There are three dice that are rolled simultaneously. Let's say that these are $a$, $b$, and $c$ points that fell on the first, second and third dice, respectively. There are 7 events:
- $A: a = b = c$,
- $B: a + b <c$, under condition $a = b$,
- $C: a + b <c$, under conditions $a <b$ and $b <c$,
- $D: a + b> c$, under conditions $b = c$ and $a <c$
- $E: a + b> c$, under conditions $a = b$ and $b <c$,
- $F: a + b = c$,
- $J: a + b> c$, under conditions $a <b$ and $b <c$.
Question. How can you prove that these seven events form a full events group? Is it possible to show it visually, for example with an Euler diagram?
My attempt. I found the number of possibilities for all events:
$m(A) = 6$, $m(B) = 18$, $m(C) = 42$, $m(D) = 45$, $m(E) = 18$, $m(F) = 45$, $m(J) = 42$.
The possibilities sum is $6+18+42+45+18+45+42=216$ and it equals to the possible outcomes $6^3 = 216$ for three dice.
I did not check your counting, but to your original question you can prove the $7$ events cover all possibilities this way:
First, I assume you sorted the three results so that $a \le b \le c$. (Otherwise, you're missing a whole bunch of cases.)
I would do the case analysis this way: $a+b$ can be $>$ or $=$ or $< c$:
Case 1: $a+b > c$. There are $4$ ways to resolve the two $\le$ signs in $a \le b \le c$:
Case 1.1: $a + b > c \cap a < b < c$: your event $J$
Case 1.2: $a + b > c \cap a = b < c$: your event $E$
Case 1.3: $a + b > c \cap a < b = c$: your event $D$
Case 1.4: $a + b > c \cap a = b = c$: your event $A$
Case 2: $a + b = c$: your event $F$
Case 3: $a + b < c$: This implies $b < c$, so we only need 2 subcases:
Case 3.1: $a + b < c \cap a < b < c$: your event $C$
Case 3.2: $a + b < c \cap a = b < c$: your event $B$