# 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

Definition: Probability

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

$P(E \text{ or } F) = P(E) + P(F)$

## The Difference Between "and" and "or"

If $$E$$ and $$F$$ are events then we use the terminology

$E \text{ and } F$

to mean all outcomes that belong to both $$E$$ and $$F$$.

We use the terminology

$E \text{ or } F$

to mean all outcomes that belong to either $$E$$ or $$F$$.

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 $$B$$

Example $$\PageIndex{1}$$

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: 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 the following definitions hold.

Definition: Independence

For independent Events

$P(E|F) = P(E) \label{1a}$

Equivalently, we can say that $$E$$ and $$F$$ are independent if

Definition: The Multiplication Rule

For Independent Events

$P(E \text{ and } F) = P(E)P(F) \label{1b}$

Example $$\PageIndex{2}$$

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 $$F$$

Hence

$P(F) = 5/36$

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.

Exercise $$\PageIndex{2}$$

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

$P(E \text{ or } F) = P(E) + P(F) - P(E \text{ and } F) \label{2}$

Example $$\PageIndex{3}$$

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

Solution

We let

• $$E$$ = the event that a heart is selected
• $$F$$ = the event that a face card is selected

then

$P(E) = \dfrac{1}{4}$

and

$P(F) = \dfrac{3}{13}$

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

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

### Contributors

This page titled Probability and Independence is shared under a not declared license and was authored, remixed, and/or curated by Larry Green.