The Multiplication Rule: $P(A \cap B) = P(A) \cdot P(B|A)$.
The Addition Rule: $P(A \cup B) = P(A)+P(B)-P(A \cap B)$.
Example: Felicity attends Southwestern Oregon Community College.
The probability that Felicity enrolls in a math class is $0.20.$ Call this event $M.$
The probability that she enrolls in a writing class is $0.64.$ Call this event $W.$
The probability that she enrolls in a writing class GIVEN that she enrolls in a math class is $0.28.$
Find the following probabilities: $P(M \cap W)$, $P(M \cup W)$, and $P(M|W).$
$$
\begin{array}{lll}
\displaystyle P(M \cap W)&\displaystyle=P(M)\cdot P(W|M) &\mbox{by the Multiplication Rule}\\
\displaystyle &\displaystyle=0.20\cdot 0.28 &\mbox{}\\
\displaystyle &\displaystyle=0.056 &\mbox{}\\
\end{array}
$$
$$ \begin{array}{lll} \displaystyle P(M \cup W)&\displaystyle=P(M)+P(W)-P(M \cap W) &\mbox{by the Addition Rule}\\ \displaystyle &\displaystyle=0.20+0.64-0.056 &\mbox{from given info and previous calculation}\\ \displaystyle &\displaystyle=0.784 &\mbox{}\\ \end{array} $$
$$ \begin{array}{lll} \displaystyle P(M|W)&\displaystyle=\frac{P(M \cap W)}{P(W)} &\mbox{definition of conditional probability}\\ \displaystyle &\displaystyle=\frac{0.056}{0.064} &\mbox{from given info and previous calculation}\\ \displaystyle &\displaystyle=0.875 &\mbox{}\\ \end{array} $$
$$ \begin{array}{lll} \displaystyle P(M \cup W)&\displaystyle=P(M)+P(W)-P(M \cap W) &\mbox{by the Addition Rule}\\ \displaystyle &\displaystyle=0.20+0.64-0.056 &\mbox{from given info and previous calculation}\\ \displaystyle &\displaystyle=0.784 &\mbox{}\\ \end{array} $$
$$ \begin{array}{lll} \displaystyle P(M|W)&\displaystyle=\frac{P(M \cap W)}{P(W)} &\mbox{definition of conditional probability}\\ \displaystyle &\displaystyle=\frac{0.056}{0.064} &\mbox{from given info and previous calculation}\\ \displaystyle &\displaystyle=0.875 &\mbox{}\\ \end{array} $$
Some Background: The Multiplication Rule $$P(A \cap B) = P(A)\cdot P(B|A)$$ Notice that this is simply a rearrangement of our definition of conditional probability of the event $B$ given $A$. $$\displaystyle P(B|A)=\frac{P(B \cap A)}{P(A)}.$$
Also Notice: if $A$ and $B$ are independent events that $$P(A \cap B) = P(A)\cdot P(B|A)=P(A) \cdot P(B)$$ since $\displaystyle P(B|A)=P(B)$ by definition.
Some Background: The Addition Rule $$P(A \cup B) = P(A)+P(B)-P(A \cap B)$$
Also Notice: if $A$ and $B$ are mutually exclusive events that $$P(A \cup B) = P(A)+P(B)-P(A \cap B)=P(A)+P(B)$$ since $\displaystyle P(A \cap B)=0$ by definition.
Another Example
A school has $200$ seniors of whom $120$ will be going to college next year.
Another $40$ will be going directly to work.
The remainder are taking a gap year.
Of the seniors going to college, $60$ play sports.
Of the seniors going directly to work, $20$ play sports.
Of the seniors taking a gap year, $3$ play sports.
What is the probability that a randomly chosen senior is taking a gap year or plays sports?
We organize the given information into a contingency table.
$$
\begin{array}{c|c|c|c|c}
&\mbox{College}&\mbox{Work}&\mbox{Gap}&\\\hline
\mbox{Sports}&\color{magenta}{60}&\color{magenta}{20}&\color{magenta}{3}&\\\hline
\mbox{No Sports}&&&&\\\hline
\mbox{}&\color{magenta}{120}&\color{magenta}{40}&&\color{magenta}{200}\\
\end{array}
$$
We now fill in the rest of the table.
$$
\begin{array}{c|c|c|c|c}
&\mbox{College}&\mbox{Work}&\mbox{Gap}&\\\hline
\mbox{Sports}&\color{magenta}{60}&\color{magenta}{20}&\color{magenta}{3}&83\\\hline
\mbox{No Sports}&60&20&37&117\\\hline
\mbox{}&\color{magenta}{120}&\color{magenta}{40}&40&\color{magenta}{200}\\
\end{array}
$$
Letting $G$ be the event "taking gap year" and $S$ the event "plays sports," we may now compute the probability of
of "taking gap year OR plays sports."
$$
\begin{array}{lll}
\displaystyle P(G\cup S)&\displaystyle= P(G)+P(S)-P(G\cap S)&\mbox{using the addition rule}\\
\displaystyle &\displaystyle=\frac{40}{200}+\frac{83}{200}-\frac{3}{200} &\mbox{}\\
\displaystyle &\displaystyle=\frac{120}{200}&\mbox{}\\
\displaystyle &\displaystyle=0.60&\mbox{}\\
\end{array}
$$