Experiment

©蘇國賢 2000社會統計（上） Page 1

Experiment Experiment

• An experiment is any activity from which an outcome, measurement, or result is obtained. 任何求結果的過程或活動皆可稱為「試驗」。

• When the outcomes cannot be predicted with certainty, the experiment is a random experiment. 當結果無法事先預測時，稱此試驗為「隨機試驗」。

定義定義


Example of Experiments Example of Experiments

• 1. Measuring the lifetime (time to failure) of a given product

• 2. Inspecting an item to determine whether it is defective

• 3. Recording the income of a bank employee

• 4. Recording the balance in an individual's checking account

實例實例


Basic Outcomes and Sample SpaceBasic Outcomes and Sample Space

• The set of all possible basic outcomes for a given experiment is called the sample space. 隨機實驗的所有可能結果稱為樣本空間，一般以 S或 Ω 表示。

• Each possible outcome of a random experiment is called a basic outcome (or a sample point, an element in the sample space). 隨機實驗中所得到的任何可能的個別結果稱之為「基本結果」（或樣本點、或簡稱樣本），以小寫 oi表示。

定義定義


Basic Outcomes and Sample SpaceBasic Outcomes and Sample Space

• 某家公司在六個不同的城市有辦公室，某新進員工將被分派到其中的一個辦公室上班，則所有可能的分派為：

• o1= 台北 o2= 桃園 o3= 台中 • o4= 台南 o5= 高雄 o6= 屏東 • 分派的樣本空間為：• S = { o1 , o2 , o3 , o4 ,o5 ,o6 }• 如果這名員工被分派至台南上班，我們說該試驗的結果為 o4 = 台南，即該試驗的 o4結果發生了。

實例實例


Venn DiagramsVenn Diagrams

• S = { o1 , o2 , o3 , o4 ,o5 ,o6 }

o1 o2 o3

o4 o5 o6

‧‧‧

‧‧‧

定義定義


EventEvent 事件事件• An event is a specific collection of basic outco

mes, that is, a set containing one or more of the basic outcomes from the sample space. We say that an event occurs if any one of the basic outcomes in the event occurs.

• 事件為包含樣本空間中一個以上樣本元素（基本結果）的子集合，即樣本空間的任何部分集合 (subset) ，一般以英文大寫字母表示。

定義定義


EventEvent 事件事件

• 上例中，該員工被分派至南台灣的事件為：• A = {o4 ,o5 ,o6 }

• 如果該員工被分派至台南上班，我們可以說 A 事件發生了。

• 若 B 事件為該員工被分派至直轄市上班，則

• B = {o1 ,o5}

實例實例定義定義


Venn DiagramsVenn Diagrams

• S = { o1 , o2 , o3 , o4 ,o5 ,o6 }

o1 o2 o3

o4 o5 o6

‧‧‧

‧‧‧

定義定義

A 事件

B 事件



• 寫出丟骰子一次所得數目的樣本空間：• S = {1, 2, 3, 4, 5, 6}

• 寫出得到偶數的事件：• A = {2,4,6}

• 寫出數字大於 2 的事件：• B = {3,4,5,6}

實例實例



• 丟銅板得到 H 正面或 T 反面，寫出丟一銅板三次的樣本空間：

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

• 設 A 表示第一次出現正面的事件，則• A = {HHH, HHT, HTH, HTT}

實例實例


Assigning Probabilities to EventsAssigning Probabilities to Events事件的機率事件的機率

• There are two types of random experiments, those that can be repeated over and over again under essentially identical conditions and those that are unique and cannot be repeated.

定義定義



• A numerical measure that indicates the likelihood of a specific outcome in a repeatable random experiment is called an objective probability, whereas the probability associated with a specific outcome of a unique and nonrepeatable random experiment is called a subjective probability.

定義定義



• There are three different approaches to assigning probabilities to basic outcomes:

• 1. The relative frequency approach

• 2. The equally likely approach

• 3. The subjective approach

定義定義


The Relative Frequency Approach The Relative Frequency Approach 相對次數（後天）機率理論相對次數（後天）機率理論

• Let fA be the number of occurrences, or frequency of occurrence, of event A in n repeated identical trials. The probability that A occurs is the limit of the ratio fA/n as the number of trials n becomes infinitely large.

n

fAP A

n lim)(

定義定義


The Relative Frequency Approach The Relative Frequency Approach 相對次數（後天）機率理論相對次數（後天）機率理論

• 當一試驗在完全相同的情境中重複試驗，某事件A發生的機率可以定義為：當試驗在相同情境下重複無限次時， A 所發生的次數與試驗總次數之比。

• 由於這種機率為長期試驗的結果，因此又稱之為後天機率、相對次數。由於這種機率為歸納多次試驗的結果所得，故又稱為統計機率或經驗機率。

n

fAP A

n lim)(

定義定義


後天機率理論的缺陷後天機率理論的缺陷

• (1) Because we can never replicate an experiment an infinite number of times, it is impossible to determine the limit of the ratio fA/n as n approaches infinity.

• (2) We can never be sure that we have repeated an experiment under identical conditions.

定義定義


後天機率理論的缺陷後天機率理論的缺陷

• When we use the relative frequency approach, we use the observed ratio fA/n to approximate the theoretical probability that event A occurs. That is, we assume that P(A) fA/n when n is sufficiently large.

定義定義


The Equally Likely ApproachThe Equally Likely Approach古典（先天）機率理論古典（先天）機率理論

• Suppose that an experiment must result in one of n equally likely outcomes. Then each possible basic outcome is considered to have probability 1/n of occurring on any replication of the experiment.

• 一樣本空間有 n 個樣本點（基本結果），且每一個樣本點發生的機會皆相等。則在任何重複試驗中，每一個樣本點發生的機率為 1/n 。若事件 A 的樣本點個數為 nA，則 A 發生的機率為：

• P(A) = nA/n ，符合某條件結果的個數與總結果數之比。

定義定義


The Equally Likely ApproachThe Equally Likely Approach古典（先天）機率理論古典（先天）機率理論

• 直一骰子兩次，求點數和等於 6 的機率？• E = { (1,5) (2,4) (3,3) (4,2) (5,1)}

• P(E) = 5/36

• 其他先天機率：丟銅板、抽籤、彩券等。• 先天機率最重要的問題：所有樣本點發生的機率是否相同？

實例實例


Objective ProbabilityObjective Probability 客觀機率客觀機率

• A probability obtained by using a relative frequency approach or an equally likely approach is called an objective probability.

定義定義


The Subjective ApproachThe Subjective Approach主觀機率主觀機率

• 如果樣本點出現的可能性不等，則無法求事件的先天機率。若試驗無法進行，或無法多次重複，則相對次數無法計算，事件的後天機率也無法計算。

• 在此情況下，我們經常仰賴主觀機率，即人們對於某事件發生的主觀評價，或可以看成是「專家」的意見。

• A subjective probability is a number in the interval [0, 1] that reflects a person's degree of belief that an event will occur.

定義定義


OddsOdds 可能性可能性

• 有時候人們以「發生」與「不發生」的比值來陳述他們對於事件發生機率的意見。

• 如經常聽到某人說某賭局、競賽、或遊戲有 3:1的勝算，則表示某人有 75% 贏的機率。

• If the odds in favor of event A occurring are a to b, then

定義定義

ba

aAP

)(


Which approach is bestWhich approach is best ？？

• The nature of the problem determines which approach is best.

• Problems with an underlying symmetry, such as coin, dice, and card problems, are especially suited to the equally likely approach.

• Problems for which we have large samples of data based on many replications of an experiment are especially suited to the relative frequency approach.

• Problems that occur only once, such as a sporting event, are especially suited to the subjective approach.

定義定義


Which approach is bestWhich approach is best ？？

• 已知懷單胎，生男的機率為何？• (1) 依先天機率：• 性別有男女兩類，故生男的機率為 1/2• (2) 依後天機率：• 台灣地區男性人口佔 52% ，故生男的機率為 52

%• (3) 主觀機率：• 由於飲食習慣的改變，現代人大量攝取動物性蛋白質，使生男的機率增加為 60%

實例實例


Set TheorySet Theory

• Subset 子集合• An event A is contained in another event B if every outcom

e that belongs to the subset defining the event A also belongs to the subset defining the event B.

• A = {2,4,6} B={2,3,4,5,6}

• A B, A is a subset of B

• If A B and B A, then A = B

• If A B and B C, then A C

• Empty Set or Null Set 空集合 Ø

• For any event A, Ø A S,

定義定義


Operation of Set Theory: Operation of Set Theory: UnionsUnions聯集聯集

• Unions聯集• Let A and B be two events in

the sample space S. Their union, denoted A U B. is the event composed of all basic outcomes in S that belong to at least one of the two events A or B. Hence, the union A U B occurs if either A or B (or both) occurs.

定義定義

AS

B


Operation of Set Theory: Operation of Set Theory: UnionsUnions聯集聯集

• Unions聯集 : The union of n events A1,A2,…,An is defined to be the event that contains all outcomes which belong to at least one of these n events.

定義定義

AS

B

nini AAAAAA 43211


Operation of Set Theory: Operation of Set Theory: IntersectionIntersection 交集交集

• Intersection交集 :• Let A and B be two events in the sampl

e space S. The intersection of A and B, denoted A B. is the event composed of all basic outcomes in S that belong to both A and B. Hence, the intersection A B occurs if both A and B occur.

定義定義

AS

B


Operation of Set Theory: Operation of Set Theory: IntersectionIntersection 交集交集

• Intersection交集 : • The intersection of n events, A1,

…An is defined to be the event that contains the outcomes which are common to all these n events.

定義定義

n

nini

AAAAA

AAAAAA

4321

43211

AS

B


Complement of an EventComplement of an Event

• Let A denote some event in the sample space S. The complement of A (A 的餘事件） , denoted by Ac, represents the event composed of all basic outcomes in S that do not belong to A.

定義定義

AAc

S


Complement has the following properties:Complement has the following properties:

• (Ac)c =A

• A Ac = S

• Øc = S

• Sc = Ø

• A Ac = Ø

定義定義

AAc

S


Complement has the following properties:Complement has the following properties:

• (A B)∪ c =Ac ∩Bc

• (A ∩ B)c =Ac ∪ Bc

定義定義

A

Ac

S

BBc

Ii

ci

c

Iii AA

)(

Ii

ci

c

Iii AA

)(

• P(A) = P(A ∩ B) +P(A ∩ Bc)• P(Ac ∩ Bc) = 1 － P(A ∪ B)• P(Ac B∪ c) = 1 － P(A ∩ B)


Mutually Exclusive Events Mutually Exclusive Events (Disjoint Events)(Disjoint Events)

• Let A and B be two events in a sample space S. If A and B have no basic outcomes in common, then they are said to be mutually exclusive. If A and B are mutually exclusive events, we write (A B) = Ø, where Ø denotes the empty set. P(A B) = 0.


Some basic rules of probabilitySome basic rules of probability

• Probability of a basic outcome:• For each basic outcome oi, 0 P(oi) 1. • Probability of an event:• Let event A = { o1 , o2 , o3 , o4 ,o5 ,…ok }, where o1 , o2 , o3 , o4

,o5 ,…ok are k different basic outcomes. The probability of any event A is the sum of the probabilities of the basic outcomes in A. That is,

• P(A) = P(o1) + P(o2) + P(o3) + P(o4) + P(o5) +…+P(ok) = AP(oi)

• where AP(oi) means to obtain the sum over all basic outcomes in event A.

定義定義


Some basic rules of probabilitySome basic rules of probability

• Probability of an Event

• For any event A, 0 P(A) 1.

• Probability of a sample space:• Let event S = { o1 , o2 , o3 , o4 ,o5 ,…on } represent t

he sample space of an experiment. The probability of S is

• P(S) = sP(oi) = 1

定義定義


Definition of ProbabilityDefinition of Probability

• Axiom 1 公理 1: For any event A, P(A) 0 事件 A 發生的機率為實數

• Axiom 2: P(S) = 1.

• Axiom 3: For any infinite sequence of disjoint events (互斥事件） A1, A2, …

11

)()(i

ii

i APAP


Definition of ProbabilityDefinition of Probability

• A probability distribution , or simply a probability, on a sample space S is a specification of numbers P(A) which satisfy Axioms 1,2, and 3.

• 設有一試驗其樣本空間為 S ，對 S 中之任一事件 A指定一值 P(A) ，若 P(.)滿足上述三個公理，則稱 P （ . ）為一機率測度，且稱 P(A) 為事件 A 的機率。


Theorem 1:Theorem 1: 零事件的機率零事件的機率

• 零事件的機率等於零• Pr(Ø)=0


Theorem 2:Theorem 2:

• For any finite sequence of n disjoint events A1,A2,…,An

n

ii

n

ii APAP

11

)()(

11

)()(i

ii

i APAP

與公理 3的不同處


Theorem 2:Theorem 2:

• Proof. 假設一組無窮盡的事件序列 A1, A2, A3,…, 其中 A1… An 為互斥事件 , A

i = , i > n

0)(1

n

iiAP

11

)()(i

ii

i APAP

111

)()()(ni

i

n

ii

ii APAPAP


Theorem 3:Theorem 3:餘事件的機率餘事件的機率Probability of the complement of an eventProbability of the complement of an event

• Let Ac denote the complement of A. Then P(Ac) = 1 – P(A).

• Proof: • Since A and Ac are disjoint events and A Ac = S, • it follows from Theorem 2 that P(S) = P(A) +

P(Ac). • Since P(S) =1 by Axiom 2, • then P(Ac) = 1 – P(A).

定義定義


Theorem 4:Theorem 4: 機率的範圍機率的範圍

• For any event A, 0 P(A) 1.≦ ≦• Proof.

• 從公理 1 得知 P(A) 0.≧• 如果 P(A) > 1

• 則 P(Ac) < 0 →違反公理 1

• 所以 P(A) 1≦

定義定義


Theorem 5Theorem 5

• If A B, then P(A) P(B)≦• Proof.• B = A BAc

• P(B) = P(A) P(BAc )• P(BAc ) 0≧• P(A) P(B)≦

定義定義

A

BAcB


Theorem 6: Theorem 6:

• P(A B) = P(A) + P(B) – P(AB)

• Proof:• P(A B) = P(ABc) + P(AB) + P(AcB)• P(A) = P(ABc) + P(AB)• P(B) = P(AcB)+ P(AB) A B

ABC AB ACB


Theorem 6: Theorem 6:

• P(A1 A∪ 2 A∪ 3)

= P(A1) + P(A2) + P(A3)

– [ P(A1 ∩ A2 ) + P(A2 ∩ A3 )

+ P(A1 ∩ A3 ) ] + P(A1 ∩ A2 ∩ A3 )


例題例題• Suppose that 15% of the freshmen fail

chemistry, • 12% fail math, • and 5% fail both. • Suppose a first‑year student is picked at

random. Find the probability that the student failed at least one of the courses.

• P(A B) = P(A) + P(B) – P(AB)• = .15 + .12 － .05 =.22


Conditional ProbabilityConditional Probability條件機率條件機率

• P(AB) : The probability that some event A occurs given that some other event B has already occurred.

• If the probability of one event varies depending on whether a second event has occurred, the two events are said to be dependent.

定義定義


Conditional ProbabilityConditional Probability條件機率條件機率

• A : 大學畢業生• B ：年薪五萬元• P(B) 年薪五萬元的機率 P(B A) 大學畢業生年薪五萬的機率


Joint Probability TablesJoint Probability Tables聯合機率表聯合機率表

性別錄取拒絕 Total

男 3800 4700 8500

女 1600 2400 4000Total 5400 7100 12,500

是否錄取 row sum

column sum


Joint Probability TablesJoint Probability Tables聯合機率表聯合機率表

男生且被拒絕的機率 = 4700/12500 =.376

Marginal

性別錄取拒絕 Probability

男 0.304 0.376 0.68

女 0.128 0.192 0.32Marginal 0.432 0.568 1

Probability

是否錄取

A joint probability shows the probability that an observation will possess two (or more) characteristics simultaneously. Every joint probability must be a number in the closed interval [0,1] and the sum of all joint probabilities must be 1.


Marginal ProbabilityMarginal Probability

男生佔全體人數的 6

8%

P(錄取） = P( 男錄取 ) + P( 女錄取 )

= .304 + .128 =.432

Marginal


男 0.304 0.376 0.68

女 0.128 0.192 0.32Marginal 0.432 0.568 1

Probability

是否錄取


Marginal ProbabilityMarginal Probability

Marginal


男 0.304 0.376 0.68

女 0.128 0.192 0.32Marginal 0.432 0.568 1

Probability

是否錄取

如何知道組織是否有性別歧視 -- 即女生的錄取率是否顯著的低於男性？


Conditional ProbabilityConditional Probability

• 已知事件 B 發生時，事件 A 發生的條件機率：= A 與 B 的交集比上事件 B 發生的機率

AS

B

0)( provided

)(

)( )(

BP

BP

BAPBAP

定義定義



• 我們想要知道下列兩種機率是否有別？• P(拒絕男生 ) = P( 已知申請人為男性，申請人被拒絕）

• P(拒絕女生 ) = P( 已知申請人為女性，申請人被拒絕）



)(

)( )(

男生男生被拒絕男生被拒絕

P

PP

Marginal


男 0.304 0.376 0.68

女 0.128 0.192 0.32Marginal 0.432 0.568 1

Probability

是否錄取

553.680.

376.



)(

)( )(

男生男生被拒絕男生被拒絕

f

fP

553.

500,8

700,4

性別錄取拒絕 Total

男 3800 4700 8500

女 1600 2400 4000Total 5400 7100 12,500

是否錄取



)( 男生被拒絕P

Marginal


男 0.304 0.376 0.68

女 0.128 0.192 0.32Marginal 0.432 0.568 1

Probability

是否錄取

553.680.

376. )( 女生被拒絕P 600.

320.

192.

為什麼男生的拒絕率看起來好像比較高 ?

<


Multiplicative law of probabilityMultiplicative law of probability

AS

B

)(

)( )(

BP

BAPBAP )()()( BAPBPBAP

)(

)( )(

AP

BAPABP )()()( ABPAPBAP

定義定義



• 例題：已知一家電公司當日進貨 60 台電視機中有 5 台有毛病。如果第二天僅賣出兩台電視，求賣出去的兩台都有毛病的機率？

• A = {賣出的第一台有毛病 }• B = {賣出的第二台有毛病 }• A∩B= 兩台都有毛病

)()()( ABPAPBAP 0056.177

1

59

4

60

5

實例實例



• 例題：續上題，如果第二天僅賣出兩台電視，求第一台有毛病，第二台沒有毛病的機率？

• A = {賣出的第一台有毛病 }• C = {賣出的第二台沒有毛病 }• P(C|A) = 55/59

)()()( ACPAPCAP

0777.59

55

60

5

實例實例


IndependenceIndependence 獨立獨立• 如果事件 A 發生的機率與事件 B 是否發生無關，即不受事件 B 的影響，則 A 與 B 為獨立事件：

• P(A|B) = P(A)• Event A and B are independent if and only if

)()()( BPAPBAP

定義定義

0)( BAP 互斥事件注意：


IndependenceIndependence 獨立獨立• Theorem: If two events A and B are independen

t, then the events A and Bc are also independent.• Proof.

)()()( BAPAPBAP c

定義定義

AS

B

)()()( BPAPAP

)](1)[( BPAP

)()( cBPAP ∵ A and B are independent


IndependenceIndependence 獨立獨立• Approximately 30% of the sales representatives hired by a firm quit in

less than 1 year. Suppose that two sales representatives are hired and assume that the first sales representative's behavior is independent of the second sales representative's behavior.

• (a) What is the probability that both quit within a year?

• (b) Find the probability that exactly one representative quits.

)()()(quit)both ( BPAPBAPP

)()( BAPBAP cc

09.)3)(.3(.

quits) second and staysP(first stays) second and quitsP(first

quits) oneexactly (

P

例題例題

)()()()( BPAPBPAP cc 42.)3)(.7(.)7)(.3(.


Tree DiagramsTree Diagrams 例題例題

.3

.7

A

Ac

.3

.7

B

Bc

.3

.7

B

Bc

09.)3)(.3(.)( BAP

21.)7)(.3(.)( cBAP

21.)3)(.7(.)( BAP c

49.)7)(.7(.)( cc BAP


IndependenceIndependence 獨立獨立• 一對 20歲的年輕夫婦考慮參加下列退休計畫：福利金只有在活超過 70歲才能領取。假設先生活到 70歲的機率為 .6，太太活到 70歲的機率為 .7，假設兩人死亡的機率為獨立事件，如果兩人都參加該計畫，至少一個人可以領到福利金的機率為？

)()()(

lives)on least at (cc WMPWMPWMP

P

例題例題

88.)3)(.6(.)7)(.4(.)7)(.6(. die)P(both - 1 lives)on least at ( P

88.)3)(.4(.1


Sampling with and without Sampling with and without replacementreplacement

Selecting a random sample can be viewed as a process in which we sequentially obtain one observation after another. When we sample with replacement, successive outcomes are independent:

概念概念

)()()( BAPBPBAP )()()( ABPAPBAP

When we sample without replacement, successive outcomes are not independent:

)()()( BPAPBAP


Sampling without replacementSampling without replacement

租車公司有十部表面上看起來一樣的車，但其中兩部車煞車有問題，有兩人同時來租車，請問第一人租到好車，第二人租到壞車的機率？令 A={ 第一個人租到好車 } B={ 第二人租到壞車 }

例題例題

)()()( ABPAPBAP

178.9

2

10

8


Sampling with replacementSampling with replacement

十家賭場中有 7 家有逃漏稅的情形， IRS 每年抽檢一家，請問第一年抽到沒有逃漏稅的賭場，第二年抽到逃漏稅的賭場的機率為何？

例題例題

21.10

7

10

3

)()()( BPAPBAP


樣本空間分割樣本空間分割The sample space of an experiment is partitioned into k mutually exclusive and exhaustive events A1, A2, … Ak

若 A1, A2, … Ak為樣本空間 S 的部分集合，且滿足下列條件：1. A1 A∪ 2 A∪ 3… A∪ k= S

2. Ai∩Aj =

則稱 {A1, A2, … Ak} 為樣本空間 S 的一分割 (partition)

定義定義


樣本空間分割樣本空間分割若 {A1, A2, … Ak} 為樣本空間 S 的一分割 (partition) ，且 B 為 S 中的任一其他事件，則 {A1B, A2

B,… AkB} 為事件 B 的一分割。

A1 A2

A5 A3A4

B)()()(

)(

21

21

BABABA

BAAABSB

k

k

S

為互斥事件)(,),(),( 21 BABABA k

k

jj BAPBP

1

)()(

定義定義


樣本空間分割樣本空間分割

若 {A1, A2, … Ak} 為樣本空間 S 的一分割 (partition) ，且 P(Aj)>0 ，則對任何 S 中的事件 B

A1 A2

A5 A3A4

BS

k

jj BAPBP

1

)()(

)()()(

1for 0)( if

jjj

j

ABPAPBAP

k,jAP

k

jjj ABPAPBP

1

)()()(

定義定義


樣本空間分割樣本空間分割• 某公司甲乙兩董事要競選董事缺，甲股東當選之機率為 .6 ，乙股東當選的機率為 .4 。假如甲選上，公司發展新產品的機率為 .8 ，若乙選上，公司發展新產品的機率為 .3 ，求公司發展新產品的機率？（交大）

• B = 公司發展新產品• P(甲 ) = .6 P(乙 )=.4• P(B∣甲 )=.8 P(B∣乙 )=.3 • P(B) =P(甲 ) P(B∣甲 ) + P(乙） P(B∣乙 )• =.6 ×.8 + .4 ×.3 = .6

例題例題


Bayes’ TheoremBayes’ Theorem 貝氏定理貝氏定理

令 {A1, A2, … Ak} 為樣本空間 S 的一分割 (partition) ，且 P(A

j)>0 ，令 B 為 S 中的任一事件，且 P(B)>0, 則 for i=1,…k

A1 A2

A5 A3A4

BS

k

jjj

iii

ABPAP

ABPAPBAP

1

)()(

)()()(

)(

)()(

:Proof

BP

BAPBAP i

i

定義定義



k

jjj

iii

ABPAP

ABPAPBAP

1

)()(

)()()(

Posterior probability

事後機率

Prior probability

事前機率

定義定義



• 醫院研發一種新的檢驗方法，可以測出新生兒是否有某種精神疾病，從過去的經驗得知，新生兒得這種病的機率為 .003 。如果一個新生兒患有這種疾病，則檢查的結果 98% 的機率會測出陽性反應，但有 2% 的機率會呈現陰性反應。如果新生兒沒有這種疾病，則此檢驗 99%會得出陰性結果，但有 1% 的機率會得到誤差的陽性結果。

• 一新生兒接受此項檢驗，結果得到陽性反應，此新生兒的到精神疾病的機率為何？

例題例題



• （解一）• 假設有 100,000 個嬰兒接受這個檢驗。• 根據先前資訊（ prior information)，我們知道 1

00,000 個嬰兒中有 300患有此疾病（ .3%)， 99,700 為健康兒。

• 300 個患病的嬰兒中，有 294 個（ 98%)會呈現陽性反應， 6 個呈現陰性反應。

• 在 99,700健康兒中， 98,703(99%) 會呈現陰性，997 個嬰兒會呈現陽性反應。

例題例題



•（解一）

例題例題

健康狀態陽性陰性 Total

健康 997 98703 99700患有精神病 294 6 300Total 1291 98709 100000

檢驗結果

P （患有精神病︳呈陽性反應） =

294/1291 = .2277


Bayes’ TheoremBayes’ Theorem 貝氏定理貝氏定理• （解二） A1={陽性反應} A2={陰性反應}• D1={嬰兒患有精神疾病} D2={健康嬰兒} • 已知 P(D1)=.003 → P(D2)=.997 • P(A1︱D1)=.98• P(A2︱D2)=.99 → P (A1︱D2)=.01

例題例題

)()()()(

)()(

)(

)()()(

212111

111

1

11111 DAPDPDAPDP

DAPDP

AP

DAPDPADP

2277.)01)(.997(.)98)(.003(.

)98)(.003(.)( 11

ADP



.003

.997

D1 有病

D2健康

.98

.02

A1 陽

A2 陰

.01

.99

00294.)98)(.003(.))(( 111 DADP

A1 陽

A2 陰

00006.)02)(.003(.))(( 121 DADP

00997.)01)(.997(.))(( 212 DADP

98703.)99)(.997(.))(( 222 DADP

(.00294)/(.00294+.00997) =.2277



• E1 = 工廠 1 所製造的引擎• E2 = 工廠 2 所製造的引擎• A =引擎有問題• 假設 P(A E∣ 1)=.02 P(A E∣ 2)=.03• P(E1) = .40 P(E2) = .60• 任意檢查一引擎發現有瑕疵，求此引擎來自於工廠 1 的機率為何？

例題例題



.4

.6

E1

E2

.02

.98

A

Ac

.03

.97

008.)02)(.4(.))(( 11 EAEP

Ac

392.)98)(.4(.))(( 11 EAEP c

018.)03)(.6(.))(( 22 EAEP

582.)97)(.6(.))(( 22 EAEP c

(.008)/(.008+.018) =.308

A


• 某女校一年級至四年級學生人數之比率分別為 27% ， 26% ， 24% ， 23% ，在第一學期結束時據了解各年級有固定男朋友之比率為10%, 25%, 30%, 35% 。某男生於該校校慶時認識一位尚無固定男友之女孩，剖為傾心，請問她不巧是四年級老大姐的機率為？ (交大）

• P （無男友） = .27 ×.9 + .26 ×.75 + .24 ×.70 + .23 ×.65 = .7555

• P(四年級∣無男友 ) = (.23 ×.65)/.7555 = .1979

例題例題


Fundamental rule of countingFundamental rule of counting

• 若事件 A 有 n1種可能• 事件 B 有 n2種可能• 則事件 A 與事件 B 依序發生的可能共有

（ n1 ×n2）種

概念概念


Factorial NotationFactorial Notation

• Let N be a positive integer.

• The product of all integers from 1 to N is called N factorial and is denoted N!:

N!=N(N-1)(N-2)…(3)(2)(1)

We define 0! = 1

概念概念


PermutationPermutation 排列排列

• A permutation of N different things taken R at a time, denoted NPR or PN, R is an arrangement in a specific order of any R of the N things.

概念概念


PermutationPermutation 排列排列概念概念

• NP3

N N-1 N-2

n1 n2 n3



)1()1(, knnnP kn

1)1)((

1)1)(()1()1(,

knkn

knknknnnP kn

)!(

!, kn

nP kn



• 公司面試 10 名應徵者，準備錄取三人，分任三個薪資高低不同的職位，請問有多少種錄取方式？

例題例題

7208910!7

!10

)!310(

!103,10

P


CombinationCombination 組合組合• A combination of N things taken R at a time,

denoted NCR, is an arrangement of any R of these things without regard to order.

• N各物件取 R 個，但不考慮 R 之間的順序。• 請問下列何者為排列問題？何者是組合問題？

• 某職棒球隊共有 24 名球員，要選 9 人參加比賽，共有幾種選法？

• 某職籃球隊共有 12 名球員，要選出 5 人參加比賽，共有幾種選法？


CombinationCombination 組合組合

• 將 A, B, C, D四取二排列： P4,2 = 4 ×3{A, B} {B, A}{A, C} {C, A}{A, D} {D, A}{B, C} {C, B}{B, D} {D, B}{C, D} {D, C}

組合方式有幾種？

先排列後將重複的組刪除



將 A, B, C, D四取三排列： P4,3 = 4 ×3 ×2{A, B, C} {A, C, B} {B, C, A} {B, A, C} {C, A, B} {C, B, A}{A, B, D} {A, D, B} {B, D, A} {B, A, D} {D, A, B} {D, B, A}{B, D, C} {B, C, D} {C, B, D} {C, D, B} {D, C, B} {D, B, C}{A, C, D} {A, D, C} {C, A, D} {C, D, A} {D, A, C} {D, C, A}

3 個字母有 3! = 3 ×2 =6 種排列可能，這 6 種可能在組合中只能算是一種。

46

24

!33,4

P



!,

, R

PC RN

RN

R 個元素有 R!種排列方式 !)!(

!, RRN

NC RN

)!(

!

RN

N

Experiment

Documents