## Fundamentals of Probability

Probability is the branch of mathematics that deals with the likelihood that certain outcomes will occur. There are five basic rules, or axioms, that one must understand while studying the fundamentals of probability.

### Learning Objectives

Explain the most basic and most important rules in determining the probability of an event

### Key Takeaways

#### Key Points

- Probability is a number that can be assigned to outcomes and events. It always is greater than or equal to zero, and less than or equal to one.
- The sum of the probabilities of all outcomes must equal [latex]1[/latex].
- If two events have no outcomes in common, the probability that one or the other occurs is the sum of their individual probabilities.
- The probability that an event does not occur is [latex]1[/latex] minus the probability that the event does occur.
- Two events [latex]A[/latex] and [latex]B[/latex] are independent if knowing that one occurs does not change the probability that the other occurs.

#### Key Terms

**event**: A subset of the sample space.**sample space**: The set of all outcomes of an experiment.**experiment**: Something that is done that produces measurable results, called outcomes.**outcome**: One of the individual results that can occur in an experiment.

In discrete probability, we assume a well-defined *experiment*, such as flipping a coin or rolling a die. Each individual result which could occur is called an *outcome*. The set of all outcomes is called the *sample* * space*, and any subset of the sample space is called an *event. *

For example, consider the experiment of flipping a coin two times. There are four individual outcomes, namely [latex]HH, HT, TH, TT.[/latex] The sample space is thus [latex]\{HH, HT, TH, TT\}.[/latex] The event “at least one heads occurs” would be the set [latex]\{HH, HT, TH\}.[/latex] If the coin were a normal coin, we would assign the probability of [latex]1/4[/latex] to each outcome.

In probability theory, the probability [latex]P[/latex] of some event [latex]E[/latex], denoted [latex]P(E)[/latex], is usually defined in such a way that [latex]P[/latex] satisfies a number of axioms, or rules. The most basic and most important rules are listed below.

### Probability Rules

*Probability is a number. It is always greater than or equal to zero, and less than or equal to one*. This can be written as [latex]0 \leq P(A) \leq 1[/latex]. An impossible event, or an event that never occurs, has a probability of [latex]0[/latex]. An event that always occurs has a probability of [latex]1[/latex]. An event with a probability of [latex]0.5[/latex] will occur half of the time.*The sum of the probabilities of all possibilities must equal*[latex]1[/latex]. Some outcome must occur on every trial, and the sum of all probabilities is 100%, or in this case, [latex]1[/latex]. This can be written as [latex]P(S) = 1[/latex], where [latex]S[/latex] represents the entire sample space.*If two events have no outcomes in common, the probability that one or the other occurs is the sum of their individual probabilities*. If one event occurs in [latex]30\%[/latex] of the trials, a different event occurs in [latex]20\%[/latex] of the trials, and the two cannot occur together (if they are disjoint ), then the probability that one or the other occurs is [latex]30\% + 20\% = 50\%[/latex]. This is sometimes referred to as the addition rule, and can be simplified with the following: [latex]P(A \ \text{or} \ B) = P(A)+P(B)[/latex]. The word “or” means the same thing in mathematics as the union, which uses the following symbol: [latex]\cup [/latex]. Thus when [latex]A[/latex] and [latex]B[/latex] are disjoint, we have [latex]P(A \cup B) = P(A)+P(B)[/latex].*The probability that an event does not occur is*[latex]1[/latex]*minus the probability that the event does occur*. If an event occurs in [latex]60\%[/latex] of all trials, it fails to occur in the other [latex]40\%[/latex], because [latex]100\% - 60\% = 40\%[/latex]. The probability that an event occurs and the probability that it does not occur always add up to [latex]100\%[/latex], or [latex]1[/latex]. These events are called complementary events, and this rule is sometimes called the complement rule. It can be simplified with [latex]P(A^c) = 1-P(A)[/latex], where [latex]A^c[/latex] is the complement of [latex]A[/latex].*Two events*[latex]A[/latex]*and*[latex]B[/latex]*are independent if knowing that one occurs does not change the probability that the other occurs*. This is often called the multiplication rule. If [latex]A[/latex] and [latex]B[/latex] are independent, then [latex]P(A \ \text{and} \ B) = P(A)P(B)[/latex]. The word “and” in mathematics means the same thing in mathematics as the intersection, which uses the following symbol: [latex]\cap[/latex]. Therefore when A and B are independent, we have [latex]P(A \cap B) = P(A)P(B).[/latex]

### Extension of the Example

Elaborating on our example above of flipping two coins, assign the probability [latex]1/4[/latex] to each of the [latex]4[/latex] outcomes. We consider each of the five rules above in the context of this example.

1. Note that each probability is [latex]1/4[/latex], which is between [latex]0[/latex] and [latex]1[/latex].

2. Note that the sum of all the probabilities is [latex]1[/latex], since [latex]\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=1 [/latex].

3. Suppose [latex]A[/latex] is the event exactly one head occurs, and [latex]B[/latex] is the event exactly two tails occur. Then [latex]A=\{HT,TH\}[/latex] and [latex]B=\{TT\}[/latex] are disjoint. Also, [latex]P(A \cup B) = \frac{3}{4} = \frac{2}{4}+\frac{1}{4}=P(A) + P(B).[/latex]

4. The probability that no heads occurs is [latex]1/4[/latex], which is equal to [latex]1-3/4[/latex]. So if [latex]A=\{HT, TH, HH\}[/latex] is the event that a head occurs, we have [latex]P(A^c)=\frac{1}{4}=1 - \frac{3}{4}=1-P(A).[/latex]

5. If [latex]A[/latex] is the event that the first flip is a heads and [latex]B[/latex] is the event that the second flip is a heads, then [latex]A[/latex] and [latex]B[/latex] are independent. We have [latex]A=\{HT,HH\}[/latex] and [latex]B=\{TH,HH\}[/latex] and [latex]A \cap B = \{HH\}.[/latex] Note that [latex]P(A \cap B) = \frac{1}{4} =\frac{1}{2}\cdot \frac{1}{2} = P(A)P(B).[/latex]

## Unions and Intersections

Union and intersection are two key concepts in set theory and probability.

### Learning Objectives

Give examples of the intersection and the union of two or more sets

### Key Takeaways

#### Key Points

- The union of two or more sets is the set that contains all the elements of the two or more sets. Union is denoted by the symbol [latex]\cup[/latex].
- The general probability addition rule for the union of two events states that [latex]P(A\cup B) = P(A)+P(B)-P(A \cap B)[/latex], where [latex]A \cap B[/latex] is the intersection of the two sets.
- The addition rule can be shortened if the sets are disjoint: [latex]P(A \cup B) = P(A) + P(B)[/latex]. This can even be extended to more sets if they are all disjoint: [latex]P(A \cup B \cup C) = P(A) + P(B) + P(C)[/latex].
- The intersection of two or more sets is the set of elements that are common to every set. The symbol [latex]\cap[/latex] is used to denote the intersection.
- When events are independent, we can use the multiplication rule for independent events, which states that [latex]P(A \cap B) = P(A)P(B)[/latex].

#### Key Terms

**independent**: Not contingent or dependent on something else.**disjoint**: Having no members in common; having an intersection equal to the empty set.

### Introduction

Probability uses the mathematical ideas of sets, as we have seen in the definition of both the sample space of an experiment and in the definition of an event. In order to perform basic probability calculations, we need to review the ideas from set theory related to the set operations of union, intersection, and complement.

### Union

The union of two or more sets is the set that contains all the elements of each of the sets; an element is in the union if it belongs to at least one of the sets. The symbol for union is [latex]\cup[/latex], and is associated with the word “or”, because [latex]A \cup B[/latex] is the set of all elements that are in [latex]A[/latex] or [latex]B[/latex] (or both.) To find the union of two sets, list the elements that are in either (or both) sets. In terms of a Venn Diagram, the union of sets [latex]A[/latex] and [latex]B[/latex] can be shown as two completely shaded interlocking circles.

In symbols, since the union of [latex]A[/latex] and [latex]B[/latex] contains all the points that are in [latex]A[/latex] or [latex]B[/latex] or both, the definition of the union is:

[latex]\displaystyle A \cup B = \{x: x\in A \ \text{or} \ x\in B \}[/latex]

For example, if [latex]A = \{1, 3, 5, 7\}[/latex] and [latex]B = \{1, 2, 4, 6\}[/latex], then [latex]A \cup B = \{1, 2, 3, 4, 5, 6, 7\}[/latex]. Notice that the element [latex]1[/latex] is not listed twice in the union, even though it appears in both sets [latex]A[/latex] and [latex]B[/latex]. This leads us to the general addition rule for the union of two events:

[latex]\displaystyle P(A \cup B) = P(A) + P(B) - P(A \cap B)[/latex]

Where [latex]P(A\cap B)[/latex] is the intersection of the two sets. We must subtract this out to avoid double counting of the inclusion of an element.

If sets [latex]A[/latex] and [latex]B[/latex] are disjoint, however, the event [latex]A \cap B[/latex] has no outcomes in it, and is an empty set denoted as [latex]\emptyset[/latex], which has a probability of zero. So, the above rule can be shortened for disjoint sets only:

[latex]\displaystyle P(A \cup B) = P(A)+P(B)[/latex]

This can even be extended to more sets if they are all disjoint:

[latex]\displaystyle P(A \cup B \cup C) = P(A) + P(B)+ P(C)[/latex]

### Intersection

The intersection of two or more sets is the set of elements that are common to each of the sets. An element is in the intersection if it belongs to all of the sets. The symbol for intersection is [latex]\cap[/latex], and is associated with the word “and”, because [latex]A \cap B[/latex] is the set of elements that are in [latex]A[/latex] and [latex]B[/latex] simultaneously. To find the intersection of two (or more) sets, include only those elements that are listed in both (or all) of the sets. In terms of a Venn Diagram, the intersection of two sets [latex]A[/latex] and [latex]B[/latex] can be shown at the shaded region in the middle of two interlocking circles.

In mathematical notation, the intersection of [latex]A[/latex] and [latex]B[/latex] is written as [latex]A \cap B = \{x: x \in A \ \text{and} \ x \in B\}[/latex]. For example, if [latex]A = \{1, 3, 5, 7\}[/latex] and [latex]B = \{1, 2, 4, 6\}[/latex], then [latex]A \cap B = \{1\}[/latex] because [latex]1[/latex] is the only element that appears in both sets [latex]A[/latex] and [latex]B[/latex].

When events are independent, meaning that the outcome of one event doesn’t affect the outcome of another event, we can use the multiplication rule for independent events, which states:

[latex]\displaystyle P(A \cap B)= P(A)P(B)[/latex]

For example, let’s say we were tossing a coin twice, and we want to know the probability of tossing two heads. Since the first toss doesn’t affect the second toss, the events are independent. Say is the event that the first toss is a heads and [latex]B[/latex] is the event that the second toss is a heads, then [latex]P(A \cap B) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}[/latex].

## Conditional Probability

The conditional probability of an event is the probability that an event will occur given that another event has occurred.

### Learning Objectives

Explain the significance of Bayes’ theorem in manipulating conditional probabilities

### Key Takeaways

#### Key Points

- The conditional probability [latex]P(B \vert A)[/latex] of an event [latex]B[/latex], given an event [latex]A[/latex], is defined by: [latex]P(B|A)=\frac{P(A\cap B)}{P(A)}[/latex], when [latex]P(A) > 0[/latex].
- If the knowledge that event [latex]A[/latex] occurs does not change the probability that event [latex]B[/latex] occurs, then [latex]A[/latex] and [latex]B[/latex] are independent events, and thus, [latex]P(B|A) = P(B)[/latex].
- Mathematically, Bayes’ theorem gives the relationship between the probabilities of [latex]A[/latex] and [latex]B[/latex], [latex]P(A)[/latex] and [latex]P(B)[/latex], and the conditional probabilities of [latex]A[/latex] given [latex]B[/latex] and [latex]B[/latex] given [latex]A[/latex], [latex]P(A|B)[/latex] and [latex]P(B|A)[/latex]. In its most common form, it is: [latex]P(A|B)=\frac{P(B|A)P(A)}{P(B)}[/latex].

#### Key Terms

**conditional probability**: The probability that an event will take place given the restrictive assumption that another event has taken place, or that a combination of other events has taken place**independent**: Not dependent; not contingent or depending on something else; free.

### Probability of [latex]B[/latex] Given That [latex]A[/latex] Has Occurred

Our estimation of the likelihood of an event can change if we know that some other event has occurred. For example, the probability that a rolled die shows a [latex]2[/latex] is [latex]1/6[/latex] without any other information, but if someone looks at the die and tells you that is is an even number, the probability is now [latex]1/3[/latex] that it is a [latex]2[/latex]. The notation [latex]P(B|A)[/latex] indicates a conditional probability, meaning it indicates the probability of one event under the condition that we know another event has happened. The bar “|” can be read as “given”, so that [latex]P(B|A)[/latex] is read as “the probability of [latex]B[/latex] given that [latex]A[/latex] has occurred”.

The conditional probability [latex]\displaystyle P(B|A)[/latex] of an event [latex]B[/latex], given an event [latex]A[/latex], is defined by:

[latex]\displaystyle P(B|A)=\frac{P(A\cap B)}{P(A)}[/latex]

When [latex]P(A) > 0[/latex]. Be sure to remember the distinct roles of [latex]B[/latex] and [latex]A[/latex] in this formula. The set after the bar is the one we are assuming has occurred, and its probability occurs in the denominator of the formula.

### Example

Suppose that a coin is flipped 3 times giving the sample space:

[latex]S=\{HHH, HHT, HTH, THH, TTH, THT, HTT, TTT\}[/latex]

Each individual outcome has probability [latex]1/8[/latex]. Suppose that [latex]B[/latex] is the event that at least one heads occurs and [latex]A[/latex] is the event that all [latex]3[/latex] coins are the same. Then the probability of [latex]B[/latex] given [latex]A[/latex] is [latex]1/2[/latex], since [latex]A \cap B=\{HHH\}[/latex] which has probability [latex]1/8[/latex] and [latex]A=\{HHH,TTT\}[/latex] which has probability [latex]2/8[/latex], and [latex]\frac{1/8}{2/8}=\frac{1}{2}.[/latex]

### Independence

The conditional probability [latex]P(B|A)[/latex] is not always equal to the unconditional probability [latex]P(B)[/latex]. The reason behind this is that the occurrence of event [latex]A[/latex] may provide extra information that can change the probability that event [latex]B[/latex] occurs. If the knowledge that event [latex]A[/latex] occurs does not change the probability that event [latex]B[/latex] occurs, then [latex]A[/latex] and [latex]B[/latex] are independent events, and thus, [latex]P(B|A) = P(B)[/latex].

### Bayes’ Theorem

In probability theory and statistics, Bayes’ theorem (alternatively Bayes’ law or Bayes’ rule) is a result that is of importance in the mathematical manipulation of conditional probabilities. It can be derived from the basic axioms of probability.

Mathematically, Bayes’ theorem gives the relationship between the probabilities of [latex]A[/latex] and [latex]B[/latex], [latex]P(A)[/latex] and [latex]P(B)[/latex], and the conditional probabilities of [latex]A[/latex] given [latex]B[/latex] and [latex]B[/latex] given [latex]A[/latex]. In its most common form, it is:

[latex]\displaystyle P(A|B)=\frac{P(B|A)P(A)}{P(B)}[/latex]

This may be easier to remember in this alternate symmetric form:

[latex]\displaystyle \frac{P(A|B)}{P(B|A)}=\frac{P(A)}{P(B)}[/latex]

### Example:

Suppose someone told you they had a nice conversation with someone on the train. Not knowing anything else about this conversation, the probability that they were speaking to a woman is [latex]50\%[/latex]. Now suppose they also told you that this person had long hair. It is now more likely they were speaking to a woman, since women in in this city are more likely to have long hair than men. Bayes’s theorem can be used to calculate the probability that the person is a woman.

To see how this is done, let [latex]W[/latex] represent the event that the conversation was held with a woman, and [latex]L[/latex] denote the event that the conversation was held with a long-haired person. It can be assumed that women constitute half the population for this example. So, not knowing anything else, the probability that [latex]W[/latex] occurs is [latex]P(W) = 0.5[/latex].

Suppose it is also known that [latex]75\%[/latex] of women in this city have long hair, which we denote as [latex]P(L|W) = 0.75[/latex]. Likewise, suppose it is known that [latex]25\%[/latex] of men in this city have long hair, or [latex]P(L|M) = 0.25[/latex], where [latex]M[/latex] is the complementary event of [latex]W[/latex], i.e., the event that the conversation was held with a man (assuming that every human is either a man or a woman).

Our goal is to calculate the probability that the conversation was held with a woman, given the fact that the person had long hair, or, in our notation, [latex]P(W|L)[/latex]. Using the formula for Bayes’s theorem, we have:

[latex]\displaystyle \begin{align} P(W|L) &= \frac{P(L|W)P(W)}{P(L)}\\ &= \frac{P(L|W)P(W)}{P(L|W)P(W)+P(L|M)P(M)}\\ &=\frac{0.75\cdot 0.5}{0.75\cdot 0.5+0.25\cdot 0.5}\\ &=0.75 \end{align}[/latex]

## Complementary Events

The complement of [latex]A[/latex] is the event in which [latex]A[/latex] does not occur.

### Learning Objectives

Explain an example of a complementary event

### Key Takeaways

#### Key Points

- The complement of an event [latex]A[/latex] is usually denoted as [latex]A'[/latex], [latex]A^c[/latex] or [latex]\bar{A}[/latex].
- An event and its complement are mutually exclusive, meaning that if one of the two events occurs, the other event cannot occur.
- An event and its complement are exhaustive, meaning that both events cover all possibilities.

#### Key Terms

**mutually exclusive**: describing multiple events or states of being such that the occurrence of any one implies the non-occurrence of all the others**exhaustive**: including every possible element

### What are Complementary Events?

In probability theory, the complement of any event [latex]A[/latex] is the event [latex][\text{not}\ A][/latex], i.e. the event in which [latex]A[/latex] does not occur. The event [latex]A[/latex] and its complement [latex][\text{not}\ A][/latex] are mutually exclusive and exhaustive, meaning that if one occurs, the other does not, and that both groups cover all possibilities. Generally, there is only one event [latex]B[/latex] such that [latex]A[/latex] and [latex]B[/latex] are both mutually exclusive and exhaustive; that event is the complement of [latex]A[/latex]. The complement of an event [latex]A[/latex] is usually denoted as [latex]A'[/latex], [latex]A^c[/latex] or [latex]\bar{A}[/latex].

### Simple Examples

A common example used to demonstrate complementary events is the flip of a coin. Let’s say a coin is flipped and one assumes it cannot land on its edge. It can either land on heads or on tails. There are no other possibilities (exhaustive), and both events cannot occur at the same time (mutually exclusive). Because these two events are complementary, we know that [latex]P(\text{heads}) + P(\text{tails}) = 1[/latex].

Finally, let’s examine a non-example of complementary events. If you were asked to choose any number, you might think that that number could either be prime or composite. Clearly, a number cannot be both prime and composite, so that takes care of the mutually exclusive property. However, being prime or being composite are not exhaustive because the number 1 in mathematics is designated as “unique. ”

## The Addition Rule

The addition rule states the probability of two events is the sum of the probability that either will happen minus the probability that both will happen.

### Learning Objectives

Calculate the probability of an event using the addition rule

### Key Takeaways

#### Key Points

- The addition rule is: [latex]P(A\cup B)=P(A)+P(B)-P(A\cap B).[/latex]
- The last term has been accounted for twice, once in [latex]P(A)[/latex] and once in [latex]P(B)[/latex], so it must be subtracted once so that it is not double-counted.
- If [latex]A[/latex] and [latex]B[/latex] are disjoint, then [latex]P(A\cap B)=0[/latex], so the formula becomes [latex]P(A \cup B)=P(A) + P(B).[/latex]

#### Key Terms

**probability**: The relative likelihood of an event happening.

### Addition Law

The addition law of probability (sometimes referred to as the addition rule or sum rule), states that the probability that [latex]A[/latex] or [latex]B[/latex] will occur is the sum of the probabilities that [latex]A[/latex] will happen and that [latex]B[/latex] will happen, minus the probability that both [latex]A[/latex] and [latex]B[/latex] will happen. The addition rule is summarized by the formula:

[latex]\displaystyle P(A \cup B) = P(A)+P(B)-P(A \cap B)[/latex]

Consider the following example. When drawing one card out of a deck of [latex]52[/latex] playing cards, what is the probability of getting heart or a face card (king, queen, or jack)? Let [latex]H[/latex] denote drawing a heart and [latex]F[/latex] denote drawing a face card. Since there are [latex]13[/latex] hearts and a total of [latex]12[/latex] face cards ([latex]3[/latex] of each suit: spades, hearts, diamonds and clubs), but only [latex]3[/latex] face cards of hearts, we obtain:

[latex]\displaystyle P(H) = \frac{13}{52}[/latex]

[latex]\displaystyle P(F) = \frac{12}{52}[/latex]

[latex]\displaystyle P(F \cap H) = \frac{3}{52}[/latex]

Using the addition rule, we get:

[latex]\displaystyle \begin{align} P(H\cup F)&=P(H)+P(F)-P(H\cap F)\\ &=\frac { 13 }{ 52 } +\frac { 12 }{ 52 } -\frac { 3 }{ 52 } \end{align}[/latex]

The reason for subtracting the last term is that otherwise we would be counting the middle section twice (since [latex]H[/latex] and [latex]F[/latex] overlap).

### Addition Rule for Disjoint Events

Suppose [latex]A[/latex] and [latex]B[/latex] are disjoint, their intersection is empty. Then the probability of their intersection is zero. In symbols: [latex]P(A \cap B) = 0[/latex]. The addition law then simplifies to:

[latex]P(A \cup B) = P(A) + P(B) \qquad \text{when} \qquad A \cap B = \emptyset[/latex]

The symbol [latex]\emptyset[/latex] represents the empty set, which indicates that in this case [latex]A[/latex] and [latex]B[/latex] do not have any elements in common (they do not overlap).

### Example:

Suppose a card is drawn from a deck of 52 playing cards: what is the probability of getting a king or a queen? Let [latex]A[/latex] represent the event that a king is drawn and [latex]B[/latex] represent the event that a queen is drawn. These two events are disjoint, since there are no kings that are also queens. Thus:

[latex]\displaystyle \begin{align} P(A \cup B) &= P(A) + P(B)\\&=\frac{4}{52}+\frac{4}{52}\\&=\frac{8}{52}\\&=\frac{2}{13} \end{align}[/latex]

## The Multiplication Rule

The multiplication rule states that the probability that [latex]A[/latex] and [latex]B[/latex] both occur is equal to the probability that [latex]B[/latex] occurs times the conditional probability that [latex]A[/latex] occurs given that [latex]B[/latex] occurs.

### Learning Objectives

Apply the multiplication rule to calculate the probability of both [latex]A[/latex] and [latex]B[/latex] occurring

### Key Takeaways

#### Key Points

- The multiplication rule can be written as: [latex]P(A \cap B) = P(B) \cdot P(A|B)[/latex].
- We obtain the general multiplication rule by multiplying both sides of the definition of conditional probability by the denominator.

#### Key Terms

**sample space**: The set of all possible outcomes of a game, experiment or other situation.

### The Multiplication Rule

In probability theory, the Multiplication Rule states that the probability that [latex]A[/latex] and [latex]B[/latex] occur is equal to the probability that [latex]A[/latex] occurs times the conditional probability that [latex]B[/latex] occurs, given that we know [latex]A[/latex] has already occurred. This rule can be written:

[latex]\displaystyle P(A \cap B) = P(B) \cdot P(A|B)[/latex]

Switching the role of [latex]A[/latex] and [latex]B[/latex], we can also write the rule as:

[latex]\displaystyle P(A\cap B) = P(A) \cdot P(B|A)[/latex]

We obtain the general multiplication rule by multiplying both sides of the definition of conditional probability by the denominator. That is, in the equation [latex]\displaystyle P(A|B)=\frac{P(A\cap B)}{P(B)}[/latex], if we multiply both sides by [latex]P(B)[/latex], we obtain the Multiplication Rule.

The rule is useful when we know both [latex]P(B)[/latex] and [latex]P(A|B)[/latex], or both [latex]P(A)[/latex] and [latex]P(B|A).[/latex]

### Example

Suppose that we draw two cards out of a deck of cards and let [latex]A[/latex] be the event the the first card is an ace, and [latex]B[/latex] be the event that the second card is an ace, then:

[latex]\displaystyle P(A)=\frac { 4 }{ 52 }[/latex]

And:

[latex]\displaystyle P\left( { B }|{ A } \right) =\frac { 3 }{ 51 }[/latex]

The denominator in the second equation is [latex]51[/latex] since we know a card has already been drawn. Therefore, there are [latex]51[/latex] left in total. We also know the first card was an ace, therefore:

[latex]\displaystyle \begin{align} P(A \cap B) &= P(A) \cdot P(B|A)\\ &= \frac { 4 }{ 52 } \cdot \frac { 3 }{ 51 } \\ &=0.0045 \end{align}[/latex]

### Independent Event

Note that when [latex]A[/latex] and [latex]B[/latex] are independent, we have that [latex]P(B|A)= P(B)[/latex], so the formula becomes [latex]P(A \cap B)=P(A)P(B)[/latex], which we encountered in a previous section. As an example, consider the experiment of rolling a die and flipping a coin. The probability that we get a [latex]2[/latex] on the die and a tails on the coin is [latex]\frac{1}{6}\cdot \frac{1}{2} = \frac{1}{12}[/latex], since the two events are independent.

## Independence

To say that two events are independent means that the occurrence of one does not affect the probability of the other.

### Learning Objectives

Explain the concept of independence in relation to probability theory

### Key Takeaways

#### Key Points

- Two events are independent if the following are true: [latex]P(A|B) = P(A)[/latex],[latex]P(B|A) = P(B)[/latex], and [latex]P(A \ \text{and} \ B) = P(A) \cdot P(B)[/latex].
- If any one of these conditions is true, then all of them are true.
- If events [latex]A[/latex] and [latex]B[/latex] are independent, then the chance of [latex]A[/latex] occurring does not affect the chance of [latex]B[/latex] occurring and vice versa.

#### Key Terms

**independence**: The occurrence of one event does not affect the probability of the occurrence of another.**probability theory**: The mathematical study of probability (the likelihood of occurrence of random events in order to predict the behavior of defined systems).

### Independent Events

In probability theory, to say that two events are independent means that the occurrence of one does not affect the probability that the other will occur. In other words, if events [latex]A[/latex] and [latex]B[/latex] are independent, then the chance of [latex]A[/latex] occurring does not affect the chance of [latex]B[/latex] occurring and vice versa. The concept of independence extends to dealing with collections of more than two events.

Two events are independent if any of the following are true:

- [latex]\displaystyle P(A|B) = P(A)[/latex]
- [latex]\displaystyle P(B|A) = P(B)[/latex]
- [latex]\displaystyle P(A \ \text{and} \ B) = P(A)\cdot P(B)[/latex]

To show that two events are independent, you must show only one of the conditions listed above. If any one of these conditions is true, then all of them are true.

Translating the symbols into words, the first two mathematical statements listed above say that the probability for the event with the condition is the same as the probability for the event without the condition. For independent events, the condition does not change the probability for the event. The third statement says that the probability of both independent events [latex]A[/latex] and [latex]B[/latex] occurring is the same as the probability of [latex]A[/latex] occurring, multiplied by the probability of [latex]B[/latex] occurring.

As an example, imagine you select two cards consecutively from a complete deck of playing cards. The two selections are not independent. The result of the first selection changes the remaining deck and affects the probabilities for the second selection. This is referred to as selecting “without replacement” because the first card has not been replaced into the deck before the second card is selected.

However, suppose you were to select two cards “with replacement” by returning your first card to the deck and shuffling the deck before selecting the second card. Because the deck of cards is complete for both selections, the first selection does not affect the probability of the second selection. When selecting cards with replacement, the selections are independent.

Consider a fair die role, which provides another example of independent events. If a person roles two die, the outcome of the first roll does not change the probability for the outcome of the second roll.

### Example

Two friends are playing billiards, and decide to flip a coin to determine who will play first during each round. For the first two rounds, the coin lands on heads. They decide to play a third round, and flip the coin again. What is the probability that the coin will land on heads again?

First, note that each coin flip is an independent event. The side that a coin lands on does not depend on what occurred previously.

For any coin flip, there is a [latex]{\frac{1}{2}}[/latex] chance that the coin will land on heads. Thus, the probability that the coin will land on heads during the third round is [latex]{\frac{1}{2}}[/latex].

### Example

When flipping a coin, what is the probability of getting tails [latex]5[/latex] times in a row?

Recall that each coin flip is independent, and the probability of getting tails is [latex]{\frac{1}{2}}[/latex] for any flip. Also recall that the following statement holds true for any two independent events A and B:

[latex]\displaystyle P(A \ \text{and} \ B) = P(A)\cdot P(B)[/latex]

Finally, the concept of independence extends to collections of more than [latex]2[/latex] events.

Therefore, the probability of getting tails [latex]4[/latex] times in a row is:

[latex]\displaystyle {\frac{1}{2}} \cdot {\frac{1}{2}} \cdot {\frac{1}{2}} \cdot {\frac{1}{2}} = {\frac{1}{16}} [/latex]

## Experimental Probabilities

The experimental probability is the ratio of the number of outcomes in which an event occurs to the total number of trials in an experiment.

### Learning Objectives

Calculate the empirical probability of an event based on given information

### Key Takeaways

#### Key Points

- In a general sense, experimental (or empirical) probability estimates probabilities from experience and observation.
- In simple cases, where the result of a trial only determines whether or not the specified event has occurred, modeling using a binomial distribution might be appropriate; then the empirical estimate is the maximum likelihood estimate.
- If a trial yields more information, the empirical probability can be improved upon by adopting further assumptions in the form of a statistical model. If such a model is fitted, it can be used to derive an estimate of the probability of the specified event.

#### Key Terms

**binomial distribution**: The discrete probability distribution of the number of successes in a sequence of [latex]n[/latex] independent yes/no experiments, each of which yields success with probability [latex]p[/latex].**experimental probability**: The probability that a certain outcome will occur, as determined through experiment.**discrete**: Separate; distinct; individual; non-continuous.

The experimental (or empirical) probability pertains to data taken from a number of trials. It is a probability calculated from experience, not from theory. If a sample of [latex]x[/latex] trials is observed that results in an event, [latex]e[/latex], occurring [latex]n[/latex] times, the probability of event [latex]e[/latex] is calculated by the ratio of [latex]n[/latex] to [latex]x[/latex].

[latex]\displaystyle \text{experimental probability of event} = \frac{\text{occurrences of event}}{\text{total number of trials}}[/latex]

Experimental probability contrasts theoretical probability, which is what we would expect to happen. For example, if we flip a coin [latex]10[/latex] times, we might expect it to land on heads [latex]5[/latex] times, or half of the time. We know that this is unlikely to happen in practice. If we conduct a greater number of trials, it often happens that the experimental probability becomes closer to the theoretical probability. For this reason, large sample sizes (or a greater number of trials) are generally valued.

In statistical terms, the empirical probability is an estimate of a probability. In simple cases, where the result of a trial only determines whether or not the specified event has occurred, modeling using a binomial distribution might be appropriate. A binomial distribution is the discrete probability distribution of the number of successes in a sequence of [latex]n[/latex] independent yes/no experiments. In such cases, the empirical probability is the most likely estimate.

If a trial yields more information, the empirical probability can be improved on by adopting further assumptions in the form of a statistical model: if such a model is fitted, it can be used to estimate the probability of the specified event. For example, one can easily assign a probability to each possible value in many discrete cases: when throwing a die, each of the six values [latex]1[/latex] to [latex]6[/latex] has the probability of [latex]\frac{1}{6}[/latex].

### Advantages

An advantage of estimating probabilities using empirical probabilities is that this procedure includes few assumptions. For example, consider estimating the probability among a population of men that satisfy two conditions:

- They are over six feet in height.
- They prefer strawberry jam to raspberry jam.

A direct estimate could be found by counting the number of men who satisfy both conditions to give the empirical probability of the combined condition.

An alternative estimate could be found by multiplying the proportion of men who are over six feet in height with the proportion of men who prefer strawberry jam to raspberry jam, but this estimate relies on the assumption that the two conditions are statistically independent.

### Disadvantages

A disadvantage in using empirical probabilities is that without theory to “make sense” of them, it’s easy to draw incorrect conclusions. Rolling a six-sided die one hundred times it’s entirely possible that well over [latex]\frac{1}{6}[/latex] of the rolls will land on [latex]4[/latex]. Intuitively we know that the probability of landing on any number should be equal to the probability of landing on the next. Experiments, especially those with lower sampling sizes, can suggest otherwise.

This shortcoming becomes particularly problematic when estimating probabilities which are either very close to zero, or very close to one. For example, the probability of drawing a number from between [latex]1[/latex] and [latex]1000[/latex] is [latex]\frac{1}{1000}[/latex]. If [latex]1000[/latex] draws are taken and the first number drawn is [latex]5[/latex], there are [latex]999[/latex] draws left to draw a [latex]5[/latex] again and thus have experimental data that shows double the expected likelihood of drawing a [latex]5[/latex].

In these cases, very large sample sizes would be needed in order to estimate such probabilities to a good standard of relative accuracy. Here statistical models can help, depending on the context.

For example, consider estimating the probability that the lowest of the maximum daily temperatures at a site in February in any one year is less than zero degrees Celsius. A record of such temperatures in past years could be used to estimate this probability. A model-based alternative would be to select of family of probability distributions and fit it to the data set containing the values of years past. The fitted distribution would provide an alternative estimate of the desired probability. This alternative method can provide an estimate of the probability even if all values in the record are greater than zero.