Prove that the events A, overlineAB and overlineA+B form a…
Prove that the events \[ A, \qquad \overline{A}B \qquad\text{and}\qquad \overline{A+B} \] form a complete system of events.
Explanation of Method
A family of events is a complete system when (1) any two of them are incompatible (their product is $\varnothing$) and (2) their sum is the certain event $\Omega$. In other words, in every trial exactly one of the events occurs. We first rewrite $\overline{A+B}=\overline{A}\,\overline{B}$ with De Morgan's law, then check the three products and the sum using $X\overline{X}=\varnothing$ and $X+\overline{X}=\Omega$.
Step 1: Definition of a complete system
Events $H_1, H_2, \dots, H_n$ form a complete system if
- they are pairwise incompatible: $H_iH_j=\varnothing$ for $i\neq j$ (no two can occur together), and
- they are exhaustive: $H_1+H_2+\dots+H_n=\Omega$ (at least one of them always occurs).
Together: in each trial exactly one of them occurs.
Step 2: Rewrite the third event
By De Morgan's law, $\overline{A+B}=\overline{A}\,\overline{B}$ ("neither $A$ nor $B$ occurs"). So the three events are
\[ H_1 = A, \qquad H_2 = \overline{A}B, \qquad H_3 = \overline{A}\,\overline{B}. \]
Step 3: They are pairwise incompatible
\[ H_1H_2 = A\,\overline{A}B = (A\overline{A})B = \varnothing B = \varnothing, \]
\[ H_1H_3 = A\,\overline{A}\,\overline{B} = (A\overline{A})\overline{B} = \varnothing, \]
\[ H_2H_3 = \overline{A}B\,\overline{A}\,\overline{B} = (\overline{A}\,\overline{A})(B\overline{B}) = \overline{A}\,\varnothing = \varnothing. \]
Each product contains an event multiplied by its own opposite, so it is impossible. ✓
Step 4: Their sum is the certain event
Take $\overline{A}$ out of the last two terms:
\[ A + \overline{A}B + \overline{A}\,\overline{B} = A + \overline{A}(B+\overline{B}) = A + \overline{A}\,\Omega = A + \overline{A} = \Omega. \ \checkmark \]
Both conditions hold, so $A$, $\overline{A}B$, $\overline{A+B}$ form a complete system. $\blacksquare$
- $A$
- $\overline{A}B$: $B$ without $A$
- $\overline{A+B}=\overline{A}\,\overline{B}$: neither
Step 5: Why it works — two questions
Sort any outcome by asking two questions in turn:
- Does $A$ occur? If yes, the outcome is in $H_1=A$.
- If no: does $B$ occur? If yes, the outcome is in $H_2=\overline{A}B$; if no, it is in $H_3=\overline{A}\,\overline{B}$.
Every outcome gets exactly one answer path, so it lands in exactly one of the three events.
Step 6: Example with a die
$\Omega=\{1,\dots,6\}$, $A=\{1,2\}$, $B=\{2,3,4\}$. Then
\[ A=\{1,2\}, \qquad \overline{A}B=\{3,4\}, \qquad \overline{A+B}=\overline{\{1,2,3,4\}}=\{5,6\}. \]
These three sets do not overlap and together give $\{1,\dots,6\}$. Note that the outcome $2$ belongs to both $A$ and $B$, but it is counted only in $A$. ✓
Step 7: A useful consequence
For a complete system the probabilities add up to $1$:
\[ P(A) + P(\overline{A}B) + P(\overline{A+B}) = 1. \]
Also, since $A+B = A + \overline{A}B$ with incompatible terms, $P(A+B) = P(A) + P(\overline{A}B)$ — a way to compute the probability of a sum without double counting.
Result
| Condition | Check | Holds? |
|---|---|---|
| Pairwise incompatible | $A\cdot\overline{A}B=\varnothing,\ \ A\cdot\overline{A}\,\overline{B}=\varnothing,\ \ \overline{A}B\cdot\overline{A}\,\overline{B}=\varnothing$ | ✓ |
| Sum is certain | $A+\overline{A}B+\overline{A}\,\overline{B}=A+\overline{A}(B+\overline{B})=A+\overline{A}=\Omega$ | ✓ |
| Conclusion | $A,\ \overline{A}B,\ \overline{A+B}$ form a complete system of events |