What refers to the branch of mathematics that deals with uncertainty and its a measure or estimation of how likely it is that an event will occur?

SECTION 3,1 Basic Concepts of Probability and Counting 143 Using a Tree Diagram In Exercises 63-66, a prohability experiment consists of rolling six-sided die and spinning the spinner shown at the left. The spinner is equally likely to land on each color: Use a tree diagram to find the probability of the event. Then explain whether the event can be considered unusual. 63. Event A: rolling a 5 and the spinner landing on blue 64 Event B: rolling an odd number and the spinner landing on green 65. Event C rolling number less than 6 and the spinner landing on yellow 66. Event D: not rolling a number less than 6 and the spinner landing on yellow 67. Access Code An access code consists of three digits. Each digit can be any number from 0 through 9,and each digit can be repeated (a) What is the probability of randomly selecting the correct access code on the first try? (b) What is the probability of not selecting the correct access code on the first try?

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

  11

  .
• 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

  11

  minus the probability that the event does occur.
• Two events

  AA

  and

  BB

  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

HH,HT,TH,TT.HH, HT, TH, TT.

The sample space is thus

{HH,HT,TH,TT}.\{HH, HT, TH, TT\}.

The event "at least one heads occurs" would be the set

{HH,HT,TH}.\{HH, HT, TH\}.

If the coin were a normal coin, we would assign the probability of

1/41/4

to each outcome.

In probability theory, the probability

PP

of some event

EE

, denoted

P(E)P(E)

, is usually defined in such a way that

PP

satisfies a number of axioms, or rules. The most basic and most important rules are listed below.

Probability Rules

1. 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

   0≤P(A)≤10 \leq P(A) \leq 1

   .  An impossible event, or an event that never occurs, has a probability of

   00

   . An event that always occurs has a probability of

   11

   . An event with a probability of

   0.50.5

   will occur half of the time.
2. The sum of the probabilities of all possibilities must equal

   11

   . Some outcome must occur on every trial, and the sum of all probabilities is 100%, or in this case,

   11

   . This can be written as

   P(S)=1P(S) = 1

   , where

   SS

   represents the entire sample space.
3. 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

   30%30\%

   of the trials, a different event occurs in

   20%20\%

   of the trials, and the two cannot occur together (if they are disjoint ), then the probability that one or the other occurs is

   30%+20%=50%30\% + 20\% = 50\%

   . This is sometimes referred to as the addition rule, and can be simplified with the following:

   P(AorB)=P(A)+P(B)P(A \\ \text{or} \\ B) = P(A)+P(B)

   . The word "or" means the same thing in mathematics as the union, which uses the following symbol:

   ∪\cup

   . Thus when

   AA

    and

   BB

   are disjoint, we have

   P(A∪B)=P(A)+P(B)P(A \cup B) = P(A)+P(B)

   .
4. The probability that an event does not occur is

   11

   minus the probability that the event does occur. If an event occurs in

   60%60\%

   of all trials, it fails to occur in the other

   40%40\%

   , because

   100%−60%=40%100\% - 60\% = 40\%

   . The probability that an event occurs and the probability that it does not occur always add up to

   100%100\%

   , or

   1 1

   . These events are called complementary events, and this rule is sometimes called the complement rule. It can be simplified with

   P(Ac)=1−P(A) P(A^c) = 1-P(A)

   , where

   AcA^c

   is the complement of

   AA

   .
5. Two events

   AA

   and

   BB

   are independent if knowing that one occurs does not change the probability that the other occurs. This is often called the multiplication rule. If

   AA

   and

   BB

   are independent, then

   P(AandB)=P(A)P(B)P(A \\ \text{and} \\ B) = P(A)P(B)

   . The word "and" in mathematics means the same thing in mathematics as the intersection, which uses the following symbol:

   ∩\cap

   . Therefore when A and B are independent, we have 

   P(A∩B)=P(A)P(B).P(A \cap B) = P(A)P(B).

Extension of the Example

Elaborating on our example above of flipping two coins, assign the probability

1/41/4

to each of the

44

outcomes. We consider each of the five rules above in the context of this example.

1. Note that each probability is

1/41/4

, which is between

00

and

11

.

2. Note that the sum of all the probabilities is

1 1

, since

14+14+14+14=1\frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=1

.

3. Suppose

AA

is the event exactly one head occurs, and

BB

is the event exactly two tails occur. Then

A={HT,TH}A=\{HT,TH\}

and

B={ TT}B=\{TT\}

are disjoint. Also,

P(A∪B)=34=24+14= P(A)+P(B).P(A \cup B) = \frac{3}{4} = \frac{2}{4}+\frac{1}{4}=P(A) + P(B).

4. The probability that no heads occurs is

1/41/4

, which is equal to

1−3/41-3/4

. So if

A={HT,TH,HH}A=\{HT, TH, HH\}

is the event that a head occurs, we have

P(A c)=14=1−34=1−P(A).P(A^c)=\frac{1}{4}=1 - \frac{3}{4}=1-P(A).

5. If

AA

is the event that the first flip is a heads and

BB

is the event that the second flip is a heads, then

AA

and

BB

are independent. We have

A= {HT,HH}A=\{HT,HH\}

and

B={TH,HH}B=\{TH,HH\}

and

A∩B={HH}.A \cap B = \{HH\}.

Note that

P(A∩B)=14=12⋅12=P(A)P(B).P(A \cap B) = \frac{1}{4} =\frac{1}{2}\cdot \frac{1}{2} = P(A)P(B).

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

  ∪\cup

  .
• The general probability addition rule for the union of two events states that

  P(A∪B)=P(A)+P(B)−P(A∩ B)P(A\cup B) = P(A)+P(B)-P(A \cap B)

  , where

  A∩BA \cap B

  is the intersection of the two sets.
• The addition rule can be shortened if the sets are disjoint:

  P(A∪B)=P(A)+P(B)P(A \cup B) = P(A) + P(B)

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

  P(A∪B∪C)=P(A)+P(B)+P(C) P(A \cup B \cup C) = P(A) + P(B) + P(C)

  .
• The intersection of two or more sets is the set of elements that are common to every set. The symbol

  ∩\cap

  is used to denote the intersection.
• When events are independent, we can use the multiplication rule for independent events, which states that

  P(A∩B)=P(A)P(B)P(A \cap B) = P(A)P(B)

  .

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

∪\cup

, and is associated with the word "or", because

A∪BA \cup B

is the set of all elements that are in

A A

or

BB

(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

AA

and

BB

can be shown as two completely shaded interlocking circles.

Union of Two Sets: The shaded Venn Diagram shows the union of set

AA

(the circle on left) with set

BB

(the circle on the right). It can be written shorthand as

A∪BA \cup B

.

In symbols, since the union of

AA

and

BB

contains all the points that are in

AA

or

BB

or both, the definition of the union is:

A∪B={x:x∈Aorx∈B}\displaystyle A \cup B = \{x: x\in A \\ \text{or} \\ x\in B \}

For example, if

A={1,3,5,7}A = \{1, 3, 5, 7\}

and

B={1,2,4,6}B = \{1, 2, 4, 6\}

, then

A∪B= {1,2,3,4,5,6,7}A \cup B = \{1, 2, 3, 4, 5, 6, 7\}

. Notice that the element

11

is not listed twice in the union, even though it appears in both sets

AA

and

BB

. This leads us to the general addition rule for the union of two events:

P(A∪B)= P(A)+P(B)−P(A∩B)\displaystyle P(A \cup B) = P(A) + P(B) - P(A \cap B)

Where

P(A∩B)P(A\cap B)

is the intersection of the two sets. We must subtract this out to avoid double counting of the inclusion of an element.

If sets

A A

and

BB

are disjoint, however, the event

A∩BA \cap B

has no outcomes in it, and is an empty set denoted as

∅\emptyset

, which has a probability of zero. So, the above rule can be shortened for disjoint sets only:

P(A∪B)=P(A)+P(B)\displaystyle P(A \cup B) = P(A)+P(B)

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

P(A∪B∪C)=P(A)+P(B)+P(C)\displaystyle P(A \cup B \cup C) = P(A) + P(B)+ P(C)

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

∩\cap

, and is associated with the word "and", because

A∩BA \cap B

is the set of elements that are in

AA

and

BB

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

AA

and

BB

can be shown at the shaded region in the middle of two interlocking circles.

AA

is the circle on the left, set

BB

is the circle on the right, and the intersection of

AA

and

BB

, or

A∩BA \cap B

, is the shaded portion in the middle.

In mathematical notation, the intersection of

AA

and

BB

is written as

A∩B={x:x∈Aandx∈B}A \cap B = \{x: x \in A \\ \text{and} \\ x \in B\}

. For example, if

A={1,3,5,7}A = \{1, 3, 5, 7\}

and

B={1,2,4,6}B = \{1, 2, 4, 6\}

, then

A∩B={1} A \cap B = \{1\}

because

11

is the only element that appears in both sets

AA

and

BB

.

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:

P(A∩B)=P(A)P(B)\displaystyle P(A \cap B)= P(A)P(B)

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

BB

is the event that the second toss is a heads, then

P(A∩B)=12⋅12=14P(A \cap B) = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}

.

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

  P(B∣A)P(B \vert A)

  of an event

  BB

  , given an event

  AA

  , is defined by:

  P(B∣A)=P(A∩B)P( A)P(B|A)=\frac{P(A\cap B)}{P(A)}

  , when

  P(A)>0P(A) > 0

  .
• If the knowledge that event

  AA

  occurs does not change the probability that event

  BB

  occurs, then

  AA

  and

  BB

  are independent events, and thus,

  P(B∣A)=P(B)P(B|A) = P(B)

  .
• Mathematically, Bayes' theorem gives the relationship between the probabilities of

  AA

  and

  BB

  ,

  P(A)P(A)

  and

  P(B)P(B)

  , and the conditional probabilities of

  AA

  given

  BB

  and

  BB

  given

  AA

  ,

  P(A∣B)P(A|B)

  and

  P(B∣A)P(B|A)

  . In its most common form, it is:

  P(A∣B)=P (B∣A)P(A)P(B)P(A|B)=\frac{P(B|A)P(A)}{P(B)}

  .

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 BB  Given That AA 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

2 2

is

1/61/6

without any other information, but if someone looks at the die and tells you that is is an even number, the probability is now

1/31/3

that it is a

22

. The notation

P(B∣A)P(B|A)

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

P(B∣A)P(B|A)

is read as "the probability of

BB

given that

AA

has occurred".

The conditional probability

P(B∣A )\displaystyle P(B|A)

of an event

BB

, given an event

A A

, is defined by:

P(B∣A)=P(A∩B) P(A)\displaystyle P(B|A)=\frac{P(A\cap B)}{P(A)}

When

P(A)>0P(A) > 0

. Be sure to remember the distinct roles of

BB

and

AA

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:

S={HHH,HHT,HTH,THH,TTH,THT,HTT,TTT}S=\{HHH, HHT, HTH, THH, TTH, THT, HTT, TTT\}

Each individual outcome has probability

1/81/8

. Suppose that

BB

is the event that at least one heads occurs and

AA

is the event that all

33

coins are the same. Then the probability of

BB

given

AA

is

1/21/2

, since

A∩B={HHH}A \cap B=\{HHH\}

which has probability

1/81/8

and

A={HHH,TTT}A=\{HHH,TTT\}

which has probability

2/82/8

, and

1/82/8=12.\frac{1/8}{2/8}=\frac{1}{2}.

Independence

The conditional probability

P(B∣A)P(B|A)

is not always equal to the unconditional probability

P(B)P(B)

. The reason behind this is that the occurrence of event

AA

may provide extra information that can change the probability that event

BB

occurs. If the knowledge that event

AA

occurs does not change the probability that event

BB

occurs, then

AA

and

BB

are independent events, and thus,

P(B∣A)=P(B)P(B|A) = P(B)

.

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

AA

and

BB

,

P(A)P(A)

and

P(B)P(B)

, and the conditional probabilities of

AA

given

BB

and

BB

given

AA

. In its most common form, it is:

P(A∣B)=P(B∣A)P(A)P(B)\displaystyle P(A|B)=\frac{P(B|A)P(A)}{P(B)}

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

P(A∣B)P(B∣A)=P(A) P(B)\displaystyle \frac{P(A|B)}{P(B|A)}=\frac{P(A)}{P(B)}

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

50%50\%

. 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

WW

represent the event that the conversation was held with a woman, and

LL

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

WW

occurs is

P (W)=0.5P(W) = 0.5

.

Suppose it is also known that

75%75\%

of women in this city have long hair, which we denote as

P(L∣W)=0.75P(L|W) = 0.75

. Likewise, suppose it is known that

25 %25\%

of men in this city have long hair, or

P(L∣M)=0.25P(L|M) = 0.25

, where

MM

is the complementary event of

WW

, 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,

P(W∣L)P(W|L)

. Using the formula for Bayes's theorem, we have:

P(W∣L) =P(L∣W)P(W)P(L)=P(L∣W)P(W)P(L∣W)P(W )+P(L∣M)P(M)=0.75⋅ 0.50.75⋅0.5+0.25⋅0.5=0.75\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}

Complementary Events

The complement of

AA

is the event in which

AA

does not occur.

Learning Objectives

Explain an example of a complementary event

Key Takeaways

Key Points

• The complement of an event

  A A

  is usually denoted as

  A′A'

  ,

  Ac A^c

  or

  Aˉ\bar{A}

  .
• 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

AA

is the event

[not A][\text{not}\ A]

, i.e. the event in which

AA

does not occur. The event

AA

and its complement

[not A][\text{not}\ A]

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

BB

such that

AA

and

BB

are both mutually exclusive and exhaustive; that event is the complement of

AA

. The complement of an event

AA

is usually denoted as

A′A'

,

Ac A^c

or

Aˉ\bar{A}

.

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

P(heads)+P(tails)=1P(\text{heads}) + P(\text{tails}) = 1

.

Another simple example of complementary events is picking a ball out of a bag. Let's say there are three plastic balls in a bag. One is blue and two are red. Assuming that each ball has an equal chance of being pulled out of the bag, we know that

P(blue)=13P(\text{blue}) = \frac{1}{3}

and

P(red)=23P(\text{red}) = \frac{2}{3}

. Since we can only either chose blue or red (exhaustive) and we cannot choose both at the same time (mutually exclusive), choosing blue and choosing red are complementary events, and

P(blue)+P(red)=1P(\text{blue}) + P(\text{red}) = 1

.

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: 

  P(A∪B )=P(A)+P(B)−P(A∩B).P(A\cup B)=P(A)+P(B)-P(A\cap B).

• The last term has been accounted for twice, once in

  P(A)P(A)

  and once in

  P(B)P(B)

  , so it must be subtracted once so that it is not double-counted.
• If

  AA

  and

  BB

  are disjoint, then

  P(A∩B)=0P(A\cap B)=0

  , so the formula becomes 

  P(A∪B)=P(A)+P(B).P(A \cup B)=P(A) + P(B).

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

AA

or

B B

will occur is the sum of the probabilities that

AA

will happen and that

BB

will happen, minus the probability that both

AA

and

BB

will happen. The addition rule is summarized by the formula:

P(A∪B)=P(A)+P(B) −P(A∩B)\displaystyle P(A \cup B) = P(A)+P(B)-P(A \cap B)

Consider the following example. When drawing one card out of a deck of

5252

playing cards, what is the probability of getting heart or a face card (king, queen, or jack)? Let

HH

denote drawing a heart and

FF

denote drawing a face card. Since there are

1313

hearts and a total of

1212

face cards (

33

of each suit: spades, hearts, diamonds and clubs), but only

33

face cards of hearts, we obtain:

P(H)=1352\displaystyle P(H) = \frac{13}{52}

P(F)=1252\displaystyle P(F) = \frac{12}{52}

P(F∩H)=352\displaystyle P(F \cap H) = \frac{3}{52}

Using the addition rule, we get:

P(H∪F)=P(H)+P( F)−P(H∩F)=1352+1252−352 \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}

The reason for subtracting the last term is that otherwise we would be counting the middle section twice (since

HH

and

FF

overlap).

Addition Rule for Disjoint Events

Suppose

A A

and

BB

are disjoint, their intersection is empty. Then the probability of their intersection is zero. In symbols:

P(A∩B)=0P(A \cap B) = 0

. The addition law then simplifies to:

P(A∪B)=P( A)+P(B)whenA∩B=∅P(A \cup B) = P(A) + P(B) \qquad \text{when} \qquad A \cap B = \emptyset

The symbol

∅\emptyset

represents the empty set, which indicates that in this case

AA

and

BB

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

AA

represent the event that a king is drawn and

BB

represent the event that a queen is drawn. These two events are disjoint, since there are no kings that are also queens. Thus:

P(A∪B)=P( A)+P(B)=452+452 =852=213 \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}

The Multiplication Rule

The multiplication rule states that the probability that

AA

and

BB

both occur is equal to the probability that

BB

occurs times the conditional probability that

AA

occurs given that

BB

occurs.

Learning Objectives

Apply the multiplication rule to calculate the probability of both

AA

and

BB

occurring

Key Takeaways

Key Points

• The multiplication rule can be written as:

  P(A∩B)=P(B)⋅P(A∣B)P(A \cap B) = P(B) \cdot P(A|B)

  .
• 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

AA

and

BB

occur is equal to the probability that

AA

occurs times the conditional probability that

BB

occurs, given that we know

AA

has already occurred. This rule can be written:

P(A∩B)=P(B)⋅P(A∣B)\displaystyle P(A \cap B) = P(B) \cdot P(A|B)

Switching the role of

AA

and

BB

, we can also write the rule as:

P( A∩B)=P(A)⋅P(B∣A)\displaystyle P(A\cap B) = P(A) \cdot P(B|A)

We obtain the general multiplication rule by multiplying both sides of the definition of conditional probability by the denominator. That is, in the equation

P(A∣B)=P(A∩B)P (B)\displaystyle P(A|B)=\frac{P(A\cap B)}{P(B)}

, if we multiply both sides by

P(B)P(B)

, we obtain the Multiplication Rule.

The rule is useful when we know both

P(B)P(B)

and

P(A∣B)P(A|B)

, or both

P(A)P(A)

and

P(B∣A).P(B|A).

Example

Suppose that we draw two cards out of a deck of cards and let

A A

be the event the the first card is an ace, and

BB

be the event that the second card is an ace, then:

P(A)=452\displaystyle P(A)=\frac { 4 }{ 52 }

And:

P(B∣A)=351\displaystyle P\left( { B }|{ A } \right) =\frac { 3 }{ 51 }

The denominator in the second equation is

51 51

since we know a card has already been drawn. Therefore, there are

5151

left in total. We also know the first card was an ace, therefore:

P(A∩B)=P( A)⋅P(B∣A)=452⋅351 =0.0045\displaystyle \begin{align} P(A \cap B) &= P(A) \cdot P(B|A)\\ &= \frac { 4 }{ 52 } \cdot \frac { 3 }{ 51 } \\ &=0.0045 \end{align}

Independent Event

Note that when

AA

and

BB

are independent, we have that

P(B∣A)=P(B)P(B|A)= P(B)

, so the formula becomes

P(A∩B)=P(A)P(B)P(A \cap B)=P(A)P(B)

, 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

22

on the die and a tails on the coin is

16⋅12=112 \frac{1}{6}\cdot \frac{1}{2} = \frac{1}{12}

, 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:

  P(A∣B)=P(A)P(A|B) = P(A)

  ,

  P(B∣A)=P(B)P(B|A) = P(B)

  , and

  P(A andB)=P(A)⋅P(B)P(A \\ \text{and} \\ B) = P(A) \cdot P(B)

  .
• If any one of these conditions is true, then all of them are true.
• If events

  AA

  and

  BB

  are independent, then the chance of

  AA

  occurring does not affect the chance of

  BB

  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

AA

and

BB

are independent, then the chance of

AA

occurring does not affect the chance of

BB

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:

1. P(A∣B)=P(A)\displaystyle P(A|B) = P(A)

2. P(B∣A)=P(B)\displaystyle P(B|A) = P(B)

3. P(AandB)=P(A )⋅P(B)\displaystyle P(A \\ \text{and} \\ B) = P(A)\cdot P(B)

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

AA

and

BB

occurring is the same as the probability of

AA

occurring, multiplied by the probability of

BB

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

12{\frac{1}{2}}

chance that the coin will land on heads. Thus, the probability that the coin will land on heads during the third round is

12{\frac{1}{2}}

.

Example

When flipping a coin, what is the probability of getting tails

55

times in a row?

Recall that each coin flip is independent, and the probability of getting tails is

12{\frac{1}{2}}

for any flip. Also recall that the following statement holds true for any two independent events A and B:

P(AandB)=P(A)⋅P(B)\displaystyle P(A \\ \text{and} \\ B) = P(A)\cdot P(B)

Finally, the concept of independence extends to collections of more than

22

events.

Therefore, the probability of getting tails

44

times in a row is:

12⋅12⋅12⋅1 2=116\displaystyle {\frac{1}{2}} \cdot {\frac{1}{2}} \cdot {\frac{1}{2}} \cdot {\frac{1}{2}} = {\frac{1}{16}}

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

  nn

  independent yes/no experiments, each of which yields success with probability

  pp

  .
• 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

xx

trials is observed that results in an event,

ee

, occurring

nn

times, the probability of event

ee

is calculated by the ratio of

nn

to

xx

.

experimental probability of event=occurrences of eventtotal number of trials \displaystyle \text{experimental probability of event} = \frac{\text{occurrences of event}}{\text{total number of trials}}

Experimental probability contrasts theoretical probability, which is what we would expect to happen. For example, if we flip a coin

10 10

times, we might expect it to land on heads

55

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

nn

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

11

to

66

has the probability of

16\frac{1}{6}

.

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:

1. They are over six feet in height.
2. 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

16\frac{1}{6}

of the rolls will land on

44

. 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

11

and

10001000

is

11000\frac{1}{1000}

. If

10001000

draws are taken and the first number drawn is

55

, there are

999999

draws left to draw a

55

again and thus have experimental data that shows double the expected likelihood of drawing a

55

.

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.

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What is a branch of mathematics which deals with uncertainty and a measure or estimation of how likely it is an event will occur *?

What is the branch of mathematics that deals with uncertainty?

What do you call a measure or estimation of how likely the event will occur?

Is a measure or estimation of how likely it is that an event will occur it is a number between and including 0 and 1 which can be expressed as a?