Determine the random event X if: 1^(°) A+X=A+B; 2^(°)…

Determine the random event $X$ if: \[ 1^{\circ}\ A+X=A+B; \qquad 2^{\circ}\ AB+X=(A+C)(B+C). \]

Step 1: Part $1^{\circ}$ — look outside $A$

Multiply both sides of $A+X=A+B$ by $\overline{A}$ (keep only what happens outside $A$):

\[ \overline{A}A + \overline{A}X = \overline{A}A + \overline{A}B \quad\Longrightarrow\quad \overline{A}X = \overline{A}B, \]

since $\overline{A}A=\varnothing$. So outside $A$, $X$ must coincide with $B$.

Step 2: Part $1^{\circ}$ — inside $A$ nothing is required

Split $X$ into its part outside and inside $A$: $X = \overline{A}X + AX$. Then

\[ A + X = A + \overline{A}X + AX = A + \overline{A}X, \]

because $AX\subseteq A$ is absorbed. So the part $AX$ never affects $A+X$: it can be anything.

Step 3: Part $1^{\circ}$ — the general solution

\[ \boxed{X = \overline{A}B + AY, \qquad Y \text{ an arbitrary event.}} \]

Check: $A + \overline{A}B + AY = A + \overline{A}B = A + B$, using absorption ($AY\subseteq A$) and the rule $A+\overline{A}B=A+B$ (from $A+\overline{A}B=(A+\overline{A})(A+B)=\Omega(A+B)$). ✓

No other solutions: if $X$ is any solution, Step 1 gives $\overline{A}X=\overline{A}B$, so $X=\overline{A}X+AX=\overline{A}B+AX$ — the form above with $Y=X$.

Smallest and largest solution

• $Y=\varnothing$: $\;X_{\min}=\overline{A}B$ ("$B$ occurs but $A$ does not").
• $Y=\Omega$: $\;X_{\max}=\overline{A}B + A = A+B$.

So all solutions are described by $\overline{A}B\subseteq X\subseteq A+B$. In particular $X=B$ is a solution (obviously), but in general not the only one.

Example with a die

$\Omega=\{1,\dots,6\}$, $A=\{1,2,3\}$, $B=\{3,4\}$. Then $\overline{A}B=\{4\}$ must be in $X$; $\overline{A}\,\overline{B}=\{5,6\}$ must not be in $X$; the outcomes $1,2,3$ are free. There are $2^3=8$ solutions, from $X_{\min}=\{4\}$ to $X_{\max}=\{1,2,3,4\}=A+B$; for instance $X=\{2,4\}$ gives $A+X=\{1,2,3,4\}=A+B$. ✓

Step 4: Part $2^{\circ}$ — simplify the right side

By the second distributive law, $(C+A)(C+B)=C+AB$. Or multiply out:

\[ (A+C)(B+C) = AB + AC + CB + CC = AB + AC + BC + C = AB + C, \]

because $CC=C$ and $AC\subseteq C$, $BC\subseteq C$ are absorbed by $C$. The equation becomes

\[ AB + X = AB + C. \]

Step 5: Part $2^{\circ}$ — use part $1^{\circ}$

This is exactly equation $1^{\circ}$ with $AB$ in place of $A$ and $C$ in place of $B$. The general solution is therefore

\[ \boxed{X = \overline{AB}\,C + AB\,Y, \qquad Y \text{ an arbitrary event,}} \]

where $\overline{AB}=\overline{A}+\overline{B}$.

• Smallest solution ($Y=\varnothing$): $\;X_{\min}=\overline{AB}\,C=(\overline{A}+\overline{B})C$ — "$C$ occurs, but not both $A$ and $B$".
• Largest solution ($Y=\Omega$): $\;X_{\max}=AB+C$.
• The simplest solution: $X=C$ (it lies between the two, since $\overline{AB}\,C\subseteq C\subseteq AB+C$).

Check of the smallest solution: by the rule $Z+\overline{Z}Y=Z+Y$ with $Z=AB$, $Y=C$,

\[ AB + \overline{AB}\,C = AB + C = (A+C)(B+C). \ \checkmark \]

Step 6: Summary in words

• $1^{\circ}$: when $A$ does not occur, $X$ must behave exactly like $B$; when $A$ occurs, $X$ may or may not occur.
• $2^{\circ}$: when "both $A$ and $B$" does not occur, $X$ must behave exactly like $C$; when both $A$ and $B$ occur, $X$ is free.