Originally Posted by

**andre.vignatti** Let $\displaystyle E_1, E_2, E_3$ be three events, where $\displaystyle Pr[E_1]=p_1, Pr[E_2]=p_2, Pr[E_3]=p_3$. Also, $\displaystyle p_1 + p_2 + p_3 = 1$. Using the inclusion-exclusion principle, $\displaystyle Pr[E_1 \cup E_2 \cup E_3] = p_1 + p_2 + p_3 - p_1p_2 - p_1p_3 - p_2p_3 + p_1p_2p_3 = f(p_1, p_2, p_3)$.$

Thus, I would like to prove that $\displaystyle \max f(p_1,p_2,p_3)$ subject to $\displaystyle p_1 + p_2 + p_3 = 1$ has solution $\displaystyle p_1 = p_2 = p_3 = 1/3$. In order to do this I used Lagrange multipliers, and after a few manipulations on the equations, I come up with this answer.

**THE QUESTION IS**: how can I generalize this for $\displaystyle n$ events? More specifically, I want to prove that $\displaystyle \max Pr[E_1 \cup \ldots \cup E_n]$ subject to $\displaystyle p_1 + \ldots + p_n = 1$ has solution $\displaystyle p_1 = \ldots = p_n = 1/n$.

**PS**: Some directions to solve this problem would be useful too (or a reference where it was already solved).....