IS 310 Business Statistics CSU Long Beach

1 1 Slide

Slide

IS 310 – Business StatisticsIS 310 – Business Statistics

IS 310

Business Statistic

sCSU

Long Beach

2 2 Slide

Slide


Chapter 4 Introduction to ProbabilityChapter 4 Introduction to Probability

Few events are certain to happen.Few events are certain to happen.

Probability offers a numeric measure of Probability offers a numeric measure of the likelihood of an event occurringthe likelihood of an event occurring

3 3 Slide

Slide


Probability as a Numerical MeasureProbability as a Numerical Measureof the Likelihood of Occurrenceof the Likelihood of Occurrence

00 11..55

Increasing Likelihood of OccurrenceIncreasing Likelihood of Occurrence

ProbabilitProbability:y:

The eventThe eventis veryis veryunlikelyunlikelyto occur.to occur.

The eventThe eventis veryis veryunlikelyunlikelyto occur.to occur.

The occurrenceThe occurrenceof the event isof the event is

just as likely asjust as likely asit is unlikely.it is unlikely.

The occurrenceThe occurrenceof the event isof the event is

just as likely asjust as likely asit is unlikely.it is unlikely.

The eventThe eventis almostis almostcertaincertain

to occur.to occur.

The eventThe eventis almostis almostcertaincertain

to occur.to occur.

4 4 Slide

Slide


Probability ConceptsProbability Concepts

To understand probability, one needs to To understand probability, one needs to understand the following concepts:understand the following concepts:

ExperimentExperiment Experimental outcomeExperimental outcome Sample spaceSample space Sample pointSample point

5 5 Slide

Slide


An Experiment and Its Sample SpaceAn Experiment and Its Sample Space

An An experimentexperiment is any process that generatesis any process that generates well-defined outcomes.well-defined outcomes. An An experimentexperiment is any process that generatesis any process that generates well-defined outcomes.well-defined outcomes.

The The sample spacesample space for an experiment is the set of for an experiment is the set of all experimental outcomes.all experimental outcomes. The The sample spacesample space for an experiment is the set of for an experiment is the set of all experimental outcomes.all experimental outcomes.

An experimental outcome is also called a An experimental outcome is also called a samplesample pointpoint.. An experimental outcome is also called a An experimental outcome is also called a samplesample pointpoint..

6 6 Slide

Slide


Experiment and OutcomesExperiment and Outcomes

Let’s look at the following experiments and their Let’s look at the following experiments and their outcomes.outcomes.

ExperimentExperiment OutcomeOutcome

Tossing a coin Head, tailTossing a coin Head, tail Selecting a part Defective, non-defectiveSelecting a part Defective, non-defective for inspectionfor inspection Playing ball game Win, lose, tiePlaying ball game Win, lose, tie Conducting a sales Sale, no saleConducting a sales Sale, no sale callcall

7 7 Slide

Slide


Counting OutcomesCounting Outcomes

In order to calculate probability, we must be In order to calculate probability, we must be able to count the total number of outcomes in able to count the total number of outcomes in an experiment. In simple experiments, it is an experiment. In simple experiments, it is easy to know the number of outcomes. In easy to know the number of outcomes. In complex experiments, we need some rules for complex experiments, we need some rules for counting the number of outcomes.counting the number of outcomes.

There are three rules: There are three rules: Rule for multiple-step experimentsRule for multiple-step experiments CombinationCombination Permutation Permutation

8 8 Slide

Slide


A Counting Rule for A Counting Rule for Multiple-Step ExperimentsMultiple-Step Experiments

If an experiment consists of a sequence of If an experiment consists of a sequence of kk steps steps in which there are in which there are nn11 possible results for the first step, possible results for the first step,

nn22 possible results for the second step, and so on, possible results for the second step, and so on,

then the total number of experimental outcomes isthen the total number of experimental outcomes is given by (given by (nn11)()(nn22) . . . () . . . (nnkk).).

A helpful graphical representation of a multiple-stepA helpful graphical representation of a multiple-step

experiment is a experiment is a tree diagramtree diagram..

9 9 Slide

Slide


Counting Rule for Multi-step ExperimentsCounting Rule for Multi-step Experiments

Let’s take the experiment of tossing two coins. Let’s take the experiment of tossing two coins. How many total outcomes are possible in this How many total outcomes are possible in this experiment?experiment?

The first coin has two outcomes (H, T)The first coin has two outcomes (H, T) The second coin has also two outcomes (H, T)The second coin has also two outcomes (H, T)

Total outcomes is: (2) (2) = 4Total outcomes is: (2) (2) = 4

Often, a tree diagram helps. Look at Figure Often, a tree diagram helps. Look at Figure 4.2 (10-Page 145; 11-Page 152).4.2 (10-Page 145; 11-Page 152).

10 10 Slide

Slide


Counting Rule for Multi-step ExperimentsCounting Rule for Multi-step Experiments

Let’s take another example. Let’s take another example. Kentucky Power & Light Company is starting an Kentucky Power & Light Company is starting an

expansion project. The project consists of two expansion project. The project consists of two stages: design and construction.stages: design and construction.

Design work can be completed in 2, 3 or 4 months Design work can be completed in 2, 3 or 4 months (three outcomes)(three outcomes)

Construction work can be completed in 6, 7 or 8 Construction work can be completed in 6, 7 or 8 months (three outcomes)months (three outcomes)

Total number of outcomes in this experiment is:Total number of outcomes in this experiment is: (3) (3) = 9 (3) (3) = 9 Look at Table 4.1 and Figure 4.3 (10-Page 146; Look at Table 4.1 and Figure 4.3 (10-Page 146;

11-Page 153).11-Page 153).

11 11 Slide

Slide


A second useful counting rule enables us to count theA second useful counting rule enables us to count thenumber of experimental outcomes when number of experimental outcomes when nn objects are to objects are tobe selected from a set of be selected from a set of NN objects. objects.

Counting Rule for CombinationsCounting Rule for Combinations

CN

nN

n N nnN

!

!( )!C

N

nN

n N nnN

!

!( )!

Number of Number of CombinationsCombinations of of NN Objects Taken Objects Taken nn at a Time at a Time

where: where: NN! = ! = NN((NN 1)( 1)(NN 2) . . . (2)(1) 2) . . . (2)(1) nn! = ! = nn((nn 1)( 1)(nn 2) . . . (2)(1) 2) . . . (2)(1) 0! = 10! = 1

12 12 Slide

Slide


Number of Number of PermutationsPermutations of of NN Objects Taken Objects Taken nn at a Time at a Time

where: where: NN! = ! = NN((NN 1)( 1)(NN 2) . . . (2)(1) 2) . . . (2)(1) nn! = ! = nn((nn 1)( 1)(nn 2) . . . (2)(1) 2) . . . (2)(1) 0! = 10! = 1

P nN

nN

N nnN

!!

( )!P n

N

nN

N nnN

!!

( )!

Counting Rule for PermutationsCounting Rule for Permutations

A third useful counting rule enables us to count A third useful counting rule enables us to count thethe

number of experimental outcomes when number of experimental outcomes when nn objects are toobjects are to

be selected from a set of be selected from a set of NN objects, where the objects, where the order oforder of

selection is important.selection is important.

13 13 Slide

Slide


Example ProblemsExample Problems

A home builder offers four different floor plans, A home builder offers four different floor plans, three different exterior designs and two types three different exterior designs and two types of roof. How many ways can a home buyer of roof. How many ways can a home buyer choose his/her home? choose his/her home?

4 x 3 x 2 = 244 x 3 x 2 = 24

14 14 Slide

Slide


Examples on Combinations and Examples on Combinations and PermutationsPermutations

Combinations:Combinations:

How many ways can three Republicans be chosen as members of How many ways can three Republicans be chosen as members of a committee from a group of seven Republicans?a committee from a group of seven Republicans?

7 7!7 7!

C = ----------- = 35C = ----------- = 35

3 3! (7 – 3)!3 3! (7 – 3)!

Permutations:Permutations:

Automobile license plates in a state consists of seven letters of Automobile license plates in a state consists of seven letters of the alphabets; a letter may not be repeated. How many the alphabets; a letter may not be repeated. How many license plates are possible?license plates are possible?

26 26!26 26!

P = --------- = 26 x 25 x 24 x 23 x 22 x 21 x 20 P = --------- = 26 x 25 x 24 x 23 x 22 x 21 x 20

7 (26-7)!7 (26-7)!

15 15 Slide

Slide



Automobile license plates in a state consist of five letters and Automobile license plates in a state consist of five letters and two digits. A letter or a digit may be repeated. How many two digits. A letter or a digit may be repeated. How many license plates are possible?license plates are possible?

26 26!26 26!

The first position may be filled in P = -------------- = 26The first position may be filled in P = -------------- = 26

1 (26 – 1)!1 (26 – 1)!

26 x 26 x 26 x 26 x 26 x 10 x 1026 x 26 x 26 x 26 x 26 x 10 x 10

Automobile license plates in CA consist of any combination of Automobile license plates in CA consist of any combination of seven letters and/or digits. A letter or a digit may be seven letters and/or digits. A letter or a digit may be repeated seven times. How many license plates are repeated seven times. How many license plates are possible? possible?

36 x 36 x 36 x 36 x 36 x 36 x 36 36 x 36 x 36 x 36 x 36 x 36 x 36

16 16 Slide

Slide



Ten individuals are interested in serving on a board of directors. Ten individuals are interested in serving on a board of directors. There are three positions available. How many ways can the There are three positions available. How many ways can the selection be made?selection be made?

10 10!10 10!

C = -------------- = 120C = -------------- = 120

3 3! (10 – 3)!3 3! (10 – 3)!

In a class of 40 students, there are 25 women and 15 men. A In a class of 40 students, there are 25 women and 15 men. A committee is to be selected with 5 women and three men. committee is to be selected with 5 women and three men. How many ways can this selection be made?How many ways can this selection be made?

25 15 25! 15!25 15 25! 15!

C x C = -------------- x -------------C x C = -------------- x -------------

5 3 5! (25 – 5)! 3! (15 – 3)!5 3 5! (25 – 5)! 3! (15 – 3)!

17 17 Slide

Slide



Five different scholarships are to be awarded to Five different scholarships are to be awarded to five students. How many ways can these five students. How many ways can these scholarships be awarded?scholarships be awarded?

5 5!5 5!

P = -------- = 5! = 5 x 4 x 3 x 2 = 120P = -------- = 5! = 5 x 4 x 3 x 2 = 120

5 (5 – 5)!5 (5 – 5)!

18 18 Slide

Slide


Assigning ProbabilitiesAssigning Probabilities

Classical MethodClassical MethodClassical MethodClassical Method

Relative Frequency MethodRelative Frequency MethodRelative Frequency MethodRelative Frequency Method

Subjective MethodSubjective MethodSubjective MethodSubjective Method

Assigning probabilities based on the assumptionAssigning probabilities based on the assumption of of equally likely outcomesequally likely outcomes

Assigning probabilities based on Assigning probabilities based on experimentationexperimentation or historical dataor historical data

Assigning probabilities based on Assigning probabilities based on judgmentjudgment

19 19 Slide

Slide


Classical MethodClassical Method

If an experiment has If an experiment has nn possible outcomes, this method possible outcomes, this method

would assign a probability of 1/would assign a probability of 1/nn to each outcome. to each outcome.

Experiment: Rolling a dieExperiment: Rolling a die

Sample Space: Sample Space: SS = {1, 2, 3, 4, 5, 6} = {1, 2, 3, 4, 5, 6}

Probabilities: Each sample point has aProbabilities: Each sample point has a 1/6 chance of occurring1/6 chance of occurring

ExampleExample

20 20 Slide

Slide


Relative Frequency MethodRelative Frequency Method

Assigns probabilities based on relative frequencies.Assigns probabilities based on relative frequencies. Let’s take the example (10-Page 149; 11-Page Let’s take the example (10-Page 149; 11-Page

156)156) Number of patients waiting for service in the X-ray Number of patients waiting for service in the X-ray

department of a hospital:department of a hospital: Number WaitingNumber Waiting Number of DaysNumber of Days 0 20 2 1 51 5 2 62 6 3 43 4 4 34 3 Total 20Total 20

21 21 Slide

Slide


Relative Frequency MethodRelative Frequency Method

The probability of zero patient waiting for The probability of zero patient waiting for service is:service is:

2/20 = 0.12/20 = 0.1 The probability of one patient waiting for The probability of one patient waiting for

service is:service is: 5/20 = 0. 255/20 = 0. 25 The probability of two patients waiting for The probability of two patients waiting for

service is:service is: 6/20 = 0.306/20 = 0.30

And so on.And so on.

22 22 Slide

Slide


Subjective MethodSubjective Method

When economic conditions and a company’sWhen economic conditions and a company’s circumstances change rapidly it might becircumstances change rapidly it might be inappropriate to assign probabilities based solely oninappropriate to assign probabilities based solely on historical data.historical data. We can use any data available as well as ourWe can use any data available as well as our experience and intuition, but ultimately a probabilityexperience and intuition, but ultimately a probability value should express our value should express our degree of beliefdegree of belief that the that the experimental outcome will occur.experimental outcome will occur.

The best probability estimates often are obtained byThe best probability estimates often are obtained by combining the estimates from the classical or relativecombining the estimates from the classical or relative frequency approach with the subjective estimate.frequency approach with the subjective estimate.

23 23 Slide

Slide


Sample ProblemsSample Problems

Problem # 11 (10-Page 152; 11-Page 159)Problem # 11 (10-Page 152; 11-Page 159)

a.a. 858/(858+228) = 858/1,086 = 0.79858/(858+228) = 858/1,086 = 0.79

b.b. Yes.Yes.


a.a. 55!/(5!) x (55-5)!55!/(5!) x (55-5)!

b.b. 1/[55!/(5!)x(55-50!]1/[55!/(5!)x(55-50!]

c.c. 1/[55!/(5!)x(55-5)!] x 1/[42!/(1!)x(42-1)!]1/[55!/(5!)x(55-5)!] x 1/[42!/(1!)x(42-1)!]

24 24 Slide

Slide


An An eventevent is a collection of sample points.is a collection of sample points. An An eventevent is a collection of sample points.is a collection of sample points.

The The probability of any eventprobability of any event is equal to the sum of is equal to the sum of the probabilities of the sample points in the event.the probabilities of the sample points in the event. The The probability of any eventprobability of any event is equal to the sum of is equal to the sum of the probabilities of the sample points in the event.the probabilities of the sample points in the event.

If we can identify all the sample points of anIf we can identify all the sample points of an experiment and assign a probability to each, weexperiment and assign a probability to each, we can compute the probability of an event.can compute the probability of an event.

If we can identify all the sample points of anIf we can identify all the sample points of an experiment and assign a probability to each, weexperiment and assign a probability to each, we can compute the probability of an event.can compute the probability of an event.

Events and Their ProbabilitiesEvents and Their Probabilities

25 25 Slide

Slide


Events and Their ProbabilitiesEvents and Their Probabilities

Let’s go back to the example of Kentucky Power Let’s go back to the example of Kentucky Power & Light Company.& Light Company.

Suppose, we want to find out the probability of Suppose, we want to find out the probability of completing the expansion project in 10 months completing the expansion project in 10 months or less.or less.

According to Table 4.3, the following sample According to Table 4.3, the following sample points (2,6), (2,7), (2,8), (3,6), (3,7) and (4,6) points (2,6), (2,7), (2,8), (3,6), (3,7) and (4,6) provide a completion time of 10 months or less.provide a completion time of 10 months or less.

Thus, the event C that the project can be Thus, the event C that the project can be completed in 10 months or less is:completed in 10 months or less is:

P(C) = P(2,6) + P(2,7) + P(2,8) + P(3,6) + P(3,7) P(C) = P(2,6) + P(2,7) + P(2,8) + P(3,6) + P(3,7) + P(4,6)+ P(4,6)

26 26 Slide

Slide


Sample ProblemSample Problem

Problem #20 (10-Page 157; 11-Page 164)Problem #20 (10-Page 157; 11-Page 164) Given:Given: StateState Number of CompaniesNumber of Companies New York 54New York 54 California 52California 52 Texas 48Texas 48 Illinois 33Illinois 33 Ohio 30Ohio 30

Find: P(N) = Find: P(T) =Find: P(N) = Find: P(T) = Find P(B) = Find P(B) =

27 27 Slide

Slide


Some Basic Relationships of ProbabilitySome Basic Relationships of Probability

There are some There are some basic probability relationshipsbasic probability relationships that thatcan be used to compute the probability of an eventcan be used to compute the probability of an eventwithout knowledge of all the sample point probabilities.without knowledge of all the sample point probabilities.

Complement of an EventComplement of an Event Complement of an EventComplement of an Event

Intersection of Two EventsIntersection of Two Events Intersection of Two EventsIntersection of Two Events

Mutually Exclusive EventsMutually Exclusive Events Mutually Exclusive EventsMutually Exclusive Events

Union of Two EventsUnion of Two EventsUnion of Two EventsUnion of Two Events

28 28 Slide

Slide


The complement of The complement of AA is denoted by is denoted by AAcc.. The complement of The complement of AA is denoted by is denoted by AAcc..

The The complementcomplement of event of event A A is defined to be the eventis defined to be the event consisting of all sample points that are not in consisting of all sample points that are not in A.A. The The complementcomplement of event of event A A is defined to be the eventis defined to be the event consisting of all sample points that are not in consisting of all sample points that are not in A.A.

Complement of an EventComplement of an Event

Event Event AA AAccSampleSpace SSampleSpace S

VennVennDiagraDiagra

mm

29 29 Slide

Slide


The union of events The union of events AA and and BB is denoted by is denoted by AA BB The union of events The union of events AA and and BB is denoted by is denoted by AA BB

The The unionunion of events of events AA and and BB is the event containing is the event containing all sample points that are in all sample points that are in A A oror B B or both.or both. The The unionunion of events of events AA and and BB is the event containing is the event containing all sample points that are in all sample points that are in A A oror B B or both.or both.

Union of Two EventsUnion of Two Events

SampleSpace SSampleSpace SEvent Event AA Event Event BB

30 30 Slide

Slide


The intersection of events The intersection of events AA and and BB is denoted by is denoted by AA The intersection of events The intersection of events AA and and BB is denoted by is denoted by AA

The The intersectionintersection of events of events AA and and BB is the set of all is the set of all sample points that are in bothsample points that are in both A A and and BB.. The The intersectionintersection of events of events AA and and BB is the set of all is the set of all sample points that are in bothsample points that are in both A A and and BB..


Intersection of Two EventsIntersection of Two Events

Intersection of A and BIntersection of A and B

31 31 Slide

Slide


The The addition lawaddition law provides a way to compute the provides a way to compute the probability of event probability of event A,A, or or B,B, or both or both AA and and B B occurring.occurring. The The addition lawaddition law provides a way to compute the provides a way to compute the probability of event probability of event A,A, or or B,B, or both or both AA and and B B occurring.occurring.

Addition LawAddition Law

The law is written as:The law is written as: The law is written as:The law is written as:

PP((AA BB) = ) = PP((AA) + ) + PP((BB) ) PP((AA BBPP((AA BB) = ) = PP((AA) + ) + PP((BB) ) PP((AA BB

32 32 Slide

Slide


Mutually Exclusive EventsMutually Exclusive Events

Two events are said to be Two events are said to be mutually exclusivemutually exclusive if the if the events have no sample points in common.events have no sample points in common. Two events are said to be Two events are said to be mutually exclusivemutually exclusive if the if the events have no sample points in common.events have no sample points in common.

Two events are mutually exclusive if, when one eventTwo events are mutually exclusive if, when one event occurs, the other cannot occur.occurs, the other cannot occur. Two events are mutually exclusive if, when one eventTwo events are mutually exclusive if, when one event occurs, the other cannot occur.occurs, the other cannot occur.


33 33 Slide

Slide


Mutually Exclusive EventsMutually Exclusive Events

If events If events AA and and BB are mutually exclusive, are mutually exclusive, PP((AA BB = 0. = 0. If events If events AA and and BB are mutually exclusive, are mutually exclusive, PP((AA BB = 0. = 0.

The addition law for mutually exclusive events is:The addition law for mutually exclusive events is: The addition law for mutually exclusive events is:The addition law for mutually exclusive events is:

PP((AA BB) = ) = PP((AA) + ) + PP((BB))PP((AA BB) = ) = PP((AA) + ) + PP((BB))

There is no need toThere is no need toinclude “include “ PP((AA BB””There is no need toThere is no need toinclude “include “ PP((AA BB””

34 34 Slide

Slide



Problem #24 (10-Page 162; 11-Page 169)Problem #24 (10-Page 162; 11-Page 169)

a. Let S = Experience surpassed expectationa. Let S = Experience surpassed expectation N = No responseN = No response M= Met expectationM= Met expectation F = Fell short of expectationF = Fell short of expectation P(S) = 1 – P(N) – P(M) – P(F) = 1- 0.04 – 0.65 – 0.26 = P(S) = 1 – P(N) – P(M) – P(F) = 1- 0.04 – 0.65 – 0.26 =

0.050.05

b. P(M or S) = P(M) + P(S) = 0.65 + 0.05 = 0.70b. P(M or S) = P(M) + P(S) = 0.65 + 0.05 = 0.70

35 35 Slide

Slide




a. P(B or P) = P(B) + P(P) – P(B and P)a. P(B or P) = P(B) + P(P) – P(B and P) = 0.458 + 0.54 – 0.3 = 0.698= 0.458 + 0.54 – 0.3 = 0.698

b. P(neither B nor P) = 1 – P(B or P)b. P(neither B nor P) = 1 – P(B or P) = 1 – 0.698= 1 – 0.698 = 0.302= 0.302

36 36 Slide

Slide


The probability of an event given that another eventThe probability of an event given that another event has occurred is called a has occurred is called a conditional probabilityconditional probability.. The probability of an event given that another eventThe probability of an event given that another event has occurred is called a has occurred is called a conditional probabilityconditional probability..

A conditional probability is computed as follows :A conditional probability is computed as follows : A conditional probability is computed as follows :A conditional probability is computed as follows :

The conditional probability of The conditional probability of AA given given BB is denoted is denoted by by PP((AA||BB).). The conditional probability of The conditional probability of AA given given BB is denoted is denoted by by PP((AA||BB).).

Conditional ProbabilityConditional Probability

( )( | )

( )P A B

P A BP B

( )( | )

( )P A B

P A BP B

37 37 Slide

Slide



Problem #33 (10-Page 169-170; 11-Page 176-177)Problem #33 (10-Page 169-170; 11-Page 176-177) ReasonReason Quality Cost/Conv Other TotalQuality Cost/Conv Other Total FT 0.22 0.20 0.04 0.46FT 0.22 0.20 0.04 0.46 PT 0.21 0.31 0.02 0.54PT 0.21 0.31 0.02 0.54 __________________________________________________________________________________ 0.43 0.51 0.06 1.000.43 0.51 0.06 1.00 b. School cost/convenience followed by qualityb. School cost/convenience followed by quality c. P(Q given FT) = 0.22/0.46 = 0.47c. P(Q given FT) = 0.22/0.46 = 0.47 d. P(Q given PT) = 0.21/0.54 = 0.39d. P(Q given PT) = 0.21/0.54 = 0.39 e. P(A) = 0.46 P(B) = 0.43 P(A and B) = 0.22e. P(A) = 0.46 P(B) = 0.43 P(A and B) = 0.22 Since P(A and B) not equal to P(A) * P(B), the events are Since P(A and B) not equal to P(A) * P(B), the events are NOT independentNOT independent

38 38 Slide

Slide


Multiplication LawMultiplication Law

The The multiplication lawmultiplication law provides a way to compute the provides a way to compute the probability of the intersection of two events.probability of the intersection of two events. The The multiplication lawmultiplication law provides a way to compute the provides a way to compute the probability of the intersection of two events.probability of the intersection of two events.


PP((AA BB) = ) = PP((BB))PP((AA||BB))PP((AA BB) = ) = PP((BB))PP((AA||BB))

39 39 Slide

Slide


Independent EventsIndependent Events

If the probability of event If the probability of event AA is not changed by the is not changed by the existence of event existence of event BB, we would say that events , we would say that events AA and and BB are are independentindependent..

If the probability of event If the probability of event AA is not changed by the is not changed by the existence of event existence of event BB, we would say that events , we would say that events AA and and BB are are independentindependent..

Two events Two events AA and and BB are independent if: are independent if: Two events Two events AA and and BB are independent if: are independent if:

PP((AA||BB) = ) = PP((AA))PP((AA||BB) = ) = PP((AA)) PP((BB||AA) = ) = PP((BB))PP((BB||AA) = ) = PP((BB))oror

40 40 Slide

Slide


The multiplication law also can be used as a test to seeThe multiplication law also can be used as a test to see if two events are independent.if two events are independent. The multiplication law also can be used as a test to seeThe multiplication law also can be used as a test to see if two events are independent.if two events are independent.


PP((AA BB) = ) = PP((AA))PP((BB))PP((AA BB) = ) = PP((AA))PP((BB))

Multiplication LawMultiplication Lawfor Independent Eventsfor Independent Events

41 41 Slide

Slide


Miscellaneous ProblemsMiscellaneous Problems


Rating FrequencyRating Frequency

Poor 4Poor 4

Below Average 8Below Average 8

Average 11Average 11

Above Average 14Above Average 14

Excellent 13Excellent 13

a.a. What is the probability that a randomly selected viewer will What is the probability that a randomly selected viewer will rate the new show as average or better?rate the new show as average or better?

(11 + 14 + 13) / 50 = 0.76 (11 + 14 + 13) / 50 = 0.76

b.b. What is the probability that a viewer will rate the show as What is the probability that a viewer will rate the show as below average or worse?below average or worse?

(8 + 4) / 50 = 0.24(8 + 4) / 50 = 0.24

42 42 Slide

Slide




Given: six percent employees suffered lost-time Given: six percent employees suffered lost-time accidents last year. Fifteen percent who had accidents last year. Fifteen percent who had lost-time accidents last year will experience the lost-time accidents last year will experience the same this year.same this year.

a.a. What percent employees will experience lost-What percent employees will experience lost-time accidents in both years?time accidents in both years?

(0.06) (0.15) = 0.009(0.06) (0.15) = 0.009

b. (0.06) + (0.15) = 0.21 b. (0.06) + (0.15) = 0.21

43 43 Slide

Slide



A box contains 20 units of a product of which four are defective and A box contains 20 units of a product of which four are defective and 16 are good. Four units are selected at random. Calculate the 16 are good. Four units are selected at random. Calculate the following probabilities:following probabilities:

a.a. All four units are defectiveAll four units are defective

P(4D) = P(D) . P(D). P(D). P(D) = (4/20) (3/19) (2/18) (1/17) = P(4D) = P(D) . P(D). P(D). P(D) = (4/20) (3/19) (2/18) (1/17) = 1/48451/4845

b.b. Two units are defective and two units are goodTwo units are defective and two units are good

Two defective units must come Two good units must Two defective units must come Two good units must come fromcome from

from the four defective units: the sixteen good units:from the four defective units: the sixteen good units:

4 164 16

C = 6 ways C = 120 waysC = 6 ways C = 120 ways

2 22 2

44 44 Slide

Slide


Miscellaneous Problems ContinuedMiscellaneous Problems Continued

Four units can be selected from the original 20 Four units can be selected from the original 20 units in units in

2020

C = 4845 waysC = 4845 ways

44

(6) (120)(6) (120)

P(2D and 2G) = ----------- = 144/969 P(2D and 2G) = ----------- = 144/969

48454845

45 45 Slide

Slide


Miscellaneous Problems ContinuedMiscellaneous Problems Continued

c.c. At least three units are defectiveAt least three units are defective

P(at least 3D) means P(3D or 4D) = P(3D) + P(4D)P(at least 3D) means P(3D or 4D) = P(3D) + P(4D)

4 16 44 16 4

C . C CC . C C

3 1 43 1 4

P(3D) = ----------------- P(4D) = ------------------P(3D) = ----------------- P(4D) = ------------------

20 2020 20

C CC C

4 44 4

46 46 Slide

Slide



In how many ways can a television director schedule six different In how many ways can a television director schedule six different commercials during the six time slots allocated to commercials during the six time slots allocated to commercials during the telecast of the first quarter of a commercials during the telecast of the first quarter of a football game?football game?

66

P = 6! = 720P = 6! = 720

66

Seven wooden dowels of varying lengths are arranged in a row. Seven wooden dowels of varying lengths are arranged in a row. What is the probability that the dowels will be arranged in What is the probability that the dowels will be arranged in order of size? order of size?

22

P(arranged in order of size) = -----------------P(arranged in order of size) = -----------------

7!7!

47 47 Slide

Slide


End of Chapter 4End of Chapter 4

IS 310 Business Statistics CSU Long Beach

Documents

Transcript of IS 310 Business Statistics CSU Long Beach