Among the students attending a probability theory class, we…

Among the students attending a probability theory class, we randomly select one. Let event $A$ mean that the selected student is male, event $B$ that the student does not smoke, and event $C$ that the student lives in a student dormitory.
$1^{\circ}$ Describe the event $AB\overline{C}$.
$2^{\circ}$ Under what conditions does the identity $ABC=A$ hold?
$3^{\circ}$ When is the relation $\overline{C}\subseteq B$ correct?
$4^{\circ}$ Explain when the equality $\overline{A}=B$ holds. Does it hold when all the male students present do not smoke?

Step 1: The events and their opposites

Since one student is selected from the class, the outcomes are the students, and each event is the set of students having that property. For example, $A$ is "the group of male students".

Step 2: Part $1^{\circ}$ — the event $AB\overline{C}$

A product means that all factors occur at once: $A$ and $B$ and $\overline{C}$.

\[ \boxed{AB\overline{C} = \text{"the selected student is a male non-smoker who does not live in a dormitory".}} \]

Step 3: Part $2^{\circ}$ — when is $ABC=A$?

For any events, $ABC=A$ holds exactly when $A\subseteq B$ and $A\subseteq C$ (see Problem 1): if $A$ has these inclusions, intersecting $A$ with $B$ and $C$ removes nothing; conversely, $A=ABC\subseteq B$ and $A=ABC\subseteq C$.

• $A\subseteq B$: every male student is a non-smoker;
• $A\subseteq C$: every male student lives in a dormitory.

\[ \boxed{ABC=A \iff \text{all male students in the class do not smoke and live in a dormitory.}} \]

(The condition says nothing about the female students.)

Step 4: Part $3^{\circ}$ — when is $\overline{C}\subseteq B$?

The inclusion means: every student who does not live in a dormitory is a non-smoker.

\[ \boxed{\overline{C}\subseteq B \iff \text{all students who do not live in a dormitory are non-smokers.}} \]

Equivalent forms:

• $\overline{C}\,\overline{B}=\varnothing$: there is no smoker who lives outside a dormitory;
• $\overline{B}\subseteq C$ (taking complements reverses an inclusion): every smoker lives in a dormitory.

Step 5: Part $4^{\circ}$ — when is $\overline{A}=B$?

The equality means that the group of female students and the group of non-smokers are the same group. This requires two inclusions:

• $\overline{A}\subseteq B$: every female student is a non-smoker;
• $B\subseteq\overline{A}$: every non-smoker is female, i.e. every male student smokes.

\[ \boxed{\overline{A}=B \iff \text{all female students do not smoke and all male students smoke.}} \]

In other words, a student is a non-smoker exactly when the student is female.

Step 6: Part $4^{\circ}$ — what if all male students do not smoke?

"All male students present do not smoke" means $A\subseteq B$. If also $\overline{A}=B$, then

\[ A\subseteq B=\overline{A}, \]

so $A\subseteq\overline{A}$, which is possible only for $A=\varnothing$ (a student cannot be both male and not male). Hence:

• If there is at least one male student in the class, then no: that non-smoking male student belongs to $B$ but not to $\overline{A}$, so $\overline{A}\neq B$.
• The equality could hold only in the special case where there are no male students at all in the class and all the female students do not smoke. (Then "all males do not smoke" is trivially true, and $\overline{A}=\Omega=B$.)

Example

The class has 10 male and 5 female students, and nobody smokes. Then $\overline{A}$ is the group of 5 female students, while $B$ is the group of all 15 students. The two groups are different, so $\overline{A}\neq B$.