Is it possible for two or more events to be collectively exhaustive and independent at the same time? Two or more events are collectively exhaustive if they cover entire sample space. 
Two or more events are independent if occurance or failure of one does not affect occurance or failure of other. 
Are these definitions correct? Is it possible that a given set of events is collectively exhaustive amd independent at the same time? 
 A: (2017-09-01 21:00: This answer just received a revenge downvote. Oh well...)

1. Some events $(A_i)_{1\leqslant i\leqslant n}$ are independent if and only if, for every $I\subseteq\{1,2,\ldots,n\}$, $$P\left(\bigcap_{i\in I}A_i\right)=\prod_{i\in I}P(A_i)$$
2. Two events $A$ and $B$ are independent if and only if $P(A\cap B)=P(A)P(B)$. If furthermore $A$ and $B$ are "collectively exhaustive", in the sense that $P(A\cup B)=1$, then one gets $$1=P(A\cup B)=P(A)+P(B)-P(A\cap B)=P(A)+P(B)-P(A)P(B)$$ which is not possible except if $P(A)=1$ or $P(B)=1$.
3. Likewise, if $(A_i)_{1\leqslant i\leqslant n}$ are independent and collectively exhaustive then $A=\bigcup\limits_{i=1}^{n-1}A_i$ and $B=A_n$ are independent and collectively exhaustive hence, by our preceding item, either $P(A)=1$ or $P(B)=1$. That is, either $P(A_n)=1$ or $(A_i)_{1\leqslant i\leqslant n-1}$ are independent and collectively exhaustive. Continuing with $(A_i)_{1\leqslant i\leqslant n-1}$ in the latter case, one sees that $P(A_i)=1$ for at least some index $i$.
To conclude:

The only events $(A_i)_{1\leqslant i\leqslant n}$ that are independent and collectively exhaustive are such that $P(A_i)=1$ for at least some index $i$.

