This is part of another post I did, but I think it has interest in its own right:
Let $Y =\{X_{1},X_{2}...X_{N}\}$ be a set of $N$ random quantities with assocated set of distributions $F=\{F_{1},F_{2}...F_{N}\}$ and acceptable absolute errors $\epsilon = \{\epsilon_{1},\epsilon_{2}...\epsilon_{N}\}$. You are to develop a single-point prediction set $\hat X =\{\hat X_{1},\hat X_{2}...\hat X_{N}\}$ which contains one point from the possible range of each random quantity such that you maximize $P(\sum\limits_{i=1}^{N}\textbf{1}_{|X_{i}-\hat X_{i}|\leq \epsilon_{i}}= N)$.
How should you select $\hat X =\{\hat X_{1},\hat X_{2}...\hat X_{N}\}$?
The answer proposed by Sun will work. For the sum to equal N, there is only one point in the sample space of outcomes where this happens, and that is when each indicator variable is equal to 1. Therefore, to maximimze the probability, you need to maximize the probability that each indicator equals 1. Since each variable is independent, you are maximizing the product $\prod P(X_{i} \in x_i \pm \epsilon_i)$ where $P(X_{i} \in x_i \pm \epsilon_i)=\int_{x_i-\epsilon_i/2}^{x_i+\epsilon_i/2} F_i(x)$ The product is maximize by individually maximizing each factor to get the highest probability prediction.