- 44 Views
- Uploaded on
- Presentation posted in: General

Probability theory

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Probability theory

The department of math of central south university

Probability and Statistics Course group

- Classical Probability Model
- supposeΩis the sample space of test E，if

- 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

NOTE:

(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: ①findout the test and sample point; ②count the sample points number of samples space and the sample points number of random event ; ③list and calculate.

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.

Answer：suppose

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

B shows the event

“the twice number is 3。”

so

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 both defective}，B={the two have get have one authentic ,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: ① if not changed back to taking it twice? ② be changed if one of the two samples do?

（2）the sample number of B 4×2+2×4=16，

so：P(B)=16/36=4/9

（3）the sample number of C 62-2×2=32，

Example3 A bag has 6 balls, 4 white balls, 2 red. Take the ball from the pocket twice, each time a random check. Taking into account both the ball: Back a sample: for the first time taking a ball, observed after the color back into the bag, mix after taking a ball. Do not back a sample: for the first time do not get a ball back into the bag for the second time from the rest of the ball in a ball again. For:

8

5

9

6

1

4

2

3

10

7

1）the probability of the two were taken are both white

(2) The probability to get the ball two different colors of; 3) the probability to take two balls at least one white ball

Answer：Get two goals from the bag, each is 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 at least.”。

Have taken back:

Without taking back:

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?

Answer：suppose A shows have one boy at least，H shows the some child is boy。

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

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

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

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

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

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

Example 6 A reception center in a week, had received 12 visitors, all known to receive this 12 times in Tuesday and Thursday. Asked whether the reception may be inferred that the time is required?

answer：Assuming no time for the reception of the reception center provides the visitors in the office one day a week to go in reception center, and so is possible, then the 12th to receive visitors are Tuesday, Thursday, with a probability of:

212/712=0.0000003，that is 3/10000000.

Example 7 there are n persons,and everyone has the same probabilityto be assigned to any room to a living of N rooms. ( )，Seek the probability of the following events: (1) designated n rooms have their own a person living; (2) have n just rooms, all of which lives one person .

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

In the second question in, n-room can be room N arbitrary selection, which have a total , n pairs of selected rooms, according to the foregoing discussion we can see there are n! Species distribution, so there are exactly n rooms All of which live in a person's probability of

Additive principle: there are n-type approach to complete one thing.for the i-type category there are ways mi species specific methods,.The total to

complete this matter have

methods.

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

Added: combination of notation

Whole array:

Repeated array:take m elements from n different elements ,that can repeatedly, the m elements can line up a row.the kinds of the row is

.

Array:take m elements from n different elements(without taking back ), according to a certain degree of order lined up a different array,A total of array is

Portfolio: take m elements from n different elrments(without taking back) ,the m elements line up a row ,The kinds of the row are

Repeated Portfolio: take one elements each time from n different elements,and take back,then 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

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

A short break to continue