j) + \sum_{i\neq j\neq l} P(U_i\cap U_j\cap U_l)\dots \pm P(U). 
$$
By assumptions, all $P(U_i\cap U_j)=P(U_i\cap U_j\cap U_l)=\dots =P(U)=\alpha$ and $P(U_i)=p$.  All put together we get
$$
1=kp + \alpha ( -\binom{k}{2} + \binom{k}{3}\dots \pm \binom{k}{k})=kp+\alpha (k-1).  
$$
Therefore $p=\frac{1+\alpha(k-1)}{k}$$<\frac{\alpha+k-1}{k}$.
Note a few things: it may be not possible to take such appropriate $U_i$ and I dont think we can say something at all in general (without at least some knowledge of $\Sigma$). On the other hand, I dont think that (for any $\Sigma$) we can obtain lower number than $\frac{1+\alpha(k-1)}{k}$. This is evident from the construction of the inclusion-exclusion sum. If some of them have some common intersection besides $U$, then the right hand side would be larger.
