Download Probability

7.1 Random Circumstances
Random circumstance is one in which
the outcome is unpredictable.
Case Study 1.1
Chapter 7
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.
Probability
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.
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
Random Circumstances in Alicia’s Day
Random Circumstances in Alicia’s Day
Random Circumstance 3: 1st student’s name is drawn
Alicia is selected.
Alicia is not selected.
Random Circumstance 1: Disease
status
Alicia has D.
Alicia does not have D.
Random Circumstance 4: 2nd student’s name is drawn
Alicia is selected.
Alicia is not selected.
Random Circumstance 2: Test result
Test is positive.
Test is negative.
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
Random Circumstance 5: 3rd student’s name is drawn
Alicia is selected.
Alicia is not selected.
3
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
4
7.2 Interpretations
of Probability
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.
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
2
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.
5
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
6
1
Example 7.1 Probability of Male
Determining the Relative Frequency
Probability of an Outcome
versus Female Births
Long-run relative frequency of males
born in the United States is about .512.
Method 1: Make an Assumption about the Physical World
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.
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.
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
7
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
Determining the Relative Frequency
Probability of an Outcome
Determining the Relative Frequency
Probability of an Outcome
Method 1: Make an Assumption about the Physical World
Method 2: Observe the Relative Frequency
Example 7.3 Probability Alicia has to Answer a Question
Example 7.4 The Probability of Lost Luggage
“1 in 176 passengers on U.S. airline carriers
will temporarily lose their luggage.”
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.
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.
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
9
Proportions and Percentages
as Probabilities
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
10
Estimating Probabilities
from Observed Categorical Data
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.).
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
8
11
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
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
12
2
Example 7.5 Nightlights and Myopia
The Personal Probability Interpretation
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?
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.
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
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
13
7.3 Probability Definitions
and Relationships
14
Random sample of college students.
Q: How many days do you drink alcohol
in a typical week?
•  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.
15
Assigning Probabilities to Simple Events
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}
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
16
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.
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.
Event A = last digit is a 9 = {009,019, . . . ,999}.
Since one out of ten numbers in set, P(A) = 1/10.
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.
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example 7.6 Days per Week of Drinking
•  Sample space: the collection of unique, nonoverlapping possible outcomes of a random
circumstance.
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
Restrictions on personal probabilities:
•  Must fall between 0 and 1 (or between 0 and 100%).
•  Must be coherent (consistent).
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.
17
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
18
3
Complementary Events
Mutually Exclusive 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.
Two events are mutually exclusive,
or equivalently disjoint, if they do not contain
any of the same simple events (outcomes).
Notation: AC represents the complement of A.
Example 7.2 A Simple Lottery (cont)
Note: P(A) + P(AC) = 1
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.
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
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
19
Independent and Dependent Events
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.
21
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
Example 7.3 Alicia Answering (cont)
Conditional Probabilities
Event A = Alicia is selected to answer Question 1.
Event B = Alicia is selected to answer Question 2.
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).
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.
22
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.
Knowing whether A occurred changes P(B).
Thus, the events A and B are not independent.
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
20
Example 7.7 Winning a Free Lunch
•  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.
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
23
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
24
4
Example 7.8 Probability That a Teenager
7.4
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:
Probability an Event Does Not Occur
Rule 1 (for “not the event”): P(AC) = 1 – P(A)
P(student is weekly gambler | teen is boy) = .20
Example 7.9 Probability a Stranger Does Not Share
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.
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
Basic Rules for
Finding Probabilities
25
Probability That Either
of Two Events Happen
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.
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
26
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.
Rule 2 (addition rule for “either/or”):
Rule 2a (general):
P(A or B) = P(A) + P(B) – P(A and B)
A = likes to party
B = snores
Rule 2b (for mutually exclusive events):
If A and B are mutually exclusive events,
P(A or B) = P(A) + P(B)
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
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
27
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
28
Example 7.11
Probability of Two
Boys or Two Girls in Two Births
Probability That Two or
More Events Occur Together
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):
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)
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).
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)
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
P(A) = 250/1000 = .25
P(B) = 350/1000 = .35
Probability woman has either two boys or two girls is:
P(A or B) = P(A) + P(B) = .2381 + .2621 = .5002
29
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
30
5
Example 7.8 Probability of Male
Example 7.12 Probability Two Strangers
and Gambler (cont)
Both Share Your Birth Month
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
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
Event B = weekly gambler
P(B|A) = .229
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
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.
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
31
Note: The probability that 4 unrelated strangers all share
your birth month would be (1/12)4.
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
32
Example 7.13 Alicia Answering
Determining a
Conditional Probability
If we know Alicia is picked to answer one of the
questions, what is the probability it was the first question?
Rule 4 (conditional probability):
A = Alicia selected to answer Question 1, P(A) = 1/50
P(B|A) = P(A and B)/P(A)
B = Alicia is selected to answer
any one of the questions,
P(A|B) = P(A and B)/P(B)
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
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
33
In Summary …
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
34
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.
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
35
Copyright ©2006 Brooks/Cole, a division of Thomson Learning, Inc.
36
6