At most two of the events $A$, $B$, or $C$ occur? Set theory Question: Consider a sample space $S$ and three events $A$, $B$, and $C$. For each of the following events draw a Venn diagram representation as well as a set expression.  At most two of the events $A$, $B$, or $C$ occur?
I'm having a challenge understanding the solution to this, which is:
$(A \cap B \cap C)^c = A^c \cup B^c \cup C^c$
and the following diagram:

I can see from the Venn Diagram that only two events may occur due to the intersection visually, however not so with the set expression.  Is the diagram necessary to understand this solution or is there some property of the set expression that can be used to explain the solution "At most two of the events $A$, $B$, or $C$ occur"?
 A: $x\in A\cup B$ means $x$ belongs to at least one of the sets between $A$ and $B$, might belong to both.
So $x\in A\cup B\cup C$ means $x$ belongs to at least one of the sets among $A,B,C$.
$x\in A^c \cup B^c \cup C^c$ means $x$ belongs to at least one of the sets $A^c,B^c,C^c$, that is to say,  for at least one set $A,B,C$, $x$ does not belong to it. So $x$ can belong to at most two of them.
Edit: Here's an example as asked by OP in the comments, consider the universe of tossing $3$ coins, i.e the entire sample space is $\{HHH,HHT,HTH,THH,HTT,THT,TTH,TTT\}$ and consider the events: 
$A=$ exactly one head 
$B=$ at least one tail 
$C=$ last toss is a head
so that we have the corresponding sets: 
$A=\{HTT,THT,TTH\}$ 
$B=\{HHT,HTH,THH,HTT,THT,TTH,TTT\}$ 
$A=\{HHH,HTH,THH,TTH\}$
so the event "at most two of $A,B,C$ occur" is the set of those combinations in the sample space where all three don't occur (check below)
$(A\cap B \cap C)^c= \{TTH\}^c=\{HHT,HTH,THH,HTT,THT,HHH,TTT\}$
A: At most two from $A,B,C$ is: $(A\cap B\cap C)^{\small\complement}$, that is everything except that which is in the triple intersection. (Everything inside the Venn diagram except that in the innermost overlap).
That is to say: everywhere that at least one does not occur.
$$(A^{\small\complement}\cup B^{\small\complement}\cup C^{\small\complement})$$
