Chapter 4: Probability (Cont.)

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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

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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 {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

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 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

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?

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?

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
• 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 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 Combinations

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

There are 2 ways to answer this question,

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