Skip to main content
Statistics LibreTexts

Probability and Independence

For an experiment we define an event to be any collection of possible outcomes.     

A simple event is an event that consists of exactly one outcome.  

  • "or"  means the union (i.e.  either can occur)
  • "and"  means intersection (i.e. both must occur)

Two events are mutually exclusive if they cannot occur simultaneously.  For a Venn diagram, we can tell that two events are mutually exclusive if their regions do not intersect

We define Probability  of an event \(E\) to be to be 

\[(P(E) = \dfrac{\text{number of simple  events within E}}{\text{ total number of possible outcomes}} \]  

We have the following:

  1. \(P(E)\) is always between 0 and 1.
  2. The sum of the probabilities of all simple events must be 1.
  3. \(P(E) + P(\text{not } E) = 1\)
  4. If \(E\) and \(F\) are mutually exclusive then

\[ P(E or F) = P(E) + P(F) \]

The Difference Between "and" and "or"

If \(E\) and \(F\) are events then we use the terminology

\[ E \textbf{ and } F \]

to mean all outcomes that belong to both \(E\) and \(F\).

We use the terminology

\[E \textbf{ or } F \]

to mean all outcomes that belong to either \(E\) or \(F\).

Example 1

Below is an example of two sets, \(A\) and \(B\), graphed in a Venn diagram.  

The green area represents \(A\) and \(B\) while all areas with color represent \(A\) or \(B\)


Our Women's Volleyball team is recruiting for new members. Suppose that a person inquires about the team.

  • Let \(E\) be the event that the person is female
  • Let \(F\) be the event that the person is a student

then \(E\) and \(F\) represents the qualifications for being a member of the team. Note that \(E\) or \(F\) is not enough.


We define

Definition of Conditional Probability

\[ P(E|F) = \dfrac{P(E \text{ and } F}{P(F)}\]

We read the left hand side as "The probability of event \(E\) given event \(F\) occurred."

We call two events independent if

Definition of Independence (1) :

For Independent Events

\[P(E|F)  =  P(E) \tag{1a}\]

Equivalently, we can say that \(E\) and \(F\) are independent if

Definition of Independence (2) :

For Independent Events

\[P(E \text{ and } F)  =  P(E)P(F) \tag{1b}\]

Example 1

Consider rolling two dice.  Let 

  • \(E\) be the event that the first die is a 3.
  • \(F\( be the event that the sum of the dice is an 8.

Then \(E\) and \(F\) means that we rolled a three and then we rolled a 5

This probability is 1/36 since there are 36 possible pairs and only one of them is (3,5)

We have 

\[ P(E)  =  1/6  \]

And note that (2,6),(3,5),(4,4),(5,3), and (6,2) give \(F\)


\[   P(F)  =  5/36\]

We have 

\[  P(E) P(F) = (1/6) (5/36) \]

 which is not 1/36 and we can conclude that \(E\) and \(F\) are not independent.


Test the following two events for independence:

  • \(E\) the event that the first die is a 1.
  • \(F\) the event that the sum is a 7.

A Counting Rule

For two events, \(E\) and \(F\), we always have

 \[ P(E \text{ or } F)  =  P(E) + P(F) - P(E \text{ and } F) \tag{2}\]

Example 1

Find the probability of selecting either a heart or a face card from a 52 card deck.


We let

  • \(E\)  =  the event that a heart is selected 
  • \(F\)  =  the event that a face card is selected


\[ P(E)  =  \dfrac{1}{4} \]


\[ P(F)  =  \dfrac{3}{13} \]

that is, Jack, Queen, or King out of 13 different cards of one kind.

\[ P(E \text{ and } F)  =  \dfrac{3}{52} \]

The counting rule formula (eq. 2) gives

\[  P(E \text{ or } F)  =  \dfrac{1}{4} + \dfrac{3}{13} - \dfrac{3}{52}  = \dfrac{22}{52} =  42\text{%}\]