Basic Probability Models - rasoulhekmati.yolasite.com fileBasic Probability Models Sections 2.3 & 2.4 Cathy Poliak, Ph.D. [email protected] Ofﬁce: Fleming 11C Department of Mathematics

Basic Probability ModelsSections 2.3 & 2.4

Cathy Poliak, [email protected]: Fleming 11C

Department of MathematicsUniversity of Houston

Lecture 5

Cathy Poliak, Ph.D. [email protected] Office: Fleming 11C (Department of Mathematics University of Houston )Sections 2.3 & 2.4 Lecture 5 1 / 37

Outline

1 Introduction Probability

2 Basic Rules of Probability


Winning the State Lottery

Suppose a person won the Jackpot of the State Lottery five timesin a row.

What do you think would happen?

Is it possible for a person to win the state lottery five consecutivetimes?


Randomness and Probability

We call a phenomenon random if individual outcomes areuncertain.

However, there is a regular distribution of outcomes in a largenumber of repetitions.

Chance behaviors unpredictable in the short run but has a regularand predictable pattern in the long run.

Long-run must be infinitely long to give them frequencies ofenough time to even out.


Proportion of Heads in Long Run

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

Trial A proportions

Trial B Proprotions

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

1.1

1

13

25

37

49

61

73

85

97

10

9

12

1

13

3

14

5

15

7

16

9

18

1

19

3

Trial A proportions

Trial B Proprotions


Probability

Probability is a value between zero and one, inclusive.

If probability is zero this indicates that the event will never occur.

If probability is one this indicates that the event will always occur.


Why Study Probability in Statistics?

Statistical Idea: Rare event rule for inferential statistics

If, under a given assumption, the probability of a particularobserved event is extremely small, we conclude that theassumption is probably not correct.


Some Definitions About Probability

The sample space of a random phenomenon is the set of allpossible outcomes. S is used to denote sample space.

An event is an outcome or a set of outcomes of a randomphenomenon. An event is a subset of the sample space.


Assigning probabilities

Classical method is use when all the experimental outcomes areequally likely. If n experimental outcomes are possible, aprobability of 1/n is assigned to each experimental outcome.Example: Drawing a card from a standard deck of 52 cards. Eachcard has a 1/52 probability of being selected.Relative frequency method is used when assigning probabilitiesis appropriate when data are available to estimate the proportionof the time the experimental outcome will occur if the experimentis repeated a large number of times. That is for any event E ,probability of E is

P(E) =number of possible outcomes of Etotal number of possible outcomes

=n(E)

n(S)


Example of Probabilities

Relative frequency method: An insurance company determined thenumber of accidents in a year. A sample of 100 people were surveyedto determine the number of accidents they were in a year: 0 accidents25 people, 1 accident 45 people, 2 accidents 20 people, 3 or moreaccidents 10 people. The following table shows the relative frequencyfor the outcomes.

Number of accidents Frequency (count) Relative frequency

0 25 25100 = 0.25

1 45 45100 = 0.45

2 20 20100 = 0.20

3 or more 10 10100 = 0.10

Total 100 1Cathy Poliak, Ph.D. [email protected] Office: Fleming 11C (Department of Mathematics University of Houston )Sections 2.3 & 2.4 Lecture 5 10 / 37

Example 2

If 5 marbles are drawn at random all at once from a bag containing 8white and 6 black marbles, what is the probability the 2 will be whiteand 3 will be black?


Example 3

Suppose a box contains 3 defective light bulbs and 12 good bulbs.Suppose we draw a simple random sample of 4 lightbulbs, find theprobability that one of the bulbs drawn is defective.


Example 3 continued

Suppose a box contains 3 defective light bulbs and 12 good bulbs.Suppose we draw a simple random sample of 4 lightbulbs,

1. What is the probability that none of bulbs drawn are defective?

2. What is the probability that at least one of the bulbs drawn isdefective?


Probability of an Event

The probability of an event E is calculated by summing theprobabilities of the sample points in the sample space for E .Notation: P(E)

In the insurance example, the probability that a driver has anynumber of accidents in a given year is: P(0) = 0.25, P(1) = 0.45,P(2) = 0.20, P(3 or more) = 0.10.


Rules of Probability, 1

The probability P(E) of any event E satisfies 0 ≤ P(E) ≤ 1.The closer P(E) is to 0, the less likely the event E will occur.The closer P(E) is to 1, the more likely the event E will occur.



If S is the sample space, then P(S) = 1.In the insurance example S = {0,1,2,3 or more}.

P(S) = P(0) + P(1) + P(2) + P(3 or more)

= 0.25 + 0.45 + 0.20 + 0.10= 1



Complement rule: For any event E,

P(EC) = 1− P(E)

This complement rule is an easy one to forget but it is one of thegreatest tools in probability.In the insurance example, what is the probability of that a driverhas at least one accident in a year?The phrase at least one is all number of accidents except fornone. So we are wanting P(0C)

P(0C) = 1− P(0) = 1− 0.25 = 0.75



General rule for addition: For any two events E and F

P(E or F ) = P(E ∪ F ) = P(E) + P(F )− P(E ∩ F )

In the insurance example, what is the probability that a driver hasone or two accidents in a given year?

P(1 or 2) = P(1) + P(2)− P(1 and 2) = 0.45 + 0.20− 0 = 0.65

Two events E and F are disjoint if they have no outcomes incommon. Thus P(E ∩ F ) = 0, and

P(E ∪ F ) = P(E) + P(F )

.


Example for Rules

A sports survey taken at UH shows that 48% of the respondents likedsoccer, 66% liked basketball and 38% liked hockey. Also, 30% likedsoccer and basketball, 22% liked basketball and hockey, and 28% likedsoccer and hockey. Finally, 12% liked all three sports.

1. What is the probability that a randomly selected student likesbasketball or hockey?

2. What is the probability that a randomly selected student does notlike any of these sports?



General rule for multiplication: For any two events E and F

P(E ∩ F ) = P(E)× P(F |E)

orP(E ∩ F ) = P(F )× P(E |F )

Where P(F |E) is the probability of F given that the event E hasoccurred. Similarly P(E |F ) is the probability of E given that F hasoccurred. These types of probabilities are called conditionalprobability.


Examples of Rule 5

Suppose we draw two cards from a deck of 52 fair playing cards, whatis the probability of getting an ace on the first draw and a king on thesecond draw?

Without replacement.

With replacement.


Examples of the Rules of Probability

Suppose that 55% of adults drink coffee, 25% of adults drink tea and15% drink both coffee and tea. Let C = an adult drinking coffee andT = adult drinking tea.

1. Determine P(C), P(T ), P(C ∩ T ) and P(C ∪ T ).

2. What is the probability that an adult drinks tea but not coffee?

3. What is the probability that an adult does not drink eitherbeverages?

4. What is the probability that an adult drinks tea, given that theydrink coffee?


Example 2 of the Rules of Probability

At a Ford dealership, if you select a Ford Mustang at random, theprobability it is red is P(R) = 0.40, the probability it is a convertible isP(C) = 0.14, and the probability that it is red or a convertibleP(R ∪ C) = 0.50.

1. What is the probability that a randomly selected Ford Mustang is aconvertible and red, P(C ∩ R)?

2. What is the probability that a randomly selected Ford Mustang is aconvertible but not red, P(C ∩ RC)?

3. What is the probability that a red Ford Mustang is a convertible?

4. Are the events "red" and "convertible" independent? Mutuallyexclusive (disjoint)?


Independent Events

Two events E and F are independent if the outcome of one eventdoes not changes the probability of the other event.If E and F are independent then,

P(E |F ) = P(E)andP(F |E) = P(F )


Determine Independence

Determine if A and B are independent.1. P(A) = 0.9, P(B) = 0.3, P(A ∩ B) = 0.27

2. P(A) = 0.4, P(B) = 0.6, P(A ∩ B) = 0.20


Example of Rule 5

A clothing store targets young customers (ages 18 through 22) wishesto determine whether the size of the purchases related to the methodpayment. Suppose a customer is picked at random. The following is300 customers the amount of the purchase and method payment.

Cash Credit Layaway TotalUnder $40 60 30 10 100

$40 or more 40 100 60 200Total 100 130 70 300

What is the probability that the customer paid with a credit card?


Example of Rule 5



$40 or more 40 100 60 200Total 100 130 70 300

What is the probability that the customer purchased under $40?


Example of Rule 5



$40 or more 40 100 60 200Total 100 130 70 300

What is the probability that the customer paid with credit card giventhat the purchase was under $40?


Example of Rule 5



$40 or more 40 100 60 200Total 100 130 70 300

What is the probability that the customer paid with credit card and thatthe purchase was under $40?


Example of Rule 5



$40 or more 40 100 60 200Total 100 130 70 300

Are type of payment and amount of purchase independent?


Example of General Multiplication Rule

Choose a person at random. Let A be the event that a person chosenis a women, and B the event that the person holds a managerial orprofessional job. From Government data P(A) = 0.46 and theprobability of managerial and professional jobs among women isP(B|A) = 0.32. What is the probability that a randomly chosen personis a woman and holds a managerial or professional job?


Two Frequently Asked Questions

1. When do I add and when do I multiply?I Add when finding the chance of events A or B or both happening.

P(A or B) = P(A ∪ B) = P(A) + P(B)− P(A ∩ B)

I Multiply when finding the chance that both events A and B happen.

P(A and B) = P(A ∩ B) = P(A)× P(B, given A) = P(A)P(B|A)


Two Frequently Asked Questions

2. What’s the difference between disjoint and independent?I Two events are disjoint if the occurrence of one prevents the other

from happening.P(A ∩ B) = 0

I Two events are independent if the occurrence of one does notchange the probability of the other.

P(A|B) = P(A)


#21 From Text

Thirty percent of the students at a local high school face a disciplinaryaction of some kind before they graduate. Of those "felony" students,40% go on to college. Of the ones who do not face disciplinary action60% go on to college.

1. What is the probability that a randomly selected student bothfaced a disciplinary action and went on to college? P(F ∩ C).

2. What percent of the students from the high school go on tocollege?

3. Show if events {faced disciplinary action} and {went to college} areindependent or not.


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Basic Probability Models - rasoulhekmati.yolasite.com fileBasic Probability Models Sections 2.3 & 2.4 Cathy Poliak, Ph.D. [email protected] Ofﬁce: Fleming 11C Department of Mathematics