Addition Rule of Probability

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

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.**