Download 3-5 Conditional Probability

3-1
Chapter 3
Probability
© The McGraw-Hill Companies, Inc., 2000
3-2
Outline
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3-1
3-2
3-3
3-4
3-5
Introduction
Fundamentals
Addition Rules for Probability
Multiplication Rules: Basics
Multiplication Rules: Beyond
the Basics
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3-3
Objectives
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Determine Sample Spaces and find
the probability of an event using
classical probability.
Find the probability of an event
using empirical probability.
Find the probability of compound
events using the addition rules.
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3-4
Objectives
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Find the probability of compound
events using the multiplication
rules.
Find the conditional probability of
an event.
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3-5
3-2 Fundamentals
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A probability experiment is a process
that leads to well-defined results called
outcomes.
An outcome is the result of a single trial
of a probability experiment.
NOTE: A tree diagram can be used as a
systematic way to find all possible
outcomes of a probability experiment.
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3-2 Tree Diagram for Tossing Two Coins
3-6
H
H
T
Second Toss
H
T
First Toss
T
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3-7
3-2 Sample Spaces - Examples
EXPERIM ENT SAM PLE SPACE
Toss one coin
H, T
Roll a die
1, 2, 3, 4, 5, 6
Answer a truefalse question
Toss two coins
True, False
H H , H T, TH , TT
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3-2 Formula for Classical
Probability
3-8
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Classical probability assumes that
all outcomes in the sample space
are equally likely to occur.
That is, equally likely events are
events that have the same
probability of occurring.
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3-9
3-2 Formula for Classical
Probability
The probability of any event E is
number of outcomes in E
.
total number of outcomes in the sample space
This probability is denoted by
n( E )
.
n( S )
This probability is called classical probability ,
P( E ) =
and it uses the sample space S .
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3-2 Classical Probability - Examples
3-10
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For a card drawn from an ordinary deck,
find the probability of getting (a) a queen
(b) a 6 of clubs (c) a 3 or a diamond.
Solution: (a) Since there are 4 queens
and 52 cards, P(queen) = 4/52 = 1/13.
(b) Since there is only one 6 of clubs,
then P(6 of clubs) = 1/52.
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3-11
3-2 Classical Probability - Examples
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(c) There are four 3s and 13
diamonds, but the 3 of diamonds
is counted twice in the listing.
Hence there are only 16
possibilities of drawing a 3 or a
diamond, thus
P(3 or
diamond) = 16/52 = 4/13.
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3-12
3-2 Classical Probability - Examples
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When a single die is rolled, find the
probability of getting a 9.
Solution: Since the sample space is 1, 2,
3, 4, 5, and 6, it is impossible to get a 9.
Hence, P(9) = 0/6 = 0.
NOTE: The sum of the probabilities of all
outcomes in a sample space is one.
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3-13
3-2 Complement of an Event
The complement of an event E is the set of outcomes in the
sample space that are not included in the outcomes
of event E . The complement of E is denoted by E ( E bar ).
E
E
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3-2 Complement of an Event Example
3-14
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Find the complement of each event.
Rolling a die and getting a 4.
Solution: Getting a 1, 2, 3, 5, or 6.
Selecting a letter of the alphabet and
getting a vowel.
Solution: Getting a consonant
(assume y is a consonant).
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3-2 Complement of an Event Example
3-15
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Selecting a day of the week and
getting a weekday.
Solution: Getting Saturday or
Sunday.
Selecting a one-child family and
getting a boy.
Solution: Getting a girl.
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3-16
3-2 Rule for Complementary Event
P(E )  1 P(E )
or
P(E ) = 1P(E )
or
P ( E ) + P ( E ) = 1.
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3-17
3-2 Empirical Probability
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The difference between classical and
empirical probability is that classical
probability assumes that certain
outcomes are equally likely while
empirical probability relies on actual
experience to determine the
probability of an outcome.
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3-18
3-2 Formula for Empirical
Probability
Given a frequency distribution ,
the probability of an event being
in a given class is
frequency for the class
P(E ) =
total frequencies in the distribution
f
 .
n
This probability is called the empirical
probability and is based on observation .
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3-2 Empirical Probability Example
3-19
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In a sample of 50 people, 21 had
type O blood, 22 had type A blood,
5 had type B blood, and 2 had AB
blood. Set up a frequency
distribution.
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3-20
3-2 Empirical Probability Example
Type
A
B
AB
O
Frequency
22
5
2
21
50 = n
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3-2 Empirical Probability Example
3-21
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Find the following probabilities for
the previous example.
A person has type O blood.
Solution: P(O) = f /n = 21/50.
A person has type A or type B blood.
Solution: P(A or B) = 22/50+ 5/50
= 27/50.
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3-22
3-3 Addition Rules for Probability
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Two events are mutually exclusive
if they cannot occur at the same
time (i.e., they have no outcomes
in common).
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3-23
3-3 The Addition Rules for
Probability
A and B are mutually exclusive
A
B
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3-24
3-3 Addition Rule 1
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 )
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3-25
3-3 Addition Rule 1- Example
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At a political rally, there are 20
Liberals (L), 13 Conservatives (C), and
6 NDPs (N). If a person is selected,
find the probability that he or she is
either a Conservative or an NDP.
Solution: P(C or N) = P(C) + P(N)
= 13/39 + 6/39 = 19/39.
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3-26
3-3 Addition Rule 1- Example
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A day of the week is selected at
random. Find the probability that
it is a weekend.
Solution: P(Saturday or Sunday)
= P(Saturday) + P(Sunday)
= 1/7 + 1/7 = 2/7.
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3-27
3-3 Addition Rule 2
When two events A and B
are not mutually exclusive, the
probabilityy that A or B will
occur is
P ( A or B )  P ( A) + P ( B)  P ( A and B)
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3-28
3-3 Addition Rule 2
A and B
A
(common portion)
B
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3-29
3-3 Addition Rule 2- Example
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In a hospital unit there are eight
nurses and five physicians. Seven
nurses and three physicians are
females. If a staff person is selected,
find the probability that the subject is
a nurse or a male.
The next slide has the data.
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3-30
3-3 Addition Rule 2 - Example
STAFF
STAFF
FEMALES
FEMALES
MALES
MALES
TOTAL
TOTAL
NURSES
NURSES
77
11
88
PHYSICIANS
PHYSICIANS
33
22
55
TOTAL
TOTAL
10
10
33
13
13
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3-31
3-3 Addition Rule 2 - Example
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Solution: P(nurse or male)
= P(nurse) + P(male) – P(male
nurse) = 8/13 + 3/13 – 1/13 = 10/13.
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3-32
3-3 Addition Rule 2 - Example
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On New Year’s Eve, the probability that a
person driving while intoxicated is 0.32,
the probability of a person having a
driving accident is 0.09, and the
probability of a person having a driving
accident while intoxicated is 0.06. What is
the probability of a person driving while
intoxicated or having a driving accident?
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3-33
3-3 Addition Rule 2 - Example
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Solution:
P(intoxicated or accident)
= P(intoxicated) + P(accident)
– P(intoxicated and accident)
= 0.32 + 0.09 – 0.06 = 0.35.
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3-4 The Multiplication Rules and
Conditional Probability
3-34
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Two events A and B are independent
if the fact that A occurs does not
affect the probability of B occurring.
Example: Rolling a die and getting a
6, and then rolling another die and
getting a 3 are independent events.
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3-35
3-4 Multiplication Rules
W hen two events A and B
are independent , the
probability of both
occurring is
P ( A and B )  P ( A )  P ( B ).
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3-4 Multiplication Rule 1 Example
3-36
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A card is drawn from a deck and
replaced; then a second card is
drawn. Find the probability of getting
a queen and then an ace.
Solution: Because these two events
are independent (why?), P(queen and
ace) = (4/52)(4/52) = 16/2704 = 1/169.
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3-4 Multiplication Rule 1 Example
3-37
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A Decima pole found that 46% of
Canadians 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 stress at least once a week.
Solution: Let S denote stress. Then
P(S and S and S) = (0.46)3 = 0.097.
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3-4 Multiplication Rule 1 Example
3-38
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The probability that a specific medical
test will show positive is 0.32. If four
people are tested, find the probability
that all four will show positive.
Solution: Let T denote a positive test
result. Then P(T and T and T and T) =
(0.32)4 = 0.010.
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3-5 Multiplication Rules:
Conditional Probability
3-39
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When 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,
the events are said to be dependent.
Example: Having high grades and
getting a scholarship are dependent
events.
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3-5 Multiplication Rules:
Conditional Probability
3-40
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The conditional probability of an event B
in relationship to an event A is the
probability that an event B occurs after
event A has already occurred.
The notation for the conditional
probability of B given A is P(B|A).
NOTE: This does not mean B  A.
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3-41
3-4 Multiplication Rule 2
When two events A and B
are dependent , the
probability of both
occurring is
P ( A and B )  P ( A)  P ( B| A).
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3-5 The Multiplication Rules:
Conditional Probability - Example
3-42
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In a shipment of 25 microwave ovens,
two are defective. If two ovens are
randomly selected and tested, find the
probability that both are defective if the
first one is not replaced after it has
been tested.
Solution: See next slide.
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3-5 The Multiplication Rules and
Conditional Probability - Example
3-43
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Solution: Since the events are
dependent, P(D1 and D2)
= P(D1)P(D2| D1) = (2/25)(1/24)
= 2/600 = 1/300.
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3-5 The Multiplication Rules and
Conditional Probability - Example
3-44
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The KW Insurance Company found that
53% of the residents of a city had
homeowner’s insurance with its company.
Of these clients, 27% also had automobile
insurance with the company. If a resident
is selected at random, find the probability
that the resident has both homeowner’s
and automobile insurance.
© The McGraw-Hill Companies, Inc., 2000
3-5 The Multiplication Rules and
Conditional Probability - Example
3-45
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Solution: Since the events are
dependent, P(H and A)
= P(H)P(A|H) = (0.53)(0.27)
= 0.1431.
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3-5 The Multiplication Rules and
Conditional Probability - Example
3-46
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Box 1 contains two red balls and one
blue ball. Box 2 contains three blue
balls and one red ball. A coin is tossed.
If it falls heads up, box 1 is selected
and a ball is drawn. If it falls tails up,
box 2 is selected and a ball is drawn.
Find the probability of selecting a red
ball.
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3-5 Tree Diagram for Example
3-47
P(R|B1) 2/3
P(B1) 1/2
P(B2) 1/2
Red (1/2)(2/3)
Box 1
Blue (1/2)(1/3)
P(B|B1) 1/3
P(R|B2) 1/4
Box 2
Red (1/2)(1/4)
P(B|B2) 3/4 Blue (1/2)(3/4)
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3-5 The Multiplication Rules and
Conditional Probability - Example
3-48
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Solution: P(red) = (1/2)(2/3) +
(1/2)(1/4) = 2/6 + 1/8 = 8/24 + 3/24
= 11/24.
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3-49
3-5 Conditional Probability Formula
The probability that the second event B occurs
given that the first event A has occurred can be
found by dividing the probability that both events
occurred by the probability that the first event has
occurred . The formula is
P( A and B)
.
P( B| A) =
P( A)
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3-5 Conditional Probability Example
3-50
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The probability that Sam parks in a noparking 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 0.2. On Tuesday,
Sam arrives at school and has to park in a
no-parking zone. Find the probability that
he will get a ticket.
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3-5 Conditional Probability Example
3-51
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Solution: Let N = parking in a noparking zone and T = getting a
ticket.
Then P(T|N) = [P(N and T) ]/P(N) =
0.06/0.2 = 0.30.
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3-5 Conditional Probability Example
3-52
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A recent survey asked 100 people
if they thought women in the
armed forces should be permitted
to participate in combat. The
results are shown in the table on
the next slide.
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3-53
3-5 Conditional Probability Example
Gender
Gender
Yes
Yes
No
No
Total
Total
Male
Male
32
32
18
18
50
50
Female
Female
88
42
42
50
50
Total
Total
40
40
60
60
100
100
© The McGraw-Hill Companies, Inc., 2000
3-5 Conditional Probability Example
3-54
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Find the probability that the respondent
answered “yes” given that the respondent
was a female.
Solution: Let M = respondent was a male;
F = respondent was a female;
Y = respondent answered “yes”;
N = respondent answered “no”.
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3-5 Conditional Probability Example
3-55
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P(Y|F) = [P( F and Y) ]/P(F) =
[8/100]/[50/100] = 4/25.
Find the probability that the respondent
was a male, given that the respondent
answered “no”.
Solution: P(M|N) = [P(N and M)]/P(N) =
[18/100]/[60/100] = 3/10.
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