Probability theory

Probability theory

The department of math of central south university

Probability and Statistics Course group

(1)Classical Probability Model

supposeΩis the sample space of test E ， if

① （ The limited nature ） Ωonly contains limited sa

mple;

② （ the nature of same probability ） Each of the ba

sic events have same possibility; E is said classic

al sample.

§§1.3 Classical Probability

(2) Estimate the probability of classical-definition, E for the

classical estimate, Ω for the

      sample space of E, A for any event,.The definition of

the probability of events A

( )( )

( )

n A kP A

n n

(1) the method to determine the type of classical sample (limited, and so on and over);

   (2) the calculation of the probability of classical steps:          ① find out the test and sample point;

          count the sample points number of samples space and ②the sample points number of random event ;                 list and calculate.③

NOTE:

A shows the event of“add the twice is 8”

B shows the event

Answer ： suppose

Example 1. throwing a dice twice, try to find the following for the probability of events: (1) add the twice number,is 8 (2)the twice number is 3.

“ the twice number is 3 。”

(6,2)}(5,3),(4,4),(3,5),{(2,6),A

(6,3)}(5,3),(4,3),(3,3),(2,3),{(1,3),B

36

5)( AP

6

1

36

6P(B)

so

(6,6)}(6,1),,(2,6),,(2,1),(1,6),,(1,2),{(1,1),Ω

so

Example2 the box has 6 light bulbs, 2 defective 4

authentic, take twice with having taken back , each

time taking a test for the probability of the following

events: （ 1 ） the two have get are both defective;

（ 2 ） the two have get ,one is fefective,one is auhentic

（ 3 ） the two have get have one authentic at least.

Answer ： supposeA ={the two have get are bot

h defective} ， B={the two have get have one aut

hentic ,one defective}, C={the two have get have

one authentic at least}.

（ 1 ） the total of sample is62 ， the sample number of A is 22 ，

so ： P(A)=4/36=1/9

P(C)=32/36=8/9

Thoughts: if not changed back to taking ①it twice?        be changed if one of the two ②samples do?

（ 3 ） the sample number of C 62-2×2=32 ，

so ： P(B)=16/36=4/9

（ 2 ） the sample number of B 4×2+2×4=16 ，

Example3 A bag has 6 balls, 4 white balls, 2 red. Take

the ball from the pocket twice, each time a random che

ck. Taking into account both the ball:

Back a sample: for the first time taking a ball, observe

d after the color back into the bag, mix after taking a b

all.

Do not back a sample: for the first time do not get a bal

l back into the bag for the second time from the rest of

the ball in a ball again. For:

1 ） the probability of the two were taken are b

oth white

(2) The probability to get the ball two different

colors of;

3) the probability to take two balls at least one

white ball

2 34

79

10

861

5

Answer ： Get two goals from the bag, each i

s a fundamental incident.

Suppose A= “the two have get are both white

balls ” ，

B= “the two have get have same colour ” ，

C= “the two have get have one white ball a

t least.” 。

Have taken back:

444.06

4)(

2

2

AP

556.06

24)( 2

22

BP

889.06

21)(1)( 2

2

CPCP

Without taking back:

26

24

C

CP(A) 2

6

22

24

C

CCP(B)

26

22

C

C1)CP(1P(C)

Example 4 A family has three children .The

probability of having each child is male or female is

same, how many the probability of having one boy at

least?

N(S)={HHH,HHT,HTH,THH,HTT,TTH,THT,TTT}

N(A)={HHH,HHT,HTH,THH,HTT,TTH,THT}

8

7

)(

)()(

SN

ANAP

Answer ： suppose A shows have one boy at le

ast ， H shows the some child is boy 。

Example5 take any one number from natural number 1-

200;

(1) the probability to obtain a number that can be divide

by 6;

(2) the probability to obtain the number can be divide by

8;

(3) the probability to obtain can be divisible by 6 and 8

answer:N(S)=200, N(1)=[200/6]=33,

N(2)=[200/8]=25, N(3)=[200/24]=8

The probability of (1),(2),(3) :33/200,1/8,1/25

Example 6 A reception center in a week, had received 1

2 visitors, all known to receive this 12 times in Tuesday a

nd Thursday. Asked whether the reception may be inferr

ed that the time is required?answer ： Assuming no time for the reception of the recep

tion center provides the visitors in the office one day a wee

k to go in reception center, and so is possible, then the 12t

h to receive visitors are Tuesday, Thursday, with a proba

bility of:

212/712=0.0000003 ， that is 3/10000000.

Example 7 there are n persons,and everyone has the s

ame probabilityto be assigned to any room to a livi

ng of N rooms. ( ) ， Seek

the probability of the following events: (1) designat

ed n rooms have their own

a person living; (2) have n just rooms, all of which

lives one person .

n N

Answer :Because everyone has N rooms

available, so the kind of living

is , such as they have same possibility.

In the first issue, n designated rooms have a

living person, which could total number of

individuals n the whole array of n!, Then

nN

1

!n

nP

N

In the second question in, n-room can be room N arbitra

ry selection, which have a total , n pairs of selecte

d rooms, according to the foregoing discussion we can s

ee there are n! Species distribution, so there are exactly

n rooms All of which live in a person's probability of

N

n

2

!!

( )!n n

Nn

n NP

N N N n

Added: combination of notation

Multiplication principle: there is n steps to complete one

thing, the i steps have approachs, then the total

of ways to complete this matter is

n

iim

1

im

n

iim

1

Additive principle: there are n-type approach to complet

e one thing.for the i-type category

there are ways mi species specific methods,.The total to

methods.complete this matter have

Array:take m elements from n different elements(wi

thout taking back ), according to a certain degree of

order lined up a different array,

A total of array is

)1()2)(1( mnnnnPmn

Whole array: !nPnn

Repeated array:take m elements from n different el

ements ,that can repeatedly, the m elements can lin

e up a row.the kinds of the row is mn.

Portfolio: take m elements from n different elr

ments(without taking back) ,the m elements lin

e up a row ,The kinds of the row are

)!(!!

mnmn

m

n

10

1!0

n和

r

rn 1

Repeated Portfolio : take one elements each t

ime from n different elements,and take back,t

hen take one.taking r times line up a portfolio,

is said to repeated portfolio

.the number of repeated portfolio is

For example: two groups have both 50 products, of

which both have 5 defective from these two

groups .Take one product from both groups.

(1) the number of species that the two products are

not defective?

(2)the number of species that only have one defective?

Answer: (1) use multiplication principle, the results for the

2145

145 45. CC

(3) Additive combination of theory and principle of a multiplication method for the election:

4504552.. 15

145

145

15 CCCC

A short break to continue