Chapter 3 Probability(概率 )

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

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

3

Probability and statistics

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.

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

Sample space (Sample space (SS)()(样本空间样本空间 ): ): 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(事件 )

Example 3.2Example 3.2 Genders 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

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

Example 3.3Example 3.3 Answering Three True-False QuestionsAnswering Three True-False QuestionsA 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.

So, each of the 8 outcomes is equally likely to occur. P(CCC)=P(CCI)= ... = P(III)=1/8.

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

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

Relative Frequency Method(概率的频率解释 )

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

Example2: Long-Run Relative Frequency Method

1. An accounts receivable manager knows from 1. An accounts receivable manager knows from past data that about 70 of 1000 accounts became past data that about 70 of 1000 accounts became uncollectible. uncollectible.

The manager would estimate the probability of bad The manager would estimate the probability of bad debts as 70/1000 = .07 or 7%. debts as 70/1000 = .07 or 7%.

2. Tossing a fair coin 3000 times, we can see that 2. Tossing a fair coin 3000 times, we can see that although the proportion of heads was far from 0.5 in although the proportion of heads was far from 0.5 in the first 100 tosses, it seemed to stabilize and the first 100 tosses, it seemed to stabilize and approach 0.5 as the number of tosses increased. approach 0.5 as the number of tosses increased.

Long-Run Relative Frequency Method: Example

Often we determine the probability from a random sample (Long-Run Relative Frequency Method) and apply it to the population.

Of 5528 Zhuhai residents randomly sampled,445 prefer to watch CCTV-1

Estimated Share P(CCTV-1) = 445 / 5528 = 0.0805 So the probability that any Zhuhai resident chosen at

random prefers CCTV-1 is 0.0805

Assuming total population in Zhuhai is 1,000,000 : Size of audience in the city = Population x Share

so 1,000,000 x 0.0805 = 80,500

Long-Run Relative Frequency Method: application

Subjective Probability

Using experience, intuitive judgment, or personal expertise to assess/derive a probability

May or may not have relative frequency interpretation (Some events cannot be repeated many times)

Contains a high degree of personal bias. What is the probability of your favorite basketball

or football team win the next game? (e.g. sports betting)

Subjective probability & bettingThe odds in betting reflect the subjective probability guessed by the mass.Who much are you willing to pay for a ticket which worth $10 if there was life on Mars and nothing if there was not?

Subjective probability usually reflects the mind/opinion more than the reality. Sometimes, it is used to gauge the public opinions.

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”.”.

Union of A and B, (A 和 B的并集 )

Is an event consisting of the outcomes that belong to either A or B (or both).

Intersection of A and B, (A 和 B的交集 )

Is an event consisting of the outcomes that belong to both A and B.

BA

BA

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(加法准则 )

P(A B) = P(A) + P(B) - P(A B)

B)P(A

A and B are mutually exclusive(相互排斥 ) 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

Newspaper Subscribers #1Newspaper Subscribers #1 Example 3.7Example 3.7

Define events:Define events: A = event that a randomly selected household A = event that a randomly selected household

subscribes to the subscribes 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

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

Table3.1 A Contingency Table(列联表 ) 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

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

Newspaper Subscribers #4The 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

A Mutually Exclusive CaseRecall the radio station example. The percentages of LA Recall the radio station example. The percentages of LA residents who favor each of the top 10 stations is listed in the residents who favor each of the top 10 stations is listed in the Figure. Let the name of each station, for example Figure. Let the name of each station, for example KPWRKPWR, , represent the event that the station, say KPWR, is the most represent the event that the station, say KPWR, is the most favorable radio station for a randomly selected resident.favorable radio station for a randomly selected resident.Since the survey asked each resident to Since the survey asked each resident to name the singlename the single station station that he/she listens to mostthat he/she listens to most, , the 10 events are mutually exclusive. the 10 events are mutually exclusive. Therefore, the probability that a randomly selected LA resident Therefore, the probability that a randomly selected LA resident would favor one of the top 10 stations is would favor one of the top 10 stations is P(KPWR U KLAX U …… U KSBC-FM) = P(KPWR)+P(KLAX)+……+P(KCBS-FM) = 0.08+0.064+ …….+0.036=0.508.

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”(”(条件概率条件概率 ) 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)

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.

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

Independence(独立 ) of EventsTwo 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?

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.

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

A QuestionSuppose 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

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 4 Discrete Random Variables(离散随机变量 )

Two Types of Random VariablesTwo Types of Random Variables

Discrete Probability DistributionsDiscrete Probability Distributions

The Binomial DistributionThe Binomial Distribution

The Poisson DistributionThe Poisson Distribution

Random Variables (随机变量 )A A random variablerandom variable is a variable that assumes numerical is a variable that assumes numerical values that are determined by the outcome of an values that are determined by the outcome of an experimentexperiment.

Consider a random experiment in which a coin Consider a random experiment in which a coin is tossed three times. Let is tossed three times. Let XX be the number of heads. Let be the number of heads. Let HH represent the outcome of a head and represent the outcome of a head and TT the outcome of a tail. the outcome of a tail.

The possible outcomes for such an experiment:

TTT, TTH, THT, THH, HTT, HTH, HHT, HHH

Thus the possible values of Thus the possible values of X X (number of heads) are 0,1,2,3(number of heads) are 0,1,2,3.From the definition of a random variable, X as defined in this experiment, is a random variable.

Example 4.1Example 4.1

Section 4.1 Two Types of Random Variables Discrete random variable（离散型随机变量） :

Possible values can be counted or listed

- For example, the number of TV sets sold at the store in one day. Here x could be 0, 1, 2, 3, 4 and so forth.

Continuous random variable （连续型随机变量） : May assume any numerical value in one or more intervals

- For example, the waiting time for a credit card authorization, the interest rate charged on a business loan

Example: Two Types of Random Variables

Question Random Variable x Type

Family x = Number of people in Discrete(离散 )

size family reported on tax return

 

Distance from x = Distance in miles from Continuous(连续 )

home to store home to a store

Own dog x = 1 if own no pet; Discrete

or cat = 2 if own dog(s) only;

= 3 if own cat(s) only;

= 4 if own dog(s) and

cat(s)

Properties

1. For any value x of the random variable, p(x) 0

2. The probabilities of all the events in the sample space must sum to 1, that is,

Section 4.2Discrete Probability Distributions（离散概率分布）The probability distribution of a discrete random variable is a table, graph, or formula that gives the probability associated with each possible value that the variable can assume

Notation: Denote the values of the random variable by x and the value’s associated probability by p(x)

1all

x

xp

Number of Radios （ Sold at South City in a Week）

Let Let xx be the random variable of the number of radios sold per be the random variable of the number of radios sold per week, week, xx has values has values xx = 0, 1, 2, 3, 4, 5 = 0, 1, 2, 3, 4, 5

Given sales history over past 100 weeksGiven sales history over past 100 weeks Let f be the number of weeks (of the past 100) during which Let f be the number of weeks (of the past 100) during which

xx number of radios were sold number of radios were sold Records tell us thatRecords tell us that

f(0)=3 No radios have been sold in 3 of the weeksf(1)=20 One radios has been sold in 20 of the weeksf(2)=50 Two radios have been sold in 50 of the weeksf(3)=20 Three radios have been sold in 20 of the weeksf(4)=5 Four radios have been sold in 4 of the weeksf(5)=2 Five radios have been sold in 2 of the weeksNo more than five radios were sold in any of the past 100 weeks

Example 4.2Example 4.2

Frequency distribution of sales history over past 100 weeks # Radios, x Frequency Relative Frequency Probability, p(x) 0 f(0) =3 3/100 = 0.03 p(0) = 0.03

1 f(1) =20 20/100 = 0.20 p(1) = 0.20 2 f(2) =50 0.50 p(2) = 0.50 3 f(3) =20 0.20 p(3) = 0.20

4 f(4) =5 0.05 p(4) = 0.05 5 f(5) = 2 0.02 P(5) = 0.02

100 1.00 1.00 Interpret the relative frequenciesInterpret the relative frequencies as probabilitiesas probabilities

So for any value So for any value xx, f(, f(xx)/)/nn = = pp((xx)) Assuming that sales Assuming that sales remain stable over timeremain stable over time

What is the chance that two radios will be sold in a week? P(x = 2) = 0.50

What is the chance that fewer than 2 radios will be sold in a week? p(x < 2) = p(x = 0 or x = 1)

= p(x = 0) + p(x = 1)

= 0.03 + 0.20 = 0.23 What is the chance that three or more radios will be sold in a

week? p(x ≥ 3) = p(x = 3, 4, or 5)

= p(x = 3) + p(x = 4) + p(x = 5)

= 0.20 + 0.05 + 0.02 = 0.27

Using the addition rule for the mutuallyexclusive values ofthe random variable.

The The mean(mean( 均值均值 )) or or expected valueexpected value of a discrete of a discrete random variable random variable XX is: is:

Expected Value of a Discrete Random Variable

xAll

X xpx

is the value expected to occur in the long run and on average

Number of Radios How many radios should be expected to be sold in a

week? Calculate the expected value of the number of radios

sold, XRadios, x Probability, p(x) x p(x) 0 p(0) = 0.03 0 0.03 = 0.00

1 p(1) = 0.20 1 0.20 = 0.20

2 p(2) = 0.50 2 0.50 = 1.00

3 p(3) = 0.20 3 0.20 = 0.60

4 p(4) = 0.05 4 0.05 = 0.20

5 p(5) = 0.02 5 0.02 = 0.10 1.00 2.10

• On average, expect to sell 2.1 radios per week

Example 4.3Example 4.3

The variance of a discrete random variable is:

Variance and Standard Deviation

xAll

XX xpx 22

The standard deviation is the square root of the variance

2XX

• The variance is the average of the squared deviations of the different values of the random variable from the expected value

• The variance and standard deviation measure the spread of the values of the random variable from their expected value

Calculate the variance and standard deviation of the number of radios sold at Sound City in a week

89.02 XVarianceStandard deviation

94340890 ..X

Radios, x Probability, p(x) (x - X)2 p(x)

0 p(0) = 0.03 (0 – 2.1)2 (0.03) = 0.1323

1 p(1) = 0.20 (1 – 2.1)2 (0.20) = 0.2420

2 p(2) = 0.50 (2 – 2.1)2 (0.50) = 0.0050

3 p(3) = 0.20 (3 – 2.1)2 (0.20) = 0.1620

4 p(4) = 0.05 (4 – 2.1)2 (0.05) = 0.1805

5 p(5) = 0.02 (5 – 2.1)2 (0.02) = 0.1682

1.00 0.8900

Example 4.4Example 4.4 Number of Radios

The Binomial Distribution(二项分布 )

The Binomial Experiment:1. Experiment consists of n identical trials2. Each trial results in either “success” or “failure”3. Probability of success, p, is constant from trial to trial4. Trials are independent

If If xx is the total number of successes in is the total number of successes in nn trials of a binomial trials of a binomial experiment, then experiment, then xx is a is a binomial random variablebinomial random variable

Note: The probability of failure, q, is 1 – p and is constant from trial to trial

The Binomial Distribution #2

x-nxqp

x-nx

n =xp

!!

!

For a binomial random variable x, the probability of x successes in n trials is given by the binomial distribution:

• NoteNote: : nn! is read as “! is read as “nn factorial” and factorial” and nn! = ! = nn ×× ( (nn-1) -1) ×× ( (nn-2) -2) ×× ... ... ×× 1 1– For example, 5! = 5 For example, 5! = 5 4 4 3 3 2 2 1 = 120 1 = 120

• Also, 0! =1Also, 0! =1• Factorials are not defined for negative numbers or fractionsFactorials are not defined for negative numbers or fractions

The Binomial Distribution #3• What does the equation mean?

– The equation for the binomial distribution consists of the product of two factors

x-nxqp

x-nx

n =xp

!!

!

Number of ways to get x successes and (n–x) failures in n trials

The chance of getting x successes and (n–x) failures in a particular arrangement

Incidence of Nausea The company claims that, at most, 10 percentage of all patients treated with Phe-Mycin would experience nausea as a side effect of taking the drug.

x = number of patients who will experience nausea following treatment with Phe-Mycin out of the 4 patients tested

Find the probability that 2 of the 4 patients treated will experience nausea

Given: n = 4, p = 0.1, with x = 2Then: q = 1 – p = 1 – 0.1 = 0.9

0486.09.01.06

9.01.0!24!2

!42

22

242

xp

Example 4.5Example 4.5

Binomial Distribution (n = 4, p = 0.1)

x 0.05 0.1 0.15 … 0.500 0.8145 0.6561 0.5220 … 0.0625 41 0.1715 0.2916 0.3685 … 0.2500 32 0.0135 0.0486 0.0975 … 0.3750 23 0.0005 0.0036 0.0115 … 0.2500 14 0.0000 0.0001 0.0005 … 0.0625 0

0.95 0.9 0.85 … 0.50 x

Binomial Probability Table (Appendix Table A.1, P817)

values of p (.05 to .50)

values of p (.05 to .50)

Table 4.7(a) for n = 4, with x = 2 and p = 0.1

P(x = 2) = 0.0486

p = 0.1

Incidence of Nausea(after Treatment)

xx = = number of patients who will experience nausea number of patients who will experience nausea following treatment with Phe-Mycin out of the following treatment with Phe-Mycin out of the 44 patients testedpatients tested

Find the probability that at least 3 of the 4 patients treated Find the probability that at least 3 of the 4 patients treated will experience nauseawill experience nauseaSet Set x x = 3, = 3, nn = 4, = 4, pp = 0.1, so = 0.1, so qq = 1 – = 1 – pp = 1 = 1 – – 0.1 = 0.90.1 = 0.9Then:Then:

0037.00001.0036.0

43

4or 33

xpxp

xpxp Using the addition rule for the mutually exclusive values of the binomial random variable

Example 4.5Example 4.5

Rare Events Suppose at least three of four sampled patients

actually did experience nausea following treatment If If pp = 0.1 is believed, then there is a chance of only 37 in = 0.1 is believed, then there is a chance of only 37 in

10,000 of observing this result10,000 of observing this result So this is very unlikely!So this is very unlikely!

But it actually occurredBut it actually occurred So, this is very strong evidence that So, this is very strong evidence that pp does not equal 0.1 does not equal 0.1

There is very strong evidence that There is very strong evidence that pp is actually greater is actually greater than 0.1than 0.1

Several Binomial Distributions

Mean and Variance of a Binomial Random Variable

If If xx is a binomial random variable with is a binomial random variable with parametersparameters nn and and pp (so (so qq = 1 – = 1 – pp), then), then

npq

npq

np

X

X

X

deviation standard

variance

mean

2

Back to Example 4.5 Of 4 randomly selected patients, how many should be Of 4 randomly selected patients, how many should be

expected to experience nausea after treatment?expected to experience nausea after treatment?

Given: Given: nn = 4, = 4, pp = 0.1 = 0.1

Then mThen mXX = = npnp = 4 = 4 0.1 = 0.4 0.1 = 0.4

So expect 0.4 of the 4 patients to experience nauseaSo expect 0.4 of the 4 patients to experience nausea

If at least three of four patients experienced nausea, If at least three of four patients experienced nausea, this would be many more than the 0.4 that are this would be many more than the 0.4 that are expectedexpected

Binomial Distribution Binomial Distribution EXAMPLE: EXAMPLE: Pat Statsdud is registered in a statistics course and intends to Pat Statsdud is registered in a statistics course and intends to

rely on luck to pass the next quiz.rely on luck to pass the next quiz. The quiz consists on 10 multiple choice questions with 5 The quiz consists on 10 multiple choice questions with 5

possible choices for each question, only one of which is the possible choices for each question, only one of which is the correct answer.correct answer.

Pat will guess the answer to each questionPat will guess the answer to each question Find the following probabilitiesFind the following probabilities

Pat gets no answer correctPat gets no answer correct Pat gets two answer correct?Pat gets two answer correct? Pat fails the quizPat fails the quiz If all the students in Pat’s class intend to guess the If all the students in Pat’s class intend to guess the

answers to the quiz, what is the mean and the standard answers to the quiz, what is the mean and the standard deviation of the quiz mark?deviation of the quiz mark?

Solution Checking the conditions

An answer can be either correct or incorrect. There is a fixed finite number of trials (n=10) Each answer is independent of the others. The probability p of a correct answer (.20) does not

change from question to question.

Determining the binomial probabilities:Let X = the number of correct answers

1074.)80(.)20(.)!010(!0

!10)0( 0100

XP

3020.)80(.)20(.)!210(!2

!10)2( 2102

XP

Determining the binomial probabilities:Pat fails the test if the number of correct answers is less than 5, which means less than or equal to 4.

= p(0) + p(1) + p(2) + p(3) + p(4) = .1074 + .2684 + .3020 + .2013 + .0881

=.9672

P(X4

μ= np = 10(.2) = 2. σ= [np(1-p)]1/2 = [10(.2)(.8)]1/2 = 1.26

TThe mean and the standard deviation of the quiz mark?he mean and the standard deviation of the quiz mark?

The Poisson Distribution(泊松分布 )

1. The probability of occurrence is the same for any intervals of equal length

2. The occurrence in any interval is independent of an occurrence in any non-overlapping interval

Consider the number of times an event occurs over Consider the number of times an event occurs over an interval of time or space, and assume thatan interval of time or space, and assume that

If x = the number of occurrences in a specified interval, then x is a Poisson random variable

The probability of The probability of xx occurrences in the interval when occurrences in the interval when are expected is described by the are expected is described by the Poisson distributionPoisson distribution::

The Poisson Distribution Continued

!x

exp

x

Suppose is the mean or expected number of Suppose is the mean or expected number of occurrences during a specified intervaloccurrences during a specified interval

where where xx can take any of the values can take any of the values xx = 0, 1, 2, 3, … = 0, 1, 2, 3, …

and and ee = 2.71828… ( = 2.71828… (ee is the base of the natural logs) is the base of the natural logs)

Let x be the number of errors made by the ATC center during one week

Given: = 20.8 errors per year

Suppose that an air traffic control (ATC) center has been Suppose that an air traffic control (ATC) center has been averaging averaging 20.820.8 errors per year and lately the center errors per year and lately the center experiences experiences 33 errors in a week. errors in a week.

ATC Center Errors

Then: Then: = 0.4 errors per week = 0.4 errors per week

• Because there are 52 weeks per year, m for a week is: Because there are 52 weeks per year, m for a week is: = (20.8 errors= (20.8 errors//year) year) // (52 weeks (52 weeks//year) = 0.4 year) = 0.4

errorserrors//weekweek

Example 4.6Example 4.6

Find the probability that Find the probability that 33 errors errors (x =3)(x =3) will occur in a will occur in a weekweek

– Want Want pp((xx = 3) when = 0.4 = 3) when = 0.4

ATC Center Errors Continued

0072.0

!3

4.03

34.0

e

xp

6703.0

!0

4.00

04.0

e

xp

Find the probability that no errors (x = 0) will occur in a week

– Want p(x = 0) when = 0.4

Poisson Probability Table (Appendix Table A.2, P821)

x 0.1 0.2 … 0.4 … 1.000 0.9048 0.8187 … 0.6703 … 0.36791 0.0905 0.1637 … 0.2681 … 0.36792 0.0045 0.0164 … 0.0536 … 0.18393 0.0002 0.0011 … 0.0072 … 0.06134 0.0000 0.0001 … 0.0007 … 0.01535 0.0000 0.0000 … 0.0001 … 0.0031

, Mean number of Occurrences =0.4

0072.0

!3

4.03

34.0

e

xp

Poisson Distribution ( = 0.4)

Mean and Variance of a Poisson Random Variable

If x is a Poisson random variable with parameter , then

X

X

X

deviation standard

variance

mean

2

Several Poisson Distributions

Back to Example 4.6In the ATC center situation, 28.0 errors occurred on average per year

• mean = 0.4 errors/week

• standard deviation = 0.6325 errors/week.• Because = √0.4 = 0.6325

Assume that the number Assume that the number xx of errors during any span of of errors during any span of time follows a Poisson distribution for that time spantime follows a Poisson distribution for that time span

Per Per weekweek, the , the parametersparameters of the Poisson distribution of the Poisson distribution are:are:

Customers arrive at a rate of Customers arrive at a rate of 7272 per per hour. What is the probability of hour. What is the probability of 4 4 customers arriving in customers arriving in 3 3 minutes?minutes?

!x

exp

x

Poisson Distribution Example

Solution:Solution:72 per hr. = 1.2 per min.72 per hr. = 1.2 per min. = 3.6 per 3 mins.= 3.6 per 3 mins.

1912.0

!4

6.34

46.3

e

xp