copyright ©2006 brooks/cole, a division of thomson learning, inc. probability chapter 7

Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.

ProbabilityChapter 7

Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc. 2

7.1 Random CircumstancesRandom circumstance is one in which

the outcome is unpredictable.

Case Study 1.1 Alicia Has a Bad DayDoctor Visit:

Diagnostic test comes back positive for a disease (D).

Test is 95% accurate.About 1 out of 1000 women actually have D.

Statistics Class:Professor randomly selects 3 separate students at the beginning of each class to answer questions.Alicia is picked to answer the third question.


Random Circumstance 1: Disease status

Alicia has D.Alicia does not have D.

Random Circumstance 2: Test resultTest is positive.Test is negative.

Random Circumstances in Alicia’s Day


Random Circumstances in Alicia’s Day

Random Circumstance 3: 1st student’s name is drawnAlicia is selected.Alicia is not selected.

Random Circumstance 4: 2nd student’s name is drawnAlicia is selected.Alicia is not selected.

Random Circumstance 5: 3rd student’s name is drawnAlicia is selected.Alicia is not selected.


Assigning Probabilities

• A probability is a value between 0 and 1 and is written either as a fraction or as a decimal fraction.

• A probability simply is a number between 0 and 1 that is assigned to a possible outcome of a random circumstance.

• For the complete set of distinct possible outcomes of a random circumstance, the total of the assigned probabilities must equal 1.


7.2 Interpretations of Probability

The Relative Frequency Interpretation of Probability

In situations that we can imagine repeating many times, we define the probability of a specific outcome as the proportion of times it would occur over the long run -- called the relative frequency of that particular outcome.


Example 7.1 Probability of Male versus Female Births

Long-run relative frequency of males born in the United States is about .512.

Information Please Almanac (1991, p. 815).

Table provides results of simulation: the proportion is far from .512 over the first few weeks but in the long run settles down around .512.


Determining the Relative FrequencyProbability of an Outcome

Method 1: Make an Assumption about the Physical World

Example 7.2 A Simple Lottery

Choose a three-digit number between 000 and 999.

Player wins if his or her three-digit number is chosen.

Suppose the 1000 possible 3-digit numbers (000, 001, 002, . . . , 999) are equally likely.

In long run, a player should win about 1 out of 1000 times.

This does not mean a player will win exactly once in every thousand plays.



Method 1: Make an Assumption about the Physical World

Example 7.3 Probability Alicia has to Answer a Question

There are 50 student names in a bag.

If names mixed well, can assume each student is equally likely to be selected.

Probability Alicia will be selected to answer the first question is 1/50.



Method 2: Observe the Relative Frequency

Example 7.4 The Probability of Lost Luggage

“1 in 176 passengers on U.S. airline carriers will temporarily lose their luggage.”

This number is based on data collected over the long run. The probability that a randomly selected passenger on a U.S. carrier will temporarily lose luggage is 1/176 or about .006.


Proportions and Percentages as Probabilities

Ways to express the relative frequency of lost luggage:

• The proportion of passengers who lose their luggage is 1/176 or about .006.

• About 0.6% of passengers lose their luggage.• The probability that a randomly selected

passenger will lose his/her luggage is about .006.• The probability that you will lose your luggage

is about .006.

Last statement is not exactly correct – your probability depends on other factors (how late you arrive at the airport, etc.).


Estimating Probabilitiesfrom Observed Categorical Data

Assuming data are representative, the probability of a particular outcome is estimated to be the relative frequency (proportion) with which that outcome was observed.

Approximate margin of error for the estimated probability is

n1


Example 7.5 Nightlights and Myopia Revisited

Assuming these data are representative of a larger population, what is the approximate probability that someone from that population who sleeps with a nightlight in early childhood will develop some degree of myopia?

Note: 72 + 7 = 79 of the 232 nightlight users developed some degree of myopia. So we estimate the probability to be 79/232 = .34. This estimate is based on a sample of 232 people with a margin of error of about .066


The Personal Probability Interpretation

Personal probability of an event = the degree to which a given individual believes the event will happen.

Sometimes subjective probability used because the degree of belief may be different for each individual.

Restrictions on personal probabilities:• Must fall between 0 and 1 (or between 0 and 100%).• Must be coherent (consistent).


7.3 Probability Definitions and Relationships• Sample space: the collection of unique, non-

overlapping possible outcomes of a random circumstance.

• Simple event: one outcome in the sample space; a possible outcome of a random circumstance.

• Event: a collection of one or more simple events in the sample space; often written as A, B, C, and so on.


Example 7.6 Days per Week of Drinking

Random sample of college students.Q: How many days do you drink alcohol

in a typical week?

Simple Events in the Sample Space are:0 days, 1 day, 2 days, …, 7 days

Event “4 or more” is comprised of the simple events {4 days, 5 days, 6 days, 7 days}


Assigning Probabilities to Simple Events

P(A) = probability of the event A

Conditions for Valid Probabilities 1. Each probability is between 0 and 1.2. The sum of the probabilities over all

possible simple events is 1.

Equally Likely Simple EventsIf there are k simple events in the sample space and they are all equally likely, then the probability of the occurrence of each one is 1/k.


Example 7.2 A Simple Lottery (cont)

Random Circumstance: A three-digit winning lottery number is selected.

Sample Space: {000,001,002,003, . . . ,997,998,999}. There are 1000 simple events.

Probabilities for Simple Event: Probability any specific three-digit number is a winner is 1/1000. Assume all three-digit numbers are equally likely.

Event A = last digit is a 9 = {009,019, . . . ,999}. Since one out of ten numbers in set, P(A) = 1/10.

Event B = three digits are all the same = {000, 111, 222, 333, 444, 555, 666, 777, 888, 999}. Since event B contains 10 events, P(B) = 10/1000 = 1/100.


Complementary Events

Note: P(A) + P(AC) = 1

One event is the complement of another event if the two events do not contain any of the same simple events and together they cover the entiresample space.

Notation: AC represents the complement of A.

Example 7.2 A Simple Lottery (cont)A = player buying single ticket

winsAC = player does not winP(A) = 1/1000 so P(AC) = 999/1000


Mutually Exclusive Events

Two events are mutually exclusive, or equivalently disjoint, if they do not contain any of the same simple events (outcomes).

Example 7.2 A Simple Lottery (cont)

A = all three digits are the same.B = the first and last digits are

different

The events A and B are mutually exclusive (disjoint), but they are not complementary.


Independent and Dependent Events

• Two events are independent of each other if knowing that one will occur (or has occurred) does not change the probability that the other occurs.

• Two events are dependent if knowing that one will occur (or has occurred) changes the probability that the other occurs.


Example 7.7 Winning a Free Lunch

Customers put business card in restaurant glass bowl. Drawing held once a week for free lunch. You and Vanessa put a card in two consecutive weeks.

Event A = You win in week 1.Event B = Vanessa wins in week 1.Event C = Vanessa wins in week 2.

• Events A and B refer to the same random circumstance and are not independent.

• Events A and C refer to to different random circumstances and are independent.


Example 7.3 Alicia Answering (cont)

Event A = Alicia is selected to answer Question 1.Event B = Alicia is selected to answer Question 2.

• P(A) = 1/50.• If event A occurs, her name is no longer in the bag,

so P(B) = 0.• If event A does not occur, there are 49 names in the

bag (including Alicia’s name), so P(B) = 1/49.

Events A and B refer to different random circumstances, but are A and B independent events?

Knowing whether A occurred changes P(B). Thus, the events A and B are not independent.


Conditional Probabilities

Conditional probability of the event B, given that the event A occurs, is the long-run relative frequency with which event B occurs when circumstances are such that A also occurs; written as P(B|A).

P(B) = unconditional probability event B occurs.

P(B|A) = “probability of B given A” = conditional probability event B occurs given that we know A has occurred or will occur.


Example 7.8 Probability That a TeenagerGambles Depends upon Gender

Notice dependence between “weekly gambling habit” and “gender.” Knowledge of a 9th grader’s gender changes probability that he/she is a weekly gambler.

Survey: 78,564 students (9th and 12th graders)The proportions of males and females admitting they gambled at least once a week during the previous year were reported. Results for 9th grade:

P(student is weekly gambler | teen is boy) = .20

P(student is weekly gambler | teen is girl) = .05


7.4 Basic Rules forFinding Probabilities

Rule 1 (for “not the event”): P(AC) = 1 – P(A)

Example 7.9 Probability a Stranger Does Not Share Your Birth DateP(next stranger you meet will share your birthday)

= 1/365.P(next stranger you meet will not share your birthday)

= 1 – 1/365 = 364/365 = .9973.

Probability an Event Does Not Occur


Rule 2 (addition rule for “either/or”):

Rule 2a (general): P(A or B) = P(A) + P(B) – P(A and B)

Rule 2b (for mutually exclusive events): If A and B are mutually exclusive events, P(A or B) = P(A) + P(B)

Probability That Either of Two Events Happen


Example 7.10 Roommate Compatibility

Brett is off to college. There are 1000 male students. Brett hopes his roommate will not like to party and not snore.

A = likes to party P(A) = 250/1000 = .25B = snores P(B) = 350/1000 = .35

Probability Brett will be assigned a roommate who either likes to party or snores, or both is: P(A or B) = P(A) + P(B) – P(A and B) = .25 + .35 – .15 = .45So the probability his roommate is acceptable is 1 – .45 = .55

Snores? Yes No Total Likes to Yes 150 100 250

Party? No 200 550 750 350 650 1000


Rule 3 (multiplication rule for “and”):

Rule 3a (general): P(A and B) = P(A)P(B|A)

Rule 3b (for independent events): If A and B are independent events, P(A and B) = P(A)P(B)

Extension of Rule 3b (for > 2 indep events): For several independent events,P(A1 and A2 and … and An) = P(A1)P(A2)…P(An)

Probability That Two or More Events Occur Together


Example 7.11 Probability of Two Boys or Two Girls in Two Births

What is the probability that a woman who has two children has either two girls or two boys?

Recall that the probability of a boy is .512 and probability of a girl is .488. Then we have (using Rule 3b):

Event A = two girls P(A) = (.488)(.488) = .2381Event B = two boys P(B) = (.512)(.512) = .2621

Probability woman has either two boys or two girls is: P(A or B) = P(A) + P(B) = .2381 + .2621 = .5002

Note: Events A and B are mutually exclusive (disjoint).


Example 7.8 Probability of Male and Gambler (cont)

About 11% of all 9th graders are males and weekly gamblers.

For 9th graders, 22.9% of the boys and 4.5% of the girls admitted they gambled at least once a week during the previous year. The population consisted of 50.9% girls and 49.1% boys.

P(male and gambler) = P(A and B) = P(A)P(B|A) = (.491)(.229) = .1124

Event A = male Event B = weekly gambler P(A) = .491 P(B|A) = .229


Example 7.12 Probability Two Strangers Both Share Your Birth Month

Note: The probability that 4 unrelated strangers all share your birth month would be (1/12)4.

Assume all 12 birth months are equally likely.What is the probability that the next two unrelated strangers you meet both share your birth month?

P(both strangers share your birth month) = P(A and B) = P(A)P(B) = (1/12)(1/12) = .007

Event A = 1st stranger shares your birth month P(A) = 1/12 Event B = 2nd stranger shares your birth month P(B) = 1/12

Note: Events A and B are independent.


Rule 4 (conditional probability):

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

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

Determining a Conditional Probability


Example 7.13 Alicia Answering

A = Alicia selected to answer Question 1, P(A) = 1/50

B = Alicia is selected to answer any one of the questions, P(B) = 3/50

Since A is a subset of B, P(A and B) = 1/50

P(A|B) = P(A and B)/P(B) = (1/50)/(3/50) = 1/3

If we know Alicia is picked to answer one of the questions, what is the probability it was the first question?


In Summary …

• When two events are mutually exclusive and one happens, it turns the probability of the other one to 0.

• When two events are independent and one happens, it leaves the probability of the other one alone.

Students sometimes confuse the definitions of independent and mutually exclusive events.


In Summary …

When Events Are: P(A or B) is: P(A and B) is: P(A|B) is: Mutually Exclusive

P(A)+P(B) 0 0

I ndependent

P(A)+P(B)- P(A)P(B) P(A)P(B) P(A)

Any

P(A)+P(B)- P(A and B) P(A)P(B|A) P(A and B)/P(B)

copyright ©2006 brooks/cole, a division of thomson learning, inc. probability chapter 7

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