chapter 1 probability spaces 主講人 : 虞台文. content sample spaces and events event...

Chapter 1Probability Spaces

主講人 :虞台文

Content Sample Spaces and Events Event Operations Probability Spaces Conditional Probabilities Independence of Events Reliabilities Bayes’ Rule


Sample Spacesand

Events

Definitions Sample Spaces and Events

The set of all possible outcomes of a random

experiment is called the sample space,

denoted by , of that experiment.

An element is called a sample point.

A subset A is called an event.

Example 1

The set of all possible outcomes of a random experiment is called the sample space, denoted by , of that experiment.



The set of all possible outcomes of a random experiment is called the sample space, denoted by , of that experiment.



Tossing a die.

1 1,2,3,4,5,6 sample space of tossing a die

1 1,2,3A an event whose face number is less than 4

2 4,5,6A an event whose face number is lager than 3

3 1,3,5A an event whose face number is odd

4 2,4,6A an event whose face number is even

Notations

A

a sample space

an event

an sample point

( )P A

( )P

the probability for the occurrence of event A

P

Example 2

Example1. Tossing a die.


1 1,2,3A an event whose f ace number is less than 4

2 4,5,6A an event whose f ace number is lager than 3

3 1,3,5A an event whose f ace number is odd

4 2,4,6A an event whose f ace number is even







1. Suppose that the die in Example 1 is fair. Find the probability of each event.

2. Suppose that the die is unfair, and with

Find the probability of each event.

14(1) (6)P P 1

16(3) (4)P P 316(2) (5)P P

1( ) ?P A

2( ) ?P A 3( ) ?P A

4( ) ?P A

1( ) ?P A

2( ) ?P A 3( ) ?P A

4( ) ?P A

Example 2
















14(1) (6)P P 1

16(3) (4)P P 316(2) (5)P P

1( ) ?P A

2( ) ?P A 3( ) ?P A

4( ) ?P A

1( ) ?P A

2( ) ?P A 3( ) ?P A

4( ) ?P A

12

12

12

12

Example 2
















14(1) (6)P P 1

16(3) (4)P P 316(2) (5)P P

1( )P A

2( )P A 3( )P A

4( )P A

1( ) ?P A

2( ) ?P A 3( ) ?P A

4( ) ?P A

12

12

12

12

Example 2
















14(1) (6)P P 1

16(3) (4)P P 316(2) (5)P P

1( )P A

2( )P A 3( )P A

4( )P A

1( ) ?P A

2( ) ?P A 3( ) ?P A

4( ) ?P A

12

12

12

12

31 1 14 16 16 2

31 1 116 16 4 2

31 1 14 16 16 2 3 1 1 1

16 16 4 2

Example 3 Bulb Life (Months)

2 [0, )

1 [10, )B

2 [10,20)B

3 [0,10)B

4 (15,30]B

sample space of a bulb life

the life of a bulb is not less than 10 months

the life of a bulb is not less than 10 but less than 20 months

the life of a bulb is less than 10 months

the life of a bulb is larger than 15 but not larger than 30 months


2 [0, )

1 [10, )B

2 [10,20)B

3 [0,10)B

4 (15,30]B

sample space of a bulb life

the life of a bulb is not less than 10 months

the life of a bulb is not less than 10 but less than 20 months

the life of a bulb is less than 10 months

the life of a bulb is larger than 15 but not larger than 30 months

0 5 10 15 20 25 30 35 40 45 50


2 [0, )

1 [10, )B

2 [10,20)B

3 [0,10)B

4 (15,30]B

0 5 10 15 20 25 30 35 40 45 50

1( ) ?P B

2( ) ?P B

3( ) ?P B

4( ) ?P B

Example 4

Tossing three balanced coins

1. Write the sample space of this experiment.

2. Write the event A to denote that at least two coins land heads.

3. P(A) =?

Example 4




3. P(A) =?

, , , ,

, , ,

HHH HHT HTH HTT

THH THT TTH TTT

, , ,A HHH HHT HTH THH

12( )P A

,HHH ,HHT ,HTH ,HTT

,THH ,THT ,TTH TTT

Example 4




3. P(A) =?

, , , ,

, , ,

HHH HHT HTH HTT

THH THT TTH TTT

, , ,A HTT THT TTH TTT

12( )P A

,HHH ,HHT ,HTH ,HTT

,THH ,THT ,TTH TTT

The method to define a sample space is not unique, e.g.,

0 3 ,1 2 ,2 1 ,3 0H T H T H T H T


Events Operations

Event Operations

Intersection And

Union Or

Complement Not

Intersection () And

Venn Diagram

A B

AB

Example 5













1. A1 A3 = {1, 3}

2. A2 A4 = {4, 6}

3. A1 A2 =

The face number is less than 4 and odd.

The face number is larger than 3 and even.

A null event.

Example 5













1. A1 A3 = {1, 3}

2. A2 A4 = {4, 6}

3. A1 A2 =

The face number is less than 4 and odd.

The face number is larger than 3 and even.

A null event.

Remark: Two events A and B are

said to be mutually exclusive if A ∩

B = .

Example 5













1. A1 A3 = {1, 3}

2. A2 A4 = {4, 6}

3. A1 A2 =

1 3( ) ?P A A

2 4( ) ?P A A

1 2( ) ?P A A

13

13

0

Assume the die is fair.


2 [0, )

1 [10, )B

2 [10,20)B

3 [0,10)B

4 (15,30]B

0 5 10 15 20 25 30 35 40 45 50

1 3B B A null event.


2 [0, )

1 [10, )B

2 [10,20)B

3 [0,10)B

4 (15,30]B

0 5 10 15 20 25 30 35 40 45 50

2 4 (15,20)B B The bulb life is lager than 15 but less than 20 months


2 [0, )

1 [10, )B

2 [10,20)B

3 [0,10)B

4 (15,30]B

0 5 10 15 20 25 30 35 40 45 50

1 2 [10,20)B B 2B

Union () Or

Venn Diagram

A B

AB

Example 7













1. A1 A3 = {1, 2, 3, 5}

2. A2 A4 = {2, 4, 5, 6}

3. A1 A2 = 1

The face number is less than 4 or odd.

The face number is larger than 3 or even.

A universal event.


2 [0, )

1 [10, )B

2 [10,20)B

3 [0,10)B

4 (15,30]B

0 5 10 15 20 25 30 35 40 45 50

1 3 2B B A universal event.


2 [0, )

1 [10, )B

2 [10,20)B

3 [0,10)B

4 (15,30]B

0 5 10 15 20 25 30 35 40 45 50

2 4 [10,30]B B The bulb life is not less than 10 but not large than 30 months


2 [0, )

1 [10, )B

2 [10,20)B

3 [0,10)B

4 (15,30]B

0 5 10 15 20 25 30 35 40 45 50

1 2 [10, )B B 1B

Complement Not

Venn Diagram

AAc

{ | , }cA A A A

Example 9-1













1 4,5,6cA 2A

The face number is not less than 4.

Example 9-2 Bulb Life (Months)

2 [0, )

1 [10, )B

2 [10,20)B

3 [0,10)B

4 (15,30]B

0 5 10 15 20 25 30 35 40 45 50

1 [0,10)cB 3B

Laws of Event Algebra

A B C A B C

A B C A B C

1. Associative laws

2. Commutative laws

3. Distributive laws

A B B A

A B B A

A B C A B A C

A B C A B A C


A A

A A

4. Identity laws

5. Complementation laws

6. Idempotent laws

c

c

A A

A A

A A A

A A A


A

A

7. Domination laws

8. Absorption laws

9. De Morgan’s laws

A A B A

A A B A

c c c

c c c

A B A B

A B A B

More on De Morgan’s laws

cc

i ii i

cc

i ii i

A A

A A

c c c c

c c c c

A B C A B C

A B C A B C

c c c

c c c

A B A B

A B A B


Probability Spaces

-Field

A nonempty collection of subsets A is called -

field of a set provided that the following two

properties hold:

1. A

2.

cA A A A

an, 1, d2, i i ii iA i A A A A A

Example 10

Let = {1, 2, 3, 4, 5, 6}.

1. Let A1 = {A0 = , A1 = {1, 2, 3}, A2 = {4, 5, 6}, A3 = } .

Whether or not A1 forms a -field of ?

2. Let A2 = {A1 = {1, 6}, A2 = {2, 5}, A3 = {3, 4}}.

Add minimum number of subsets of into A2 such that A2 becomes a -field of .

The Axioms of Probability

A probability measure P on a -field of subset A of a set is a real-valued function having domain A satisfying the following properties:

1. P() = 1;2. P(A) 0 for all A A;3. If Ai A, i=1,2,. . . are

mutually disjoint then ( )i iii

P A P A

AA1A2 A3

A4

Definition Probability Space

A probability measure P on a -field of subset A of a set is a real-valued function having domain A satisfying the following properties:

1. P() = 1;2. P(A) 0 for all A A;3. If Ai A, i=1,2,. . . are

mutually disjoint then ( )i iiiP A P A

A probability space, denoted by (, A, P), is a set , a -field of subsets A, and a probability measure P defined on A.

Example 11

= {1, 2, 3, 4, 5, 6}.

A = {A0 = , A1 = {1, 2, 3}, A2 = {4, 5, 6}, A3 = }

P (A0) = 0, P (A1) = 1/2, P (A2) = 1/2, P (A3) = 1

(, A, P) A probability space?

1. P() = 1;2. P(A) 0 f or all A A;3. I f Ai A, i=1,2,. . . are




Example 11

= {1, 2, 3, 4, 5, 6}.

A = {A0 = , A1 = {1, 2, 3}, A2 = {4, 5, 6}, A3 = }

P (A0) = 0, P (A1) = 1/3, P (A2) = 2/3, P (A3) = 1






Example 11

= {1, 2, 3, 4, 5, 6}.

A = {A0 = , A1 = {1, 2, 3}, A2 = {4, 5, 6}, A3 = }

P (A0) = 1/3, P (A1) = 1/3, P (A2) = 1/3, P (A3) = 1






Example 11

= {1, 2, 3, 4, 5, 6}.

A = {A0 = , A1 = {1, 2, 3}, A2 = {4, 5, 6}, A3 = }

P (A0) = 0, P (A1) = 1/3, P (A2) = 1/3, P (A3) = 2/3






Theorem 1-1

For any event A, P(Ac) = 1 P(A).

Pf) , c cA A A A Facts:





( ) ( )c cP A A P A P A ( )P 1

( ) 1 ( )cP A P A

Theorem 1-2

For any two events A and B,

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

Pf) ( )cA B A A B

A BcA BA B

( ) ( )cB A B A B ( ) ( ) ( )cP A B P A P A B

( ) ( ) ( )cP B P A B P A B ( ) ( ) ( )cP A B P B P A B ( ) ( ) ( ) ( )P A B P A P B P A B

More on Theorem 1-2

A

B C

?( )P A B C

( ) ( ) ( )P A P B P C

( ) ( ) ( )P A B P B C P C A

( )P A B C

Theorem 1-3

1 21

( )n

i ni

P A P A A A

1

( )n

ii

P A

1

( )i ji j n

P A A

1

( )i j ki j k n

P A A A

1

1 2( 1) ( )nnP A A A

See the text for the proof.

Theorem 1-3

1 21

( )n

i ni

P A P A A A

1

( )n

ii

P A

1

( )i ji j n

P A A

1

( )i j ki j k n

P A A A

1

1 2( 1) ( )nnP A A A

S1

S2

S3

Sn

11 2 3

1

( 1)n

ni n

i

P A S S S S

1

1

( 1)n

kk

k

S

Example 12

Let A, B be two events of a sample space with P(A) = 1/2 , P(B) = 1/2, P(A B) = 2/3

1. Find the probabilities of the following events:(1) P(Ac) (2) P(Bc)

(3) P(A B) (4) P(Ac B)

(5) P(Ac B) (6) P(A Bc)

(7) P(Ac Bc) (8) P(Ac Bc).

2. Are A, B mutually exclusive?

A B1/3

1/61/6


Conditional Probabilities

Definition

Let A, B be two events such that P(A) > 0. Then the conditional probability of B given A, written P(B|A), is defined by

( )( | )

( )

P A BP B A

P A

A BA B

If P(A) = 0, then P(B|A) is undefined.

Example 13

Toss three balanced coins. Let A denote the event that two coins land heads, B denote that the first coin lands heads. Find P(A), P(B), P(B|A), and P(A|B).

Toss three balanced coins. Let A denote the event that two coins land heads, B denote that the first coin lands heads. Find P(A), P(B), P(B|A), and P(A|B).

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

, ,A HHT HTH THH

, , ,B HHH HHT HTH HTT

,A B HHT HTH

( ) 3 / 8P A

( ) 1/ 2P B ( ) 1/ 4P A B

( ) 1/ 4 2( | )

( ) 3 / 8 3

P A BP B A

P A

( ) 1/ 4 1( | )

( ) 1/ 2 2

P A BP A B

P B

By Product

( )( ) 0

( )( | )

( ) 0

P A BP A

P AP B A

undefined P A

( )( ) 0

( )( | )

( ) 0

P A BP B

P BP A B

undefined P B

( | ) ( ) ( ) 0

( ) ( | ) ( ) ( ) 0

0

P A B P B P B

P A B P B A P A P A

otherwise


Independence of Events

Definition

Two events A and B are independent if and only if

( ) ( ) ( )P A B P A P B

A B

Example 14

Toss one balanced coin three times. Let A denote

the event that the first two tosses land heads, B

denote that exactly one toss lands heads, and C

denote that the third toss lands tails.

1. Are A, B independent?

2. Are A, C independent?

3. Are B, C independent?

Toss one balanced coin three times. Let A denote

the event that the first two tosses land heads, B

denote that exactly one toss lands heads, and C

denote that the third toss lands tails.

1. Are A, B independent?

2. Are A, C independent?

3. Are B, C independent?

Example 14

Toss one balanced coin three times. Let A denote the event that the first two tosses land heads, B denote that exactly one toss lands heads, and C denote that the third toss lands tails.1. Are A, B independent?2. Are A, C independent?3. Are B, C independent?

Toss one balanced coin three times. Let A denote the event that the first two tosses land heads, B denote that exactly one toss lands heads, and C denote that the third toss lands tails.1. Are A, B independent?2. Are A, C independent?3. Are B, C independent?

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

,A HHH HHT

, ,B HTT THT TTH

, , ,C HHT HTT THT TTT

A B

{ }A C HHT

,B C HTT THT

( ) 1/ 4P A

( ) 3 / 8P B

( ) 1/ 2P C

( ) 0P A B

( ) 1/ 8P A C

( ) 1/ 4P B C

( ) ( )P A P B

( ) ( )P A P C

( ) ( )P B P C

Theorem 1-4

A BAc BA Bc

Ac Bc

Theorem 1-5

A B P(B|A)=P(B)

( )( | )

( )

P A BP B A

P A

( ) ( )

( )

P A P B

P APf)

Definition

n events A1, A2, . . . , An are said to be

mutually independent if and only if for any

different

k = 2, . . . , n events satisfy

1 2, , , kA A A

1 2 1 2( ) ( ) ( ) ( )k kP A A A P A P A P A

Example 15

Given the following Venn Diagram, are events A,

B, C mutually independent?

A

C

B0.06 0.24 0.18

0.060.24

0.14

0.06

A

C

B0.06 0.24 0.18

0.060.24

0.14

0.06

P(A) = 0.6, P(B) = 0.8, P(C) = 0.5

P(ABC) = 0.24 = P(A)P(B)P(C)

P(AB) = 0.48 = P(A)P(B)

P(AC) = 0.30 = P(A)P(C)

P(BC) = 0.38 P(B)P(C)

Remarks

1. P(A1 A2 · · · An) = P(A1)P(A2) · · · P(An) does not imply that A1, A2, . . . , An are pairwise independent (Example 15).

2. A1, A2, . . . , An are pairwise independent does not imply that they are mutually independent.

Example 16

Toss two dice. Let

A = “The 1st die is 1,2, or 3”

B = “The 2nd die is 4,5, or 6”

C = “The sum of two dice is 7”

Show that A, B, C are pairwise independent but not m

utually independent.

Toss two dice. Let

A = “The 1st die is 1,2, or 3”

B = “The 2nd die is 4,5, or 6”

C = “The sum of two dice is 7”

Show that A, B, C are pairwise independent but not m

utually independent.


Reliabilities

Reliabilities

Reliability of a component Let Ri denote the probability of a component in a system which is functioning properly (event Ai).

We will assume that the failure events of components in a system are mutually independent.

Reliabilities of systems 1. series systems — The entire system will fail if any one of its component fails.2. parallel systems — The entire system will fail only if all its component fail.

Reliability of a Series System (Rss)

C1C1 C2

C2 CnCnA B

("The series system functions properly")ssR P

1 2( )nP A A A

1 2( ) ( ) ( )nP A P A P A

1

n

ii

R

Reliability of a Parallel System (Rps)

C1C1

C2C2

CnCn

A B...

("The parallel system functions properly")psR P

1 21 ( )c c cnP A A A

1

1 (1 )n

ii

R

1 21 ( ) ( ) ( )c c cnP A P A P A

1 21 [1 ( )][1 ( )] [1 ( )]nP A P A P A

Example 17

Consider the following system.

C1C1

C2C2

C3C3A B

Let R1 = R2 = R3 = 0.95.Find the reliability of the system.

Example 17

Consider the f ollowing system.

C1C1

C2C2

C3C3A B



C1C1

C2C2

C3C3A B



C1C1

C2C2

C3C3A B



C1C1

C2C2

C3C3A B


1 2 3( )R P A A A

1 3 2 3( ) ( )P A A A A

1 3 2 3 1 3 2 3( ) ( ) ( ) ( )P A A P A A P A A A A

1 3 2 3 1 2 3( ) ( ) ( )P A A P A A P A A A

1 3 2 3 1 2 3( ) ( ) ( ) ( ) ( ) ( ) ( )P A P A P A P A P A P A P A

1 3 2 3 1 2 3R R R R R R R 2 2 30.95 0.95 0.95

0.947625


Bayes’ Rule

A Story The umbrella must have been made in Taiwan

Event Space A Partition of Sample Space

日德法

義台

B1 B2

B3

B4

B5

, i jB B i j

1

n

ii

B

1{ , , }nB B a partition of

Event Interests Us (A)

日德法

義台

B1 B2

B3

B4

B5

, i jB B i j

1

n

ii

B

1{ , , }nB B a partition of

A

Preliminaries

日德法

義台

B1 B2

B3

B4

B5 APrior Probabilities:

( ), 1, ,iP B i n

Likelihoods:

( | ), 1, ,iP A B i n

Law of Total Probability

日德法

義台

B1 B2

B3

B4


( ), 1, ,iP B i n

Likelihoods:

( | ), 1, ,iP A B i n

1 1( ) ( | ) ( ) ( | ) ( )n nP A P A B P B P A B P B

Law of Total Probability

日德法

義台

B1 B2

B3

B4


( ), 1, ,iP B i n

Likelihoods:

( | ), 1, ,iP A B i n

1

( ) ( | ) ( )n

j jj

P A P A B P B

Goal: Posterior Probabilities

日德法

義台

B1 B2

B3

B4


( ), 1, ,iP B i n

Likelihoods:

( | ), 1, ,iP A B i n

1

( ) ( | ) ( )n

j jj

P A P A B P B

Goal: ( | ) ,? 1, ,iP B A i n

Bayes’ Rule

Given( ), 1, ,iP B i n ( | ), 1, ,iP A B i n

1

( ) ( | ) ( )n

j jj

P A P A B P B

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

( )i i

i

P A B P BP B A i n

P A

Example 18

Suppose that the population of a certain city is 40% male and 60%

female. Suppose also that 50% of the males and 30% of the females

smoke. Find the probability that a smoker is male.

Suppose that the population of a certain city is 40% male and 60%

female. Suppose also that 50% of the males and 30% of the females

smoke. Find the probability that a smoker is male.

M : A selected person is maleW : A selected person is femaleS : A selected person who smokes

Define

( ) 0.4P M

We are givenWe are given

( | ) ( )( | )

( )

P S M P MP M S

P S

( ) ( | ) ( ) ( | ) ( )P S P S M P M P S W P W 0.5 0.4 0.3 0.6 0.38

0.5 0.4

0.38

0.526

?( | )P M S

( ) 0.6P W ( | ) 0.5P S M ( | ) 0.3P S W

M

W

S

NS

S

NS

0.4

0.6

0.5

0.5

0.3

0.7

Example 19

Consider a binary communication channel. Owing to noise, error

may occur.

For a given channel, assume a probability of 0.94 that a transmitted

0 is correctly received as a 0 and a probability of 0.91 that a

transmitted 1 is received as a 1. Further assumed a probability of

0.45 of transmitting of a 0. Determine

1. Probability that a 1 is received.


3. Probability that a 1 was transmitted, given that a 1 was received.


5. Probability of an error.


may occur.










Example 19


may occur.











may occur.










0

1

0

1

0.94

0.06

0.91

0.09

0.45

0.55

T0 : A 0 is transmittedT1 : A 1 is transmittedR0 : A 0 is receivedR1 : A 1 is received

Define

T0 R0

T1 R1

1( ) ?P R

0( ) ?P R 1 1 ?( | )P T R

0 0 ?( | )P T R ?(" ")P Error

Example 19

0

1

0

1

0.94

0.06

0.91

0.09

0.45

0.55

T0 R0

T1 R1

1 ?1. ( )P R

0 ?2. ( )P R

1 13. ( | ) ?P T R

0 04. ( | ) ?P T R

5. (" ") ?P Error

1 1 1 1 0 0( | ) ( ) ( | ) ( )P R T P T P R T P T 0.91 0.55 0.06 0.45 0.5275

1 0.5275 0.4725

1 1 1

1

( | ) ( )

( )

P R T P T

P R0.91 0.55

0.5275

0.9488

0 0 0

0

( | ) ( )

( )

P R T P T

P R0.94 0.45

0.4725

0.8952

0 1 1 0( ) ( )P T R P T R 1 0 0 0 1 1( | ) ( ) ( | ) ( )P R T P T P R T P T

0.06 0.45 0.09 0.55 0.0765

Example 20

There are n boxes each contains a white and b black balls. Now randomly select one ball from the 1st box to put it into the 2nd box, then randomly select one ball from the 2nd box to put it into the 3r

d box, . . . This procedure is continued. Let Wi and Bi denote the ev

ents that the chosen ball from the ith box being white and black, respectively. Determine

1. P(Wn) = ?;

2. P(Wn|W1) = ?;

3. limn→∞P(Wn|W1) = ?




1. P(Wn) = ?;

2. P(Wn|W1) = ?;

3. limn→∞P(Wn|W1) = ?

Example 20




1. P(Wn) = ?;

2. P(Wn|W1) = ?;

3. limn→∞P(Wn|W1) = ?




1. P(Wn) = ?;

2. P(Wn|W1) = ?;

3. limn→∞P(Wn|W1) = ?

ab

ab

ab

ab

ab

1 2 3 n1 n

. . .

Example 20 1. P(Wn) = ?;2. P(Wn|W1) = ?;3. limn→∞P(Wn|W1) = ?

1. P(Wn) = ?;2. P(Wn|W1) = ?;3. limn→∞P(Wn|W1) = ?

ab

ab

ab

ab

ab

1 2 3 n 1 n

. . .abab

abab

abab

abab

abab

1 2 3 n 1 n

. . .

1 1 1 1( ) ( | ) ( ) ( | ) ( )n n n n n n nP W P W W P W P W B P B

1( )a

P Wa b

1

1

a

a b

1

a

a b

Define pi=P(Wi).

1np 11 np

1p

np

1

1

1

1 1n n

ap

a ba

p pa b a b

1

1

1

1 1n n

ap

a ba

p pa b a b

np



ab

ab

ab

ab

ab

1 2 3 n 1 n

. . .abab

abab

abab

abab

abab

1 2 3 n 1 n

. . .

1

1

1

1 1n n

ap

a ba

p pa b a b

1

1

1

1 1n n

ap

a ba

p pa b a b

np

1n np cp d

1 2n np cp d

2 3n np cp d

3 2p cp d

2 1p cp d

.

.

.



ab

ab

ab

ab

ab

1 2 3 n 1 n

. . .abab

abab

abab

abab

abab

1 2 3 n 1 n

. . .

1

1

1

1 1n n

ap

a ba

p pa b a b

1

1

1

1 1n n

ap

a ba

p pa b a b

np

1n np cp d

1 2n np cp d

2 3n np cp d

3 2p cp d

2 1p cp d

.

.

.

21 2n ncp c p cd

2 3 22 3n nc p c p c d

3 2 33 2

n n nc p c p c d 2 1 2

2 1n n nc p c p c d

11

1

(1 )

1

nn

n

d cp c p

c

21 2n ncp c p cd

2 3 22 3n nc p c p c d

3 2 33 2

n n nc p c p c d 2 1 2

2 1n n nc p c p c d



ab

ab

ab

ab

ab

1 2 3 n 1 n

. . .abab

abab

abab

abab

abab

1 2 3 n 1 n

. . .

1

1

1

1 1n n

ap

a ba

p pa b a b

1

1

1

1 1n n

ap

a ba

p pa b a b

np

1n np cp d

.

.

.

11

1

(1 )

1

nn

n

d cp c p

c

1

1

11

1 1111 1

1

n

n

n

aa b a ba

pa b a b

a b

1

1

11

1 1111 1

1

n

n

n

aa b a ba

pa b a b

a b

a

a b



ab

ab

ab

ab

ab

1 2 3 n 1 n

. . .abab

abab

abab

abab

abab

1 2 3 n 1 n

. . .

1

1

1

1 1n n

ap

a ba

p pa b a b

1

1

1

1 1n n

ap

a ba

p pa b a b

np

1

1

11

1 1111 1

1

n

n

n

aa b a ba

pa b a b

a b

a

a b

11

1

(1 )

1

nn

n

d cp c p

c

1

1( | )nP W W

11

1

nb

a b a b

1

1

1( | )

1

n

n

b aP W W

a b a b a b



ab

ab

ab

ab

ab

1 2 3 n 1 n

. . .abab

abab

abab

abab

abab

1 2 3 n 1 n

. . .

np

1

1

1( | )

1

n

n

b aP W W

a b a b a b

1lim ( | )nn

aP W W

a b

( )nP W

chapter 1 probability spaces 主講人 : 虞台文. content sample spaces and events event...

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