Find the random event X if overlineX+A+overlineX+overlineA=B
Find the random event $X$ if \[ \overline{X+A}+\overline{X+\overline{A}}=B. \]
Explanation of Method
We use De Morgan's law $\overline{X+Y}=\overline{X}\,\overline{Y}$ to remove the bars over the sums. Both terms then contain the factor $\overline{X}$, which we take outside the bracket. What remains in the bracket is $\overline{A}+A=\Omega$, so the event $A$ disappears completely and we get $\overline{X}=B$. Taking complements gives $X$. Finally we check the answer and explain why it is the only solution.
Step 1: Apply De Morgan's law to each term
De Morgan's law says: "neither $X$ nor $Y$" is the opposite of "$X$ or $Y$":
\[ \overline{X+Y} = \overline{X}\,\overline{Y}. \]
Apply it to both terms, and use $\overline{\overline{A}}=A$ in the second one:
\[ \overline{X+A} = \overline{X}\,\overline{A}, \qquad \overline{X+\overline{A}} = \overline{X}\,\overline{\overline{A}} = \overline{X}A. \]
Step 2: Substitute into the equation
\[ \overline{X}\,\overline{A} + \overline{X}A = B. \]
Step 3: Take $\overline{X}$ outside the bracket
By the distributive law $\overline{X}\,\overline{A} + \overline{X}A = \overline{X}(\overline{A}+A)$, so
\[ \overline{X}(\overline{A}+A) = B. \]
Step 4: Simplify
An event and its opposite together cover everything: $\overline{A}+A=\Omega$. And $\overline{X}\,\Omega=\overline{X}$. Therefore
\[ \overline{X} = B. \]
Step 5: Solve for $X$
Take the complement of both sides and use $\overline{\overline{X}}=X$:
\[ \boxed{X = \overline{B}}. \]
Step 6: Check
Put $X=\overline{B}$ into the left side of the original equation:
\[ \overline{\overline{B}+A} + \overline{\overline{B}+\overline{A}} = B\overline{A} + BA = B(\overline{A}+A) = B\,\Omega = B. \ \checkmark \]
Step 7: The solution is unique
We did not guess the answer. Starting from any event $X$ that satisfies the equation, Steps 1–4 show that necessarily $\overline{X}=B$, and so $X=\overline{B}$. Therefore no other solution exists.
Step 8: What the equation says in words
- $\overline{X+A}$ = "neither $X$ nor $A$ occurs" = "$X$ does not occur and $A$ does not occur".
- $\overline{X+\overline{A}}$ = "$X$ does not occur and $A$ occurs".
The sum of the two is "$X$ does not occur, whether or not $A$ occurs" = "$X$ does not occur". The equation says this is the same as "$B$ occurs". So $X$ is exactly the opposite of $B$, and $A$ plays no role.
Step 9: Example with a die
Let $\Omega=\{1,\dots,6\}$, $A=\{1,2,3\}$, $B=\{5,6\}$. The formula gives $X=\overline{B}=\{1,2,3,4\}$. Check:
\[ X+A=\{1,2,3,4\},\ \ \overline{X+A}=\{5,6\}; \qquad X+\overline{A}=\{1,2,3,4,5,6\},\ \ \overline{X+\overline{A}}=\varnothing. \]
The sum is $\{5,6\}=B$. ✓
Step 10: Special cases
- If $B=\Omega$ (certain), then $X=\varnothing$ (impossible).
- If $B=\varnothing$ (impossible), then $X=\Omega$ (certain).
Result
| Answer | |
|---|---|
| Solution | $X=\overline{B}$ (the event opposite to $B$); it is the only solution |
| Key steps | $\overline{X}\,\overline{A}+\overline{X}A = \overline{X}(\overline{A}+A) = \overline{X} = B$ |
| Remark | the event $A$ does not influence the answer |