If A , B and C are random events, simplify the expressions…

If $A$, $B$ and $C$ are random events, simplify the expressions: \[ 1^{\circ}\ (A+B)(B+C); \qquad 2^{\circ}\ (A+B)(A+\overline{B}); \qquad 3^{\circ}\ (A+B)(A+\overline{B})(\overline{A}+B). \]

Step 1: The laws we will use

Here $\Omega$ is the certain event (the whole sample space) and $\varnothing$ the impossible event. Product means "both occur", sum means "at least one occurs", bar means "does not occur".

Why absorption holds: $AB\subseteq A$, so adding $AB$ to $A$ adds nothing ($A+AB=A$); and $A\subseteq A+B$, so intersecting $A$ with $A+B$ removes nothing ($A(A+B)=A$).

Why the second distributive law holds: for any outcome $\omega$, "$\omega\in A$ or ($\omega\in B$ and $\omega\in C$)" is true exactly when "($\omega\in A$ or $\omega\in B$) and ($\omega\in A$ or $\omega\in C$)" is true.

Step 2: Part $1^{\circ}$: $\;(A+B)(B+C)$

Method 1: multiply out

Multiply every term of the first bracket by every term of the second:

\[ (A+B)(B+C) = AB + AC + BB + BC. \]

Use $BB = B$ (idempotent law) and reorder the terms:

\[ = B + AB + BC + AC. \]

Now $AB\subseteq B$ and $BC\subseteq B$, so by absorption $B + AB = B$ and $B + BC = B$:

\[ = B + AC. \]

Method 2: second distributive law

The law $B + AC = (B+A)(B+C)$, read from right to left, gives the answer at once:

\[ (A+B)(B+C) = (B+A)(B+C) = B + AC. \]

\[ \boxed{(A+B)(B+C) = B + AC} \]

In words: "($A$ or $B$) and ($B$ or $C$)" happens exactly when $B$ happens, or, if $B$ does not happen, when both $A$ and $C$ happen.

Check with a die

$\Omega=\{1,\dots,6\}$, $A=\{1,2\}$, $B=\{2,3\}$, $C=\{1,4\}$. Then $A+B=\{1,2,3\}$, $B+C=\{1,2,3,4\}$, so $(A+B)(B+C)=\{1,2,3\}$. On the other side $AC=\{1\}$ and $B+AC=\{1,2,3\}$. ✓

Step 3: Part $2^{\circ}$: $\;(A+B)(A+\overline{B})$

Method 1: multiply out

\[ (A+B)(A+\overline{B}) = AA + A\overline{B} + BA + B\overline{B}. \]

Use $AA = A$ and $B\overline{B}=\varnothing$ (an event and its opposite never occur together):

\[ = A + A\overline{B} + AB + \varnothing = A + A\overline{B} + AB. \]

Both $A\overline{B}$ and $AB$ lie inside $A$, so they are absorbed:

\[ = A. \]

Method 2: second distributive law

\[ (A+B)(A+\overline{B}) = A + B\overline{B} = A + \varnothing = A. \]

\[ \boxed{(A+B)(A+\overline{B}) = A} \]

In words: suppose $A$ does not occur. Then the first bracket needs $B$ and the second needs "not $B$" — impossible. So the product can occur only when $A$ occurs, and if $A$ occurs, both brackets are satisfied.

Step 4: Part $3^{\circ}$: $\;(A+B)(A+\overline{B})(\overline{A}+B)$

The first two brackets are exactly part $2^{\circ}$, so they equal $A$:

\[ (A+B)(A+\overline{B})(\overline{A}+B) = A(\overline{A}+B) = A\overline{A} + AB = \varnothing + AB = AB. \]

\[ \boxed{(A+B)(A+\overline{B})(\overline{A}+B) = AB} \]

In words: the first two brackets force $A$ to occur. Once $A$ occurs, the third bracket "not $A$ or $B$" can only be satisfied by $B$. So both $A$ and $B$ occur.

Check by multiplying the last two brackets first

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

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

Step 5: Truth-table check

Write $1$ if an event occurs and $0$ if it does not. For part $1^{\circ}$ we need all eight combinations of $A$, $B$, $C$:

For parts $2^{\circ}$ and $3^{\circ}$ only $A$ and $B$ appear:

The column $(A+B)(A+\overline{B})$ equals the column $A$, and the column of expression $3^{\circ}$ equals the column $AB$. ✓