slide1 n.
Download
Skip this Video
Loading SlideShow in 5 Seconds..
Chapter 4: Probability (Cont.) PowerPoint Presentation
Download Presentation
Chapter 4: Probability (Cont.)

Loading in 2 Seconds...

play fullscreen
1 / 11

Chapter 4: Probability (Cont.) - PowerPoint PPT Presentation


  • 105 Views
  • Uploaded on

Chapter 4: Probability (Cont.). In this handout: Total probability rule Bayes’ rule Random sampling from finite population Rule of combinations. Partitions. Definition: A collection of events {S 1 , S 2 , …, S n } is a partition of a sample space S if

loader
I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.
capcha
Download Presentation

PowerPoint Slideshow about 'Chapter 4: Probability (Cont.)' - anka


Download Now An Image/Link below is provided (as is) to download presentation

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 - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript
slide1

Chapter 4: Probability (Cont.)

  • In this handout:
  • Total probability rule
  • Bayes’ rule
  • Random sampling from finite population
  • Rule of combinations
partitions
Partitions
  • Definition: A collection of events {S1, S2, …, Sn} is a partition of a sample space S if

1. S = S1 S2 …  Sn

2. S1, S2, …, Sn are mutually exclusive events.

  • Example:

Recall the example of rolling two dice.

Define S2={sum = 2}, S3={sum = 3}, …, S12={sum = 12}.

Then {S2, S3, …, S12}is a partition of the sample space.

total probability rule
Total probability rule

Assume the set of events {S1, S2, …, Sn} is a partition of a sample space S. Assume P(Si) > 0 for every i, 1 ≤ i ≤ n.

Then for any event A,

S1

S2

S3

S4

AS1

AS2

AS3

AS4

total probability rule1
Total probability rule

Example: A diagnostic test for a certain disease is known to be 95% accurate. It is also known from previous data that only 1% of the population has the disease. What is the probability that a person chosen at random will be tested positive?

Solution: Let

T+ denote the event that a person is tested positive;

T- denote the event that a person is tested negative;

D denote the event that a person has the disease.

Then

Applying the formula of total probability:

bayes rule
Bayes’ rule

Assume the set of events {S1, S2, …, Sn} is a partition of a sample space S. Assume P(Si) > 0 for every i, 1 ≤ i ≤ n.

Fix any event A. Then for any given j, 1 ≤ j ≤ n,

Example (cont.): If a person tested positive then what is the probability he has the disease?

slide6

Random sampling from finite population

When sampling from a large population, listing all the possible choices becomes a tedious job. Then a counting rule is very useful:

Note:C(N,r) is another way to denote “N choose r”.

Example: Randomly select 4 members from a group of 11 students to work on a project. How many distinct 4-person teams can be chosen?

slide7

Random sampling from finite population

Example(cont.):

  • The probability that students A,B,C,D are chosen to work on the project is 1/330.
  • Suppose the group consists of 5 juniors and 6 seniors. How many samples of 4 have exactly 3 juniors?

Think of selecting a sample as a 2-step process:

    • 1 senior can be chosen 6 different ways;
    • 3 juniors can be chosen C(5,3) different ways.

Thus, the whole sample can be chosen 6∙C(5,3) different ways.

examples on combinations
Examples on Combinations
  • Suppose that 3 cars in a production run of 40 are defective.

A sample of 4 is to be selected to be checked for defects.

Questions:

1) How many different samples can be chosen?

2) How many samples will contain

exactly one defective car?

3) What is the probability that a randomly chosen sample will contain exactly one defective car?

4) How many samples will contain

at least one defective car?

Solution:

1) C(40, 4) = (40∙39∙38∙37) / (1∙2∙3∙4) = 91,390

examples on combinations1
Examples on Combinations

2) How many samples will contain

exactly one defective car?

Think of selecting a sample as a 2-step process:

Step 1: Choose the defective cars;

Step 2: Choose the good cars.

There are C(3,1) ways to choose 1 defective car.

There are C(37,3) ways to choose 3 good cars.

Then the number of samples containing exactly 1 defective car is

C(3,1) ∙ C(37,3) = 3∙(37∙36∙35) / (1∙2∙3) = 23,310

3) What is the probability that a randomly chosen sample will contain exactly one defective car?

The probability = 23,310 / 91,390 = .255

examples on combinations2
Examples on Combinations

4) How many samples will contain at least one defective car?

There are 2 ways to answer this question,

either by the addition rule

or by the difference rule.

The solution by the difference rule is less intuitive but shorter. Let’s solve it by the difference rule.

(# of samples with ≥1 defective cars)

= (all possible samples)

– (# of samples with no defective cars) .

But (# of samples with no defective cars)

= (# of samples with only good cars) = C(37,4)=66,045

Thus, (# of samples with ≥1 defective cars)

= 91,390 – 66,045 = 25,345