Chapter 3 Probability
Chapter 3 Probability
The Concept of ProbabilityThe Concept of Probability
Sample Spaces and EventsSample Spaces and Events
Some Elementary Probability Some Elementary Probability RulesRules
Conditional Probability and Conditional Probability and Independence Independence
Chapter 3 Probability
Section 3.1 The Concept of Probability
An An experimentexperiment is any process of observation with an uncertain is any process of observation with an uncertain outcome.outcome.--- On any single trial of the experiment, one and only one of the --- On any single trial of the experiment, one and only one of the possible outcomes will occur.possible outcomes will occur.
The possible outcomes for an experiment are called the The possible outcomes for an experiment are called the experimental outcomesexperimental outcomes
ProbabilityProbability is a measure of the chance that an experimental is a measure of the chance that an experimental outcome will occur when an experiment is carried outoutcome will occur when an experiment is carried out
Chapter 3 Probability
Roll a die. The experimental outcomes are 1, 2, 3, 4, 5, Roll a die. The experimental outcomes are 1, 2, 3, 4, 5, and 6.and 6.
Example 3.1Example 3.1
An OutcomeOutcome is the particular result of an experiment.
An EventAn Event is the is the collection of one or collection of one or more outcomes of more outcomes of an experiment.an experiment.
Possible outcomesPossible outcomes: The : The numbers 1, 2, 3, 4, 5, 6 numbers 1, 2, 3, 4, 5, 6
One possible eventOne possible event: The : The occurrence of an even occurrence of an even number. That is, we collect number. That is, we collect the outcomes 2, 4, and 6.the outcomes 2, 4, and 6.
Chapter 3 Probability
Regardless of the method used, probabilities must be Regardless of the method used, probabilities must be assigned to the experimental outcomes so that two assigned to the experimental outcomes so that two conditions are met: conditions are met:
ConditionsConditions
1.1. 0 0 P(E) P(E) 1 1
such that:such that:If If EE can never occur, then can never occur, then P(E) = 0P(E) = 0If If EE is certain to occur, then is certain to occur, then P(E) = 1P(E) = 1
2.2. The probabilities of all the experimental outcomes must The probabilities of all the experimental outcomes must sum to sum to 11
Chapter 3 Probability
Sample space (Sample space (SS): ): The sample space is defined as the set of all possible The sample space is defined as the set of all possible outcomes of an experiment.outcomes of an experiment.
e.g. All 6 faces of a die:e.g. All 6 faces of a die:
e.g. All 52 cards of a bridge deck:e.g. All 52 cards of a bridge deck:
Section 3.2 Sample Spaces and Events
Chapter 3 Probability
Example 3.2Example 3.2Genders of Two Children
Let: B be the outcome that child is boy.Let: B be the outcome that child is boy. G be the outcome that child is girl.G be the outcome that child is girl.
Sample space Sample space S:S:SS = {BB, BG, GB, GG} = {BB, BG, GB, GG}
If B and G are equally If B and G are equally likely , thenlikely , then
P(B) = P(G) = ½ P(B) = P(G) = ½
andand
P(BB) = P(BG) = P(GB) = P(BB) = P(BG) = P(GB) = P(GG) = ¼P(GG) = ¼
Chapter 3 Probability
AnAn eventevent is a set of sample space outcomes is a set of sample space outcomes..
Recall example 3.2: Genders of Two ChildrenRecall example 3.2: Genders of Two Children
EventsEvents
P(one boy and one girl) = P(BG) + P(GB) = ¼ + ¼ = ½.
P(at least one girl) =P(BG) + P(GB) + P(GG) = ¼ + ¼ + ¼ = ¾.
Note:Note: Experimental Outcomes: BB, BG, GB, GGExperimental Outcomes: BB, BG, GB, GG
All outcomes equally likely: P(BB) = … = P(GG) = ¼All outcomes equally likely: P(BB) = … = P(GG) = ¼
Chapter 3 Probability
Example 3.3Example 3.3Answering Three True-False QuestionsAnswering Three True-False Questions
A student takes a quiz that consists of three true-false questions. Let C and I denote answering a question correctly and incorrectly, respectively.
The graph on the next slide shows the sample space outcomes for the experiment. The sample space consists of 8 outcomes: CCC CCI CIC CII ICC ICI IIC III
Suppose the student is totally unprepared for the quiz and has to blindly guess the answers. That is, the student has a 50-50 chance of correctly answering each question.
Chapter 3 Probability
So, each of the 8 outcomes is equally likely to occur. P(CCC)=P(CCI)= ... = P(III)=1/8.
Chapter 3 Probability
Probabilities: Equally Likely Outcomes Probabilities: Equally Likely Outcomes
If the sample space outcomesIf the sample space outcomes (or experimental (or experimental outcomes) outcomes) are all equally likelyare all equally likely, then the , then the probability that an event will occur is equal to probability that an event will occur is equal to the ratio:the ratio:
outcomes ofnumber totalThe
event the tocorrespond that outcomes ofnumber the
Chapter 3 Probability
The probability of an eventThe probability of an event is also equal the sum of the is also equal the sum of the probabilities of the sample space outcomes that correspond to probabilities of the sample space outcomes that correspond to the event.the event.
The probability that the student will get exactly two questions The probability that the student will get exactly two questions correct iscorrect is
P(CCI) + P(CIC) + P(ICC)P(CCI) + P(CIC) + P(ICC) = 1/8 + 1/8 + 1/8 = 3/8. = 1/8 + 1/8 + 1/8 = 3/8.
The probability that the student will get at least two questions The probability that the student will get at least two questions correct iscorrect is
P(CCC)P(CCC) + + P(CCI) + P(CIC) + P(ICC)P(CCI) + P(CIC) + P(ICC) = 1/8 + 1/8 + 1/8 + 1/8 = 1/8 + 1/8 + 1/8 + 1/8 = 1/2.= 1/2.
Example 3.4Example 3.4
Basic Computation of ProbabilitiesBasic Computation of Probabilities
Chapter 3 Probability
Relative Frequency Method
LetLet E E be an outcome of an experiment. be an outcome of an experiment.If the experiment is performed many times, If the experiment is performed many times, P(E)P(E) is the is the relative frequency of relative frequency of EE..P(E)P(E) is the percentage of times is the percentage of times EE occurs in many repetitions occurs in many repetitions of the experimentof the experiment..Use sampled or historical data to calculate probabilities.Use sampled or historical data to calculate probabilities.
Suppose that of Suppose that of 10001000 randomly selected consumers, randomly selected consumers, 140140 preferred brand preferred brand XX..The probability of randomly picking a person who prefers The probability of randomly picking a person who prefers brand brand XX is is 140/1000 = 0.14 or 14%.140/1000 = 0.14 or 14%.
Example 3.5Example 3.5
Chapter 3 Probability
Section 3.3 Some Elementary Probability Rules
The complement of an event A is the set of all sample space outcomes not in A. Further, P(A).-1=)AP(
A
These figures are “These figures are “Venn diagramsVenn diagrams”.”.
Chapter 3 Probability
Union of A and B,
Is an event consisting of the outcomes that belong to either A or B (or both).
Intersection of A and B,
Is an event consisting of the outcomes that belong to both A and B.
BA
BA
Chapter 3 Probability
The probability that A or B (the union of A and B) will occur is
where is the “joint” probability of A and B, i.e., both occurring.
The Addition Rule
P(A B) = P(A) + P(B) - P(A B)
B)P(A
A and B are mutually exclusive if they have no sample space outcomes in common, or equivalently, if 0.=B)P(A
If A and B are mutually exclusive, thenIf A and B are mutually exclusive, then
.B)B)=P(A)+P(P(A
Chapter 3 Probability
Newspaper Subscribers #1Newspaper Subscribers #1 Example 3.6Example 3.6
Define events:Define events: A = event that a randomly selected household subscribes A = event that a randomly selected household subscribes
to the to the Atlantic Journal.Atlantic Journal. B = event that a randomly selected household subscribes B = event that a randomly selected household subscribes
to the to the Beacon News.Beacon News. Given:Given:
total number in city, N = 1,000,000total number in city, N = 1,000,000 number subscribing to A, N(A) = 650,000number subscribing to A, N(A) = 650,000 number subscribing to B, N(B) = 500,000number subscribing to B, N(B) = 500,000 number subscribing to both, N(Anumber subscribing to both, N(A∩∩B) = 250,000B) = 250,000
Chapter 3 Probability
Newspaper Subscribers #2
Use the relative frequency method to assign probabilities
25.0000,000,1
000,250
50.0000,000,1
000,500
65.0000,000,1
000,650
BAP
BP
AP
Chapter 3 Probability
Table3.1 A Contingency Table Subscription Data for the Atlantic Journal and the Beacon News
Events
Subscribes to Beacon News, B
Does Not Subscribe to Beacon News, Total
Subscribes to Atlantic Journal, A 250,000 400,000 650,000
Does not Subscribes to Atlantic Journal, 250,000 100,000 350,000
Total 500,000 500,000 1,000,000
Chapter 3 Probability
Newspaper Subscribers #3
Refer to the contingency table in Table 3.1 for all probabilities
For example, the chance that a household does For example, the chance that a household does not subscribe to either newspapernot subscribe to either newspaper
Calculate , so from middle row Calculate , so from middle row and middle column of and middle column of Table 3.1Table 3.1,,
BAP
.10.0000,000,1
000,100BAP
Chapter 3 Probability
Newspaper Subscribers #4
The chance that a household subscribes to either newspaper:The chance that a household subscribes to either newspaper:
Note that if the joint probability was not subtracted, then Note that if the joint probability was not subtracted, then
we would have gotten 1.15, greater than 1, which is we would have gotten 1.15, greater than 1, which is
absurd.absurd.
Note: The subtraction avoids double counting the joint Note: The subtraction avoids double counting the joint probability.probability.
.90.0
25.050.065.0
)()(
BAPBB)=P(A)+PP(A
Chapter 3 Probability
A Mutually Exclusive CaseIf A and B are mutually exclusive, thenIf A and B are mutually exclusive, then
.B)B)=P(A)+P(P(AExample 3.7: Consider randomly selecting a card from a standarddeck of 52 playing cards, and define the eventsJ= the randomly selected card is jack.Q=the randomly selected card is queen.
P(J Q)=?∪
Since there are four jacks, four queens, we have P(J)=4/52 andP(Q)=4/52. Furthermore, since there is no card that is both a jackAnd a queen, the events J and Q are mutually exclusive and thusP(J∩Q)=0. So we have
P(J Q)=P(J)+P(Q)=∪13
2
52
8
Chapter 3 Probability
Interpretation: Restrict the sample space to just event B. The conditional probability is the chance of event A occurring in this new sample space.
Section 3.4 Conditional Probability and Independence
The probability of an event A, given that the event B The probability of an event A, given that the event B has occurred, is called the “has occurred, is called the “conditional probabilityconditional probability of A given Bof A given B” and is denoted as” and is denoted as
Further,Further,
Assume that P(B) is greater than 0Assume that P(B) is greater than 0.
P(A|B) =P(A B)
P(B)
P(A|B)
Chapter 3 Probability
Similarly, if A occurred, then what is the chance of B occurring?
To answer this question, we need to introduce the probability of event B, given that the event A has occurred, i.e., the conditional probability of B given A, denoted by P(B|A).
P(A)
B)P(A=A)|P(B
Assume that P(A) is greater than 0.Assume that P(A) is greater than 0.
Chapter 3 Probability
Newspaper SubscribersNewspaper Subscribers
Given that the households that subscribe to the Given that the households that subscribe to the Atlantic JournalAtlantic Journal, what is the chance that they also , what is the chance that they also subscribe to the subscribe to the Beacon NewsBeacon News??
Calculate P(B|A), whereCalculate P(B|A), where
.3846065.0
25.0
|
.
AP
BAPABP
Chapter 3 Probability
Example 7
Major Male Female Total
Accounting 170 110 280
Finance 120 100 220
Marketing 160 70 230
Management 150 120 270
Total 600 400 1000
Example: The Dean of the School of Business at Owens University collected the following information about undergraduate students in her college:
Chapter 3 Probability
P(A|F)=
If a student is selected at random, what is the probability that the student is a female (F) accounting major (A)?
P(A and F) =.
Given that the student is a female, what is the probability that she is an accounting major?
1000
110
275.0
1000400
1000110
)(
) and (
FP
FAP
Chapter 3 Probability
Independence of Events
Two events A and B are said to be independent if and only if P(A|B) = P(A) or, equivalently, P(B|A) = P(B).
That is, if the chance of event A occurring is not That is, if the chance of event A occurring is not influenced by whether the event B occurs and influenced by whether the event B occurs and vice versa; or if the occurrences of the events A vice versa; or if the occurrences of the events A and B have nothing to do with each other, then A and B have nothing to do with each other, then A and B are independent.and B are independent. In fact if one of the above two equations holds, In fact if one of the above two equations holds, so does the other, why?so does the other, why?
Chapter 3 Probability
Newspaper Subscribers
Given that the households that subscribe to the Atlantic Journal subscribers, what is the chance that they also subscribe to the Beacon News? If independent, the P(B|A) = P(B).
Is P(B|A) = P(B)? Know that P(B) = 0.50. Just calculated that P(B|A) = 0.3846. 0.50 ≠ 0.3846, so P(B|A) ≠ P(B).
B is not independent of A. A and B are said to be dependent.
Chapter 3 Probability
The Multiplication RuleThe The joint probabilityjoint probability that that AA and and BB (the intersection (the intersection of of AA and and BB) will occur is) will occur is
.P(A|B)P(B)=
P(B|A)P(A)=B)P(A
If A and B are independent, then the probability that A and B (the intersection of A and B) will occur is
.P(A) P(B)P(B) P(A)=B)P(A
Chapter 3 Probability
A Question
Suppose in the following contingency table, where the numbers represent probabilities, some data are lost.
1.Can you recover the missing data?
2.Are events R and C independent?
C C Total
R .4 .6
R .3 Total .5 1.00
Chapter 3 Probability
C C Total R .4 .2 .6 R .1 .3 .4
Total .5 .5 1.00
)CP(R )P(R
)CP()CRP(
dependent. are and events the
)()()(
3.05.06.0)()(
4.0)( As
CR
CPRPCRP
CPRP
CRP
Contingency Tables
Chapter 3 Probability
P(A or B) = P(A) + P(B) - P(A and B)
General Addition Rule:
Special Rule of Addition: If A and B are mutually exclusive
P(A or B) = P(A) + P(B)
Complement RuleComplement RuleP(A) + P(~A) = 1 or P(A) = 1 - P(~A).
General Rule of MultiplicationGeneral Rule of Multiplication
P(A and B) = P(A)P(B|A) or P(A and B) = P(B)P(A|B)
Special Rule of Multiplication: If A and B are independentSpecial Rule of Multiplication: If A and B are independent
P(A and B) = P(A)P(B)