### Calculate probabilities using the addition and multiplication rules

$$P\left( A\cup B \right) =P\left( A \right) +P\left( B \right) -P\left( A\cap B \right)$$

The addition rule of probability states that the probability of event $$A$$ or event $$B$$ occurring is the probability of event $$A$$ occurring plus the probability of event $$B$$ occurring minus the probability of both event $$A$$ and event $$B$$ occurring. As shown in the diagrams below, adding the probability of $$A$$ and the probability of $$B$$ together double counts the intersection of $$A$$ and $$B\left( A\cap B \right)$$ which is the reason it must be subtracted from the sum of $$A$$ and $$B$$.  If events $$A$$ and $$B$$ are mutually exclusive, then

$$P\left( A\cup B \right) =P\left( A \right) +P\left( B \right)$$

Because $$P\left( A\cap B \right) =\emptyset$$. Example

Consider the experiment drawing one card from a standard 52 card deck. Let event $$A$$ be drawing a heart. Let event $$B$$ be drawing a king. What is the probability of drawing a heart or drawing a king?

\begin{align*} & P \left( A \right) ={ 13 }/{ 52 } \\ & P \left(B \right) = {4}/{52} \\ & P \left(A \cap B \right) = {1}/{52} \quad \left(the\quad king\quad of\quad hearts \right) \\ & P \left(A \cup B \right) = {13}/{52} + {4}/{52} – {1}/{52} = {16}/{52} \\ \end{align*}

Consider the same experiment (drawing one card from standard 52 card deck). What is the probability of drawing either a heart or a spade?

These events are mutually exclusive. Therefore,

\begin{align*} P \left(A \cup B \right) &= P \left(A \right) + P \left(B \right) \\ & = {\frac {1}{4}} + {\frac {1}{4}} = {\frac {1}{2}} \\ \end{align*}

Multiplication Rule of Probability

$$P\left( A\cap B \right) =P\left( A \right) \ast P\left( B|A \right)$$

The multiplication rule of probability states that the probability of events $$A$$ and $$B$$ both occurring is the probability of event $$A$$ occurring times the probability of event $$B$$ occurring, given that event $$A$$ has already occurred.

If events $$A$$ and $$B$$ are independent, then

$$P \left(A \cap B \right) = P \left(A \right) \ast P \left(B \right)$$

Because the outcome of event Adoes not impact the outcome of event $$B$$.

Example

Consider the experiment flipping a coin 2 times. What is the probability both flips result in heads?

These events are independent. The result of the second flip is not dependent on the first flip. Therefore,

\begin{align*} P \left(A \cap B \right) & = P \left(A \right) \ast P \left(B \right) \\ & = {\frac {1}{2} } \ast {\frac {1}{2} } = {\frac {1}{4} } \\ \end{align*}

Now consider the experiment drawing 2 cards consecutively from a standard 52 card deck. What is the probability of drawing a king and then a queen?

These events are not independent. There is one fewer card in the deck after the first card is drawn.

\begin{align*} & P \left(A \cap B \right) = P \left(A \right) \ast P \left(B|A \right) \\ & P \left(A \right) = {4}/{52} \\ & P \left(B \right) = {4}/{51} \\ & P \left(A \cap B \right) = {4}/{52} \ast {4}/{51} \\ \end{align*}

Learning Outcome

1.b Topic: General Probability – Calculate probabilities using the addition and multiplication rules.