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## Chapter 4

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### Chapter 4

Probability and Counting Rules

Sample Spaces and Probability

- Probability experiment– a chance process that leads to well-defined results called outcomes (e.g. flipping a coin, drawing a card).
- Outcome – the result of a single trial of a probability experiment.
- Sample space– the set of all possible outcomes of a probability experiment.

Charts and Tree Diagrams

- Most common ways to display a sample space is either through a CHART or a TREE DIAGRAM.
- Chart – organized table of possible outcomes.
- Tree Diagram– uses line segments coming from starting point to each option of outcome.

Tree Diagram

- Use a tree diagram to find the sample space for the gender of three children in a family.

Classical Probability

- Assumes that all outcomes in the sample space are equally likely to occur.
- The probability of any event E is found by dividing the number of outcomes in E by the total number of outcomes in the sample space [formula: P(E) = n(E) / n(S)]

Example 1 (classical probability)

- For a card drawn from an ordinary deck, find the probability of getting a queen

OR P(queen)?

- There are 4 queens in a deck and 52 cards in a deck so,

P(queen) = 4/52 = 1/13 = .077 = 7.7%

Example 2 (classical probability)

- If a family has three children, find the probability that all the children are girls

or P(all three children girls) or P(GGG)?

- There is only one option that is all girls and eight possible outcomes in the sample space so,

P(GGG) = 1/8 = 0.125 = 12.5%

Example 3 (classical probability)

- Choose one of the fifty states at random.
- What is the probability that it begins with M?
- Michigan, Massachusetts, Maryland, Minnesota, Mississippi, Missouri, Montana, Maine
- = 8/50 = 4/25 = 0.16 = 16%
- What is the probability that it DOESN’T begin with a vowel?
- = 38/50 = 19/25 = 0.76 = 76%

Events and Complements

- Event- set of outcomes of a probability experiment.
- Complement of an event – the set of outcomes NOT included in the event.

Probability Rule #1

- The probability of any event E or P(E) is a number (either a fraction or decimal) between and including 0 and 1.
- Denoted: 0 ≤ P(E) ≤ 1

Probability Rule #2

- If an event E cannot occur (i.e. the event contains no members in the sample space), its probability is 0.
- P(E) = 0 if event cannot occur.

Probability Rule #3

- If an event E is certain, then the probability is 1.
- P(E) = 1 if event is certain to occur.

Probability Rule #4

- The sum of the probabilities of all of the outcomes in a sample space is 1.
- All P(E) in one sample space add to 1.00 or 100%.

Example (complement rule)

- Sixty-nine percent of adults favor gun licensing in general. Choose one adult at random. What is the probability that the selected adult doesn’t believe in gun licensing?
- NOTE: 69% = 0.69
- P(adult doesn’t believe in gun licensing)

= 1 – 0.69 = 0.31 = 31%

Mutually Exclusive (ME) Events

- Mutually exclusive events – two events that CANNOT occur at the same time (i.e. no outcomes in common).

Example (mutually exclusive)

- A single die is rolled. Which of the following events are mutually exclusive?
- Getting an odd number and getting an even number.
- Getting 3 and an odd number.
- Getting an odd number and a number less than 4.
- Getting a number greater than 4 and a number less than 4.

Example (cont.)

- a and d ARE mutually exclusive by definition. That means that the two outcomes cannot occur at the same time.
- b and c are NOT mutually exclusive because they can occur at the same time.

ME Addition Rule

- When two events A and B are mutually exclusive, the probability that A or B will occur is

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

Example (ME Addition)

- At a political rally, there are 20 Republicans, 13 Democrats, and 6 Independents. If a person is selected at random find the probability that he or she is either Democrat OR Independent.
- Find using ME addition
- 13/39 + 6/39 = 19/39 = 0.487 = 48.7%
- Find using complement rule
- 1 – 20/39 = 39/39 – 20/39 = 19/39 = 0.487 = 48.7%

Example (Not ME)

- A single card is drawn form a deck. Find the probability that it is a king or a club.

P(king) = 4/52 & P(club) = 13/52

But what about the king of clubs?

Example (cont.)

- We counted the king of clubs in both groups so we need to subtract off the P(king and clubs).
- So, the answer is:

P(king or club) = P(king) + P(club) – P(king and club)

= 4/52 + 13/52 – 1/52

= 16/52 = 4/13 = 0.308 = 30.8%

Independent Events

- Independent events – two events A and B with the fact that A occurring does NOT affect the probability of B occurring.
- E.g. a coin is tossed then a die is rolled.

Multiplication Rule #1

- When two events are independent, the probability of both occurring is:

P(A and B) = P(A) ∙ P(B)

Example (Multiplication Rule #1)

- A Harris poll found that 46% of Americans say they suffer great stress at least once a week. If three people are selected at random, find the probability that all three will say that they suffer great stress at least once a week.
- Let S = stress.
- P(S and S and S) = P(S) ∙ P(S) ∙ P(S) = (.46) (.46) (.46)

= 0.097 = 9.7%

Dependent Events

- Dependent events – the outcome or occurrence of the first event affects the outcome or occurrence of the second event in such a way that the probability is changed.
- E.g. drawing a card from a deck and not replacing it and drawing another card.

Conditional Probability

- Conditional probability (P(B|A)) – the probability that event B occurs after event A has already occurred.
- E.g. drawing a card from a deck and not replacing it and drawing another card.
- Sometimes you can figure out P(B|A) given the circumstances but other times you need a formula

P(B|A) = P(A and B) / P(A)

Example (Conditional Probability)

- The probability that Sam parks in a no-parking zone AND gets a parking ticket is 0.06, and the probability that Sam cannot find a legal parking space and has to park in the no-parking zone is 0.20. On Tuesday, Sam arrives and has to park in a no-parking zone. Find the probability that he will get a parking ticket.

Example cont.

- Let

N = parking in a no-parking zone

T = getting a ticket

- We know: P(N and T) = 0.06 and P(N) = 0.20

P(T|N) = P(N and T) / P(N) = 0.06/0.20

= 0.30 = 30%

Multiplication Rule #2

- When two events are dependent, the probability of both occurring is:

P(A and B) = P(A) ∙ P(B|A)

Example (Multiplication Rule #2)

- The World Wide Insurance Company found that 53% of the residents of a city had homeowner’s insurance (H) with the company. Of these clients, 27% also had automobile insurance (A) with the company. If a resident is selected at random, find the probability that the resident has both homeowner’s AND automobile insurance with World Wide Insurance Company.
- P (H and A) = P(H) ∙ P(A|H) = (.53)(.27) = .1431

Counting Rules – (Use your calculator!)

- Factorials (0! = 1)

n! = n(n-1)(n-2) ∙∙∙ 1

- Permutations – n objects in a specific order r at a time.

nPr = n!/(n-r)!

- Combinations – n objects in no distinct order r at a time.

nCr = n!/(n-r)!r!

Example (Factorials)

- Suppose a business owner has a choice of five locations in which to establish her business. She decides to rank each location according to certain criteria, such as price of the store and parking facilities. How many different ways can she rank the five locations?
- 5! = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120 different possible rankings (5 for the first, 4 for the second, etc.)

Example (Permutations)

- How many different ways can a chairperson and an assistant chairperson be selected for a research project if there are seven scientists available?
- Order is important and there are7 things chosen 2 at a time
- nPr = n!/(n-r)! = 7P2 = 7! / 5! = 42

Example 1 (Combinations)

- A bicycle shop owner has 12 mountain bicycles in the showroom. The owner wishes to select 5 of them to display at a bicycle show. How many different ways can a group of 5 be selected?
- Order is not important and there are 12 things chosen 5 at a time.
- nCr = n!/(n-r)!r! = 12C5 = 12! / 7!5! = 792

Example 2 (Combinations)

- In a club there are 7 women and 5 men. A committee of 3 women and 2 men is to be chosen. How many different possibilities are there?
- Need to choose men from the men and women from the women and multiply.
- 7C3 ∙ 5C2 = 350

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