Transcript P - Cengage

Chapter 7
Probability
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
7.1 Random Circumstances
Random circumstance is one in which
the outcome is unpredictable.
Case Study 1.1
Alicia Has a Bad Day
Doctor 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.
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Random Circumstances in Alicia’s Day
Random Circumstance 1: Disease
status
Alicia has D.
Alicia does not have D.
Random Circumstance 2: Test result
Test is positive.
Test is negative.
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Random Circumstances in Alicia’s Day
Random Circumstance 3: 1st student’s name is drawn
Alicia is selected.
Alicia is not selected.
Random Circumstance 4: 2nd student’s name is drawn
Alicia is selected.
Alicia is not selected.
Random Circumstance 5: 3rd student’s name is drawn
Alicia is selected.
Alicia is not selected.
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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.
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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.
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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.
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Determining the Relative Frequency
Probability 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.
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Determining the Relative Frequency
Probability of an Outcome
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.
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Determining the Relative Frequency
Probability of an Outcome
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. So
the probability that a randomly selected passenger on a
U.S. carrier will temporarily lose luggage is 1/176 or
about .006.
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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.).
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Estimating Probabilities
from 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 1
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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
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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.
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7.3 Probability Definitions
and Relationships
• Sample space: the collection of unique,
nonoverlapping 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.
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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}
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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 Events
If 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.
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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.
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Complementary Events
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 entire
sample space.
Notation: AC represents the complement of A.
Note: P(A) + P(AC) = 1
Example 7.2 A Simple Lottery (cont)
A = player buying single ticket wins
AC = player does not win
P(A) = 1/1000 so P(AC) = 999/1000
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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.
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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.
The definitions can apply either …
to events within the same random circumstance or
to events from two separate random circumstances.
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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.
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Example 7.3 Alicia Answering (cont)
Event A = Alicia is selected to answer Question 1.
Event B = Alicia is selected to answer Question 2.
Events A and B refer to different random circumstances,
but are A and B independent events?
• 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.
Knowing whether A occurred changes P(B).
Thus, the events A and B are not independent.
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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.
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Example 7.8 Probability That a Teenager
Gambles Depends upon Gender
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
Notice dependence between “weekly gambling habit”
and “gender.” Knowledge of a 9th grader’s gender
changes probability that he/she is a weekly gambler.
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7.4 Basic Rules for
Finding Probabilities
Probability an Event Does Not Occur
Rule 1 (for “not the event”): P(AC) = 1 – P(A)
Example 7.9 Probability a Stranger
Does Not Share Your Birth Date
P(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.
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Probability That Either
of Two Events Happen
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)
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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.
Likes to Yes
Party? No
A = likes to party
B = snores
Snores?
Yes
No
150
100
200
550
350
650
Total
250
750
1000
P(A) = 250/1000 = .25
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 = .45
So the probability his roommate is acceptable is 1 – .45 = .55
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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) = .2381
Event B = two boys P(B) = (.512)(.512) = .2621
Note: Events A and B are mutually exclusive (disjoint).
Probability woman has either two boys or two girls is:
P(A or B) = P(A) + P(B) = .2381 + .2621 = .5002
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Probability That Two or
More Events Occur Together
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)
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Example 7.8 Probability of Male
and Gambler (cont)
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.
Event A = male
P(A) = .491
Event B = weekly gambler
P(B|A) = .229
P(male and gambler) = P(A and B)
= P(A)P(B|A) = (.491)(.229) = .1124
About 11% of all 9th graders are males and weekly gamblers.
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Example 7.12 Probability Two Strangers
Both Share Your Birth Month
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?
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.
P(both strangers share your birth month)
= P(A and B) = P(A)P(B) = (1/12)(1/12) = .007
Note: The probability that 4 unrelated strangers all share
your birth month would be (1/12)4.
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Determining a
Conditional Probability
Rule 4 (conditional probability):
P(B|A) = P(A and B)/P(A)
P(A|B) = P(A and B)/P(B)
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Example 7.13 Alicia Answering
If we know Alicia is picked to answer one of the
questions, what is the probability it was the first question?
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
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In Summary …
Students sometimes confuse the definitions of
independent and mutually exclusive events.
• 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.
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In Summary …
When Events Are:
Mutually
Exclusive
Independent
Any
P(A or B) is:
P(A)+P(B)
P(A and B) is:
0
P(A|B) is:
0
P(A)+P(B)-P(A)P(B)
P(A)P(B)
P(A)
P(A)+P(B)-P(A and B)
P(A)P(B|A)
P(A and B)/P(B)
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Sampling with and without Replacement
• A sample is drawn with replacement if
individuals are returned to the eligible pool
for each selection.
• A sample is drawn without replacement if
sampled individuals are not eligible for
subsequent selection.
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7.5 Strategies for Finding
Complicated Probabilities
Example 7.2 Winning the Lottery
Event A = winning number is 956. What is P(A)?
Method 1: With physical assumption that all 1000
possibilities are equally likely, P(A) = 1/1000.
Method 2: Define three events,
B1 = 1st digit is 9, B2 = 2nd digit is 5, B3 = 3rd digit is 6
Event A occurs if and only if all 3 of these events occur.
Note: P(B1) = P(B2) = P(B3) = 1/10. Since these events
are all independent, we have P(A) = (1/10)3 = 1/1000.
* Can be more than one way to find a probability.
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Hints and Advice
for Finding Probabilities
• P(A and B): define event in physical terms and see if
know probability. Else try multiplication rule (Rule 3).
• Series of independent events all happen: multiply
all individual probabilities (Extension of Rule 3b)
• One of a collection of mutually exclusive events
happens: add all individual probabilities (Rule 2b
extended).
• Check if probability of complement easier,
then subtract it from 1 (applying Rule 1).
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Hints and Advice for Finding Probabilities
• None of a collection of mutually exclusive events
happens: find probability one happens, then subtract
that from 1.
• Conditional probability: define event in physical
terms and see if know probability. Else try Rule 4 or
next bullet as well.
• Know P(B|A) but want P(A|B): Use Rule 3a to find
P(B) = P(A and B) + P(AC and B), then use Rule 4.
P( A and B)
P( A | B) 
P( B | A) P( A)  P( B | AC ) P( AC )
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Steps for Finding Probabilities
Step 1: List each separate random circumstance
involved in the problem.
Step 2: List the possible outcomes for each
random circumstance.
Step 3: Assign whatever probabilities you can
with the knowledge you have.
Step 4: Specify the event for which you want to
determine the probability.
Step 5: Determine which of the probabilities from
step 3 and which probability rules can be
combined to find the probability of interest.
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Example 7.17 Alicia Is Probably Healthy
What is the probability that Alicia has the disease
given that the test was positive?
Steps 1 to 3: Random circumstances, outcomes, probabilities.
Random circumstance 1: Alicia’s disease status
Possible Outcomes: A = disease; AC = no disease
Probabilities: P(A) = 1/1000 = .001; P(AC) = .999
Random circumstance 2: Alicia’s test results
Possible Outcomes: B = test is positive, BC = test is negative
Probabilities:
P(B|A) = .95 (positive test given disease)
P(BC|A) = .05 (negative test given disease)
P(B|AC) = .05 (positive test given no disease)
P(BC|AC) = .95 (negative test given no disease)
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Example 7.17 Alicia Healthy? (cont)
Step 4: Specify event you want to determine the probability.
P(disease | positive test) = P(A|B).
Step 5: Determine which probabilities and probability rules
can be combined to find the probability of interest.
Note we have P(B|A) and we want P(A|B).
Hints tell us to use P(B) = P(A and B) + P(AC and B).
Note P(A and B) = P(B|A)P(A), similarly for P(AC and B). So …
P(AC and B) = P(B|AC)P(AC) = (.05)(.999) = .04995
P(A and B) = P(B|A)P(A) = (.95)(.001) = .00095
P(B) = .04995 + .00095 = .0509
P( A and B) .00095
P( A | B) 

 .019
P( B)
.0509
There is less than a 2% chance that Alicia has the disease,
even though her test was positive.
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Two-Way Table:
“Hypothetical Hundred Thousand”
Example 7.8
Teens and Gambling (cont)
Sample of 9th grade teens: 49.1% boys, 50.9% girls.
Results: 22.9% of boys and 4.5% of girls admitted
they gambled at least once a week during previous year.
Start with hypothetical 100,000 teens …
(.491)(100,000) = 49,100 boys and thus 50,900 girls
Of the 49,100 boys, (.229)(49,100) = 11,244
would be weekly gamblers.
Of the 50,900 girls, (.045)(50,900) = 2,291
would be weekly gamblers.
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Example 7.8 Teens and Gambling (cont)
Weekly Gambler Not Weekly Gambler Total
Boy
11,244
37,856
49,100
Girl
2,291
48,609
50,900
Total
13,535
86,465
100,000
P(boy and gambler) = 11,244/100,000 = .1124
P(boy | gambler) = 11,244/13,535 = .8307
P(gambler) = 13,535/100,000 = .13535
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Tree Diagrams
Step 1: Determine first random circumstance in sequence, and
create first set of branches for possible outcomes. Create one
branch for each outcome, write probability on branch.
Step 2: Determine next random circumstance and append branches
for possible outcomes to each branch in step 1.
Write associated conditional probabilities on branches.
Step 3: Continue this process for as many steps as necessary.
Step 4: To determine the probability of following any particular
sequence of branches, multiply the probabilities on those
branches. This is an application of Rule 3a.
Step 5: To determine the probability of any collection of sequences
of branches, add the individual probabilities for those sequences,
as found in step 4. This is an application of Rule 2b.
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Example 7.18 Alicia’s Possible Fates
P(Alicia has D and has a positive test) = .00095.
P(test is positive) = .00095 + .04995 = .0509.
P(Alicia has D | positive test) = .00095/.0509 = .019
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Example 7.8 Teens and Gambling (cont)
P(boy and gambler) = (.491)(.229) = .1124
P(girl and not gambler) = (.509)(.955) = .4861
P(gambler) = .1124 + .0229 = .1353
P(boy | gambler) = .1124/.1353 = .8307
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7.6 Using Simulation to
Estimate Probabilities
Some probabilities so difficult or timeconsuming to calculate – easier to simulate.
If you simulate the random circumstance n times
and the outcome of interest occurs in x out of
those n times, then the estimated probability
for the outcome of interest is x/n.
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Example 7.19 Getting All the Prizes
Cereal boxes each contain one of four prizes.
Any box is equally likely to contain each of the four prizes.
If buy 6 boxes, what is the probability you get all 4 prizes?
Shown above are 50 simulations of generating a set of 6 digits,
each equally likely to be 1, 2, 3, or 4. There are 19 bold
outcomes in which all 4 prizes were collected. The estimated
probability is 19/50 = .38. (Actual probability is .3809.)
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7.7 Coincidences & Intuitive
Judgments about Probability
Confusion of the Inverse
Example: Diagnostic Testing
Confuse the conditional probability “have the disease”
given “a positive test result” -- P(Disease | Positive),
with the conditional probability of “a positive test result”
given “have the disease” -- P(Positive | Disease), also
known as the sensitivity of the test.
Often forget to incorporate the base rate for a disease.
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Specific People versus
Random Individuals
The chance that your marriage will end in divorce is 50%.
Does this statement apply to you personally?
If you have had a terrific marriage for 30 years, your
probability of ending in divorce is surely less than 50%.
Two correct ways to express the aggregate divorce statistics:
• In long run, about 50% of marriages end in divorce.
• At the beginning of a randomly selected marriage,
the probability it will end in divorce is about .50.
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Coincidences
A coincidence is a surprising concurrence of events,
perceived as meaningfully related, with no apparent
causal connection.
Example 7.23 Winning the Lottery Twice
In 1986, Ms. Adams won the NJ lottery twice in a short time
period. NYT claimed odds of one person winning the top prize
twice were about 1 in 17 trillion. Then in 1988, Mr. Humphries
won the PA lottery twice.
1 in 17 trillion = probability that a specific individual who
plays the lottery exactly twice will win both times.
Millions of people play the lottery. It is not surprising
that someone, somewhere, someday would win twice.
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The Gambler’s Fallacy
The gambler’s fallacy is the misperception of
applying a long-run frequency in the short-run.
• Primarily applies to independent events.
• Independent chance events have no memory.
Example:
Making ten bad gambles in a row doesn’t change
the probability that the next gamble will also be bad.
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