Show that the event…
Show that the event \[ (A+B)(A+\overline{B})(\overline{A}+B)(\overline{A}+\overline{B}) \] is impossible.
Explanation of Method
An event is impossible if it equals $\varnothing$. We group the four brackets in pairs with a common term and use $(X+Y)(X+Z)=X+YZ$ (see Problem 25): the pairs collapse to $A$ and $\overline{A}$, whose product is $\varnothing$. We confirm this with a second grouping and with the observation that each bracket rules out one of the four possible combinations of $A$ and $B$ — so all combinations are ruled out.
Step 1: Method 1 — group by the common term $A$ or $\overline{A}$
Using $(X+Y)(X+Z)=X+YZ$ and $B\overline{B}=\varnothing$:
\[ (A+B)(A+\overline{B}) = A + B\overline{B} = A, \qquad (\overline{A}+B)(\overline{A}+\overline{B}) = \overline{A} + B\overline{B} = \overline{A}. \]
Hence
\[ (A+B)(A+\overline{B})(\overline{A}+B)(\overline{A}+\overline{B}) = A\overline{A} = \varnothing. \]
\[ \boxed{\text{The event is impossible.}} \]
In words: the first two brackets force "$A$ occurs", the last two force "$A$ does not occur". Both at once cannot happen.
Step 2: Method 2 — group by $B$ or $\overline{B}$
\[ (A+B)(\overline{A}+B) = B + A\overline{A} = B, \qquad (A+\overline{B})(\overline{A}+\overline{B}) = \overline{B} + A\overline{A} = \overline{B}, \]
so the product is $B\overline{B}=\varnothing$. ✓
Step 3: Method 3 — each bracket rules out one case
For any outcome, exactly one of the four combinations happens. Each bracket fails in exactly one combination:
| Combination | $A+B$ | $A+\overline{B}$ | $\overline{A}+B$ | $\overline{A}+\overline{B}$ | All four brackets? |
|---|---|---|---|---|---|
| $A$ and $B$ both occur ($AB$) | yes | yes | yes | no | no |
| only $A$ ($A\overline{B}$) | yes | yes | no | yes | no |
| only $B$ ($\overline{A}B$) | yes | no | yes | yes | no |
| neither ($\overline{A}\,\overline{B}$) | no | yes | yes | yes | no |
No combination satisfies all four brackets, so no outcome belongs to the product: it is $\varnothing$. ✓
Step 4: Connection with Problem 25
Problem 25 showed that the sums of these pairs are certain ($A+\overline{A}=\Omega$, $B+\overline{B}=\Omega$). Here we take the product of the same pairs and get $A\overline{A}=\varnothing$ — the two problems are mirror images of each other.
Result
| Grouping | Result |
|---|---|
| $[(A+B)(A+\overline{B})]\cdot[(\overline{A}+B)(\overline{A}+\overline{B})]$ | $A\cdot\overline{A}=\varnothing$ |
| $[(A+B)(\overline{A}+B)]\cdot[(A+\overline{B})(\overline{A}+\overline{B})]$ | $B\cdot\overline{B}=\varnothing$ |
| Conclusion | the event is impossible |