Let A , B and C be random events. Explain the meaning of the…

Let $A$, $B$ and $C$ be random events. Explain the meaning of the equalities \[ 1^{\circ}\quad ABC = A; \qquad\qquad 2^{\circ}\quad A + B + C = A. \]

Step 1: Notation and basic facts

All events live inside one sample space $\Omega$ — the set of all possible outcomes $\omega$ of the experiment. An event is a subset of $\Omega$; we say the event occurs if the outcome of the experiment belongs to it.

To prove that two events are equal we must show two inclusions. We will constantly use two facts that are true for any events:

Fact 1 (an intersection is inside each factor). If all three events occur, then in particular each one of them occurs:

\[ ABC \subseteq A, \qquad ABC \subseteq B, \qquad ABC \subseteq C. \]

Fact 2 (a union contains each term). If one of the events occurs, then "at least one of them occurs" is certainly true:

\[ A \subseteq A+B+C, \qquad B \subseteq A+B+C, \qquad C \subseteq A+B+C. \]

Step 2: Equality $1^{\circ}$: $\;ABC = A$

Reading the equality

The left side, $ABC$, is the event "$A$, $B$ and $C$ all occur". The right side is the event "$A$ occurs". The equality says that these two events are the same: knowing that $A$ occurred is already enough to know that all three occurred. That can only be true if $B$ and $C$ "come for free" whenever $A$ happens. Let us prove this precisely.

Claim

\[ ABC = A \quad\Longleftrightarrow\quad A\subseteq B \ \text{ and } \ A\subseteq C \quad\Longleftrightarrow\quad A \subseteq BC. \]

Proof of "$\Rightarrow$"

Assume $ABC = A$. By Fact 1, $ABC \subseteq B$. Replacing $ABC$ by $A$ (they are equal) gives $A\subseteq B$. In exactly the same way, $ABC\subseteq C$ gives $A\subseteq C$.

Proof of "$\Leftarrow$"

Assume $A\subseteq B$ and $A\subseteq C$. We must show two inclusions.

(i) $ABC\subseteq A$ — this is always true by Fact 1.

(ii) $A\subseteq ABC$: take any outcome $\omega\in A$. Because $A\subseteq B$, also $\omega\in B$; because $A\subseteq C$, also $\omega\in C$. So $\omega$ belongs to all three events, i.e. $\omega\in ABC$.

Both inclusions hold, therefore $ABC = A$. $\blacksquare$

Meaning

$ABC = A$ means that event $A$ implies both $B$ and $C$: whenever $A$ occurs, $B$ and $C$ necessarily occur as well. Geometrically, the set $A$ lies entirely inside the common part of $B$ and $C$.

Example with a die

Roll a fair die once: $\Omega=\{1,2,3,4,5,6\}$. Let

\[ A = \{\text{a six}\} = \{6\}, \qquad B = \{\text{an even number}\} = \{2,4,6\}, \qquad C = \{\text{a number greater than }3\} = \{4,5,6\}. \]

Then $BC = \{4,6\}$ and

\[ ABC = \{6\}\cap\{2,4,6\}\cap\{4,5,6\} = \{6\} = A. \]

This agrees with the claim: a six is always even and always greater than $3$, so $A\subseteq B$ and $A\subseteq C$.

Counter-check. If instead $A=\{5\}$ ("a five"), then $A\not\subseteq B$ (5 is odd), and indeed $ABC = \varnothing \neq A$. The equality fails exactly when the inclusion fails.

Everyday example

$A$ = "a student gets the grade 10 (the maximum)", $B$ = "the student passed the exam", $C$ = "the student attended the exam". Getting a 10 implies both passing and attending, so $ABC = A$.

Step 3: Equality $2^{\circ}$: $\;A + B + C = A$

Reading the equality

The left side, $A+B+C$, is the event "at least one of $A$, $B$, $C$ occurs". The right side is "$A$ occurs". The equality says: whenever any of the three events occurs, it is in fact $A$ that occurs. So $B$ and $C$ can never happen "on their own", without $A$.

Claim

\[ A+B+C = A \quad\Longleftrightarrow\quad B\subseteq A \ \text{ and } \ C\subseteq A \quad\Longleftrightarrow\quad B+C \subseteq A. \]

Proof of "$\Rightarrow$"

Assume $A+B+C = A$. By Fact 2, $B\subseteq A+B+C$. Replacing $A+B+C$ by $A$ gives $B\subseteq A$. In the same way, $C\subseteq A+B+C = A$.

Proof of "$\Leftarrow$"

Assume $B\subseteq A$ and $C\subseteq A$. Again we show two inclusions.

(i) $A\subseteq A+B+C$ — always true by Fact 2.

(ii) $A+B+C\subseteq A$: take any outcome $\omega\in A+B+C$. Then $\omega$ lies in $A$, in $B$ or in $C$.

• If $\omega\in A$ — done.
• If $\omega\in B$ — since $B\subseteq A$, also $\omega\in A$.
• If $\omega\in C$ — since $C\subseteq A$, also $\omega\in A$.

In every case $\omega\in A$. Both inclusions hold, therefore $A+B+C = A$. $\blacksquare$

Meaning

$A+B+C = A$ means that each of the events $B$ and $C$ implies $A$: whenever $B$ occurs, or $C$ occurs, the event $A$ occurs as well. Geometrically, both sets $B$ and $C$ lie entirely inside $A$, so adding them to $A$ adds nothing new.

Example with a die

Roll a die: $\Omega=\{1,2,3,4,5,6\}$. Let

\[ A = \{\text{an even number}\} = \{2,4,6\}, \qquad B = \{2,4\}, \qquad C = \{4,6\}. \]

Then

\[ A+B+C = \{2,4,6\}\cup\{2,4\}\cup\{4,6\} = \{2,4,6\} = A. \]

Indeed both $B$ and $C$ contain only even numbers, so $B\subseteq A$ and $C\subseteq A$. (Notice that $B$ and $C$ overlap in $\{4\}$ — that is allowed.)

Counter-check. If $B = \{1\}$, then $B\not\subseteq A$ and $A+B+C = \{1,2,4,6\}\neq A$.

Everyday example

$A$ = "the student passed the course", $B$ = "the student got grade 10", $C$ = "the student got grade 9". Getting a 10 or getting a 9 each implies passing, so $A+B+C = A$. (There are also other ways to pass, e.g. grade 6 — that is fine, $A$ may be larger than $B+C$.)

Step 4: Why the two equalities are "mirror images"

In $1^{\circ}$ the event $A$ is the smallest one (it sits inside $B$ and $C$); in $2^{\circ}$ it is the largest one (it contains $B$ and $C$). This is no accident. Take complements ("the event does not occur", written $\overline{A}$) and use De Morgan's law $\overline{ABC} = \overline{A}+\overline{B}+\overline{C}$:

\[ ABC = A \quad\Longleftrightarrow\quad \overline{A}+\overline{B}+\overline{C} = \overline{A}. \]

The right-hand equality is exactly of type $2^{\circ}$ for the complements, so it means $\overline{B}\subseteq\overline{A}$ and $\overline{C}\subseteq\overline{A}$: "if $B$ does not occur, then $A$ does not occur" — which is just another way of saying $A\subseteq B$. Both points of view give the same answer.

Step 5: Common mistakes to avoid

• $ABC = A$ does not mean $A=B=C$. The events $B$ and $C$ may be much larger than $A$ (in the die example, $B$ has three outcomes while $A$ has one).
• $A+B+C = A$ does not mean $B$ and $C$ are equal to $A$, nor that they are disjoint. They only have to fit inside $A$; they may even be the impossible event $\varnothing$.
• Do not mix up the direction of the inclusion. In $1^{\circ}$: $A\subseteq B$, $A\subseteq C$ ("$A$ is inside"). In $2^{\circ}$: $B\subseteq A$, $C\subseteq A$ ("$A$ is outside").
• Both equalities hold at the same time only if $A\subseteq B\subseteq A$ and $A\subseteq C\subseteq A$, i.e. only when $A=B=C$.