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5-1
A Survey of
Concepts
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
5-2
When you have completed this chapter, you will be able to:
1
Explain the terms random experiment,
outcome, sample space, permutations,
and combinations.
2
Define probability.
3
Describe the classical, empirical, and subjective
approaches to probability.
4
Explain and calculate conditional probability
and joint probability.
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
5-3
5
Calculate probability using the rules of
addition and rules of multiplication.
6
Use a tree diagram to organize and
compute probabilities.
7
Calculate a probability using Bayes’ theorem.
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Types of Statistics
Descriptive
Methods of…
collecting
organizing
presenting
and
analyzing data
5-4
Inferential
Science of…
making inferences
about a population,
based on sample
information.
Emphasis now to be on this!
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Terminology
5-5
Probability
…is a measure of the
likelihood that an event in the future will happen!
 It can only assume a value between 0 and 1.
 A value near zero means the event is not
likely happen; near one means it is likely..
 There are three definitions of probability:
classical, empirical, and subjective
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Terminology
Random Experiment
…is a process
repetitive in nature
the outcome of any trial is uncertain
well-defined set of possible outcomes
each outcome has a probability
associated with it
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
5-6
Terminology
…is a particular result of a
random experiment.
... is the collection or set of all
the possible outcomes of a
random experiment.
…is the collection of one
or more
outcomes of an experiment.
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
5-7
Approaches to Assigning Probability
5-8
Subjective
…probability is based on whatever information is available
Objective
Classical Probability
… is based on the
assumption that the
outcomes of an experiment
are equally likely
Probability
of an Event
=
Empirical Probability
… applies when the number
of times the event happens
is divided by the number of
observations
NUMBER of favourable outcomes
Total NUMBER of possible outcomes
Examples
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
S ubjective
Probability
5-9
…. refers to the chance of occurrence
assigned to an event
by a particular individual
It is not computed objectively,
i.e., not from prior knowledge or from actual data…
…that the Toronto Maple Leafs will win the Stanley
Cup next season!
…that you will arrive to class on time tomorrow!
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
E Pmpirical
robability
5 - 10
Students measure the contents of their soft
drink cans… 10 cans are underfilled,
32 are filled correctly and
8 are overfilled
When the contents of the next can is measured,
what is the probability that it is… (a) filled correctly?
P(C) = 32 / 50 = 64%
…(b) not filled correctly?
P(~C) = 1 – P(C) = 1 - .64 = 36%
This is called the Complement of C
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Random Experiment
5 - 11
The experiment is rolling the die...once!
The possible outcomes are the numbers…
1 2 3 4 5 6
An event is the occurrence of an even number
i.e. we collect the outcomes 2, 4, and 6.
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Tree Diagrams
5 - 12
This is a useful device to show all the possible outcomes
of the experiment
and their corresponding probabilities
Consider the random experiment of flipping a coin twice.
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Tree Diagrams
Origin
First
Flip
Second
Flip
H
H HH
T HT
H TH
T
T TT
5 - 13
Expressed as:
P(HH)= 0.25
P(HT)= 0.25
Simple Events
P(TH)= 0.25
P(TT)= 0.25
1.00
New
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5 - 14
Tree Diagrams
Origin
Menu
Appetizer:
Soup or
Juice
Entrée:
Beef
Turkey
Fish
Dessert:
Pie
Ice Cream
Appetizer
Entrée
Beef
Soup
Turkey
Fish
Beef
Juice
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Turkey
Fish
Dessert
Pie
Ice Cream
Pie
Ice Cream
Pie
Ice Cream
Pie
Ice Cream
Pie
Ice Cream
5 - 15
Tree Diagrams
How many complete dinners are there?
12
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5 - 16
Tree Diagrams
How many dinners include beef?
1.
2.
4
3.
4.
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5 - 17
Tree Diagrams
What is the probability that a complete dinner will
include…
Juice?
6/12
Turkey?
4/12
Both beef
and soup?
2/12
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
See next
slide…
5 - 18
M * N Rule
If one thing can be done in M ways,
and if after this is done, something else
can be done in N ways,
then both things can be done in a
total of M*N different ways in that stated order!
Legend:
Appetizer
Entrée
Dessert
Refer back to tree diagram example:
# different meals =
2 * 3 * 2 = 12
# meals with beef =
2 * 1 * 2 = 4
# meals with juice =
1 * 3 * 2 = 6
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
5 - 19
When getting dressed, you have a
choice between wearing one of:
3 pairs of shoes
2 pairs of pants
5 shirts
Find the number of
different “outfits” possible
3 * 2 * 5 = 30
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
5 - 20
robability
What is the probability of
drawing a red Ace
from a deck of well-shuffled cards?
P( Red Ace) = 2/52
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5 - 21
robability
Deck = 52 Cards
4 Suits
Clubs Diamonds Hearts
Spades
13 cards in each
Key steps
1.
Using
robabilityAnalysis
Determine….the Outcomes that Meet Our
Condition
2.
List….all Possible Outcomes
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5 - 22
robability
P = probability …of getting four(4) aces
Deck = 52 Cards(the Population)
4 Suits
Clubs Diamonds Hearts Spades
13 cards in each
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
4 Suits
x
13 cards
5 - 23
robability
Deck = 52 Cards
4 Suits (13 cards in each)
Hearts
Spades Diamonds Clubs
Each Suit has a…….
‘Honours’
cards
Scenarios
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5 - 24
Deck = 52 Cards
4 Suits (13 cards in each)
Scenarios
Hearts
1. Draw an Ace
Spades Diamonds Clubs
Condition Outcomes
All Possible Outcomes
2. Draw a Black Ace Condition Outcomes
All Possible Outcomes
3. Draw a Red Card
Condition Outcomes
All Possible Outcomes
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
4
52
2
52
26
52
5 - 25
Deck = 52 Cards
Scenarios
4 Suits (13 cards in each)
Hearts
Spades Diamonds Clubs
4. Drawing…a Red Condition Outcomes
Card or a Queen All Possible Outcomes
-or-
26
52
+
2
52
=
P(Red) + P(Queen) - P (Red Queen)
=
26
+
4
52
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-
2
=
28
52
28
52
Deck = 52 Cards
5 - 26
4 Suits (13 cards in each)
Scenarios
Hearts
Spades Diamonds Clubs
What is the probability of drawing a Jack or
a King from a deck of well-shuffled cards?
= 4/52
= 4/52
P( Jack or King) = 4/52 + 4/52 = 8/52
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Deck = 52 Cards
5 - 27
4 Suits (13 cards in each)
Scenarios
Hearts
Spades Diamonds Clubs
What is the probability of drawing one card that
is both a Jack and a King from a deck of wellshuffled cards?
These are MUTUALLY EXCLUSIVE events,
i.e. they can’t both happen at the same time!
P( Jack and King)
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
=0
Deck = 52 Cards
5 - 28
4 Suits (13 cards in each)
Scenarios
Hearts
Spades Diamonds Clubs
What is the probability of drawing one card that
is both BLACK and a King from a deck of wellshuffled cards?
P( Black and King) =2/52
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Deck = 52 Cards
5 - 29
4 Suits (13 cards in each)
Scenarios
Hearts
Spades Diamonds Clubs
What is the probability of drawing a card that is
either BLACK or a King from a deck of wellshuffled cards?
Formula P(A or B) = P (A) + P(B) –P(Both)
This is called the Addition Rule
P( Black or King) = 26/52 + 4/52 - 2/52 = 28/52
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Deck = 52 Cards
5 - 30
4 Suits (13 cards in each)
Scenarios
Hearts
Spades Diamonds Clubs
What is the probability of drawing a King given
that you have drawn a BLACK card?
Our sample space is now just the BLACK cards
P(King|Black ) = 2/26
This is called a CONDITIONAL probability
Alternate solution
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Deck = 52 Cards
5 - 31
4 Suits (13 cards in each)
Scenarios
Hearts
Spades Diamonds Clubs
What is the probability of drawing a King given
that you have drawn a BLACK card?
P (Both)
Formula P(A|B) =
P(Given)
= (2/52) / (26/52)
= (2/52) * (52/26)
= 2/26
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Deck = 52 Cards
5 - 32
4 Suits (13 cards in each)
Scenarios
Hearts
Spades Diamonds Clubs
What is the probability of drawing a King of Clubs
given that you have drawn a BLACK card?
P (Both)
Formula P(A|B) =
P(Given)
P(King of Clubs|Black ) = (1/52) / (26/52)
= (1/52) * (52/26)
= 1/26
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Deck = 52 Cards
5 - 33
4 Suits (13 cards in each)
Scenarios
Hearts
Spades Diamonds Clubs
What is the probability of drawing a King of Clubs
given that you have drawn a CLUB?
P(King of Clubs given Club)
= P(King of Clubs|Club)
= P(1/52) / (13/52)
= (1/52) * (52/13)
= 1/13
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
5 - 34
Reading
Probabilities
from a Table
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Reading Probabilities
from a Table
5 - 35
A survey of undergraduate students in the School of
Business Management at Eton College revealed the
following regarding the gender and majors of the students:
Gender
Male
Female
Accounting International
HR
TOTAL
150
175
150
160
50
65
350
400
325
310
115
750
What is the Probability of selecting a female student?
400/750 = 53.33%
More
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Reading Probabilities
from a Table
5 - 36
What is the Probability of selecting a Human
Resources or International major?
Gender
Accounting International
Male
Female
HR
TOTAL
150
175
150
160
50
65
350
400
325
310
115
750
P(HR or I) = P(HR) + P(I)
= 115/750 + 310/750 = 425/750
= 56.67%
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
More
Reading Probabilities
from a Table
5 - 37
What is the Probability of selecting a Female
or International major?
Gender
Accounting International
Male
Female
HR
TOTAL
150
175
150
160
50
65
350
400
325
310
115
750
P(F or I) = P(F) + P(I) – P(F and I)
= 400/750 + 310/750 – 160/750
= 550/750 = 73.33%
More
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Reading Probabilities
from a Table
5 - 38
What is the Probability of selecting a Female
Accounting student?
Gender
Male
Female
Accounting International
HR
TOTAL
150
175
150
160
50
65
350
400
325
310
115
750
P(F and A) = 175/750 = 23.33%
More
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Reading Probabilities
from a Table
5 - 39
What is the Probability of selecting a Female,
given that the person selected
is an International major?
Gender
Male
Female
Accounting International
HR
TOTAL
150
175
150
160
50
65
350
400
325
310
115
750
P(F|I) = 160/310 = 51.6%
Alternative Solution
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Reading Probabilities
from a Table
What is the Probability of selecting a Female,
given that the person selected
is an International major?
P(Both)
Formula P(A|B) =
P(Given)
P(F|I) = P(F and I) / P(I)
= (160/750) / (310/750)
= 160/310
= 51.6%
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
5 - 40
Reading Probabilities
from a Table
5 - 41
What is the Probability of selecting an
International major, given that the person
selected is a Female?
Gender
Male
Female
Accounting International
HR
TOTAL
150
175
150
160
50
65
350
400
325
310
115
750
P(I|F) =
160/400 = 40%
More
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Reading Probabilities
from a Table
and
…between F given I
…I given F!
Notice the significant difference:
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5 - 42
Terminology
5 - 43
Independent Events
Events are independent if the occurrence of
one event does not affect the probability of the other
Each flip is
independent of the other!
Flip once
Flip twice
Consider the random
experiment of flipping a coin twice.
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Find the probability
of flipping
2 Heads in a row
P(2H) = .5*.5
= .25 or 25%
Terminology
5 - 44
Independent Events
Draw three cards with replacement
i.e., draw one card,
look at it,
put it back,
and repeat twice more.
Each draw is independent of the other
Find the probability of drawing 3 Queens in a row:
P(3Q) = 4/52 * 4/52 *4/52 = 0.00046 = most unlikely!
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Independent Events
Consider 2 events:
 Drawing a RED card from a deck of cards
 Drawing a HEART from a deck of cards
Are these two events considered to be independent?
If two events, A and B are independent, then
P(A|B) = P(A)
P(Red) = 26/52 = 1/2
P(Red|Heart) = 13/13 = 1
Therefore these are NOT independent events!
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5 - 45
5 - 46
ayes’
heorem
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5 - 47
ayes’
heorem
…is a method for revising a probability
given additional information!
Formula
P(A1|B) =
P(A1 ) P(B|A1 )
P(A1 ) P(B|A1)+ P(A2 )P(B|A2 )
Example
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5 - 48
ayes’
heorem
Duff Cola Company recently received several
complaints that their bottles are under-filled.
A complaint was received today but the production
manager is unable to identify which of the two
Springfield plants (A or B) filled this bottle.
What is the probability that the underfilled bottle came from plant A?
A
B
% of Total Production
% of Underfilled Bottles
55
45
3
4
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
5 - 49
ayes’
heorem
What is the probability that the underfilled bottle came from plant A?
A
B
1
2
% of Total Production
% of Underfilled Bottles
55
45
3
4
List the
P(plant A) = .55
Probabilities given
P(plant B) = .45
Input values into
formula and compute
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
P(Underfilled -A) = .03
P(Underfilled -B) = .04
5 - 50
ayes’
heorem
What is the probability that the underfilled bottle came from plant A?
1
2
List the
P(plant A) = .55
Probabilities given
P(plant B) = .45
Input values into
formula and compute
P(A1 ) P(B|A1 )
P(A1 |B) =
P(A1 )P(B|A1 )+ P(A2 ) P(B|A2 )
.55(.03)
=
.55(.03) + .45(.04)
= .4783
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
P(Underfilled/A) = .03
P(Underfilled/B) = .04
The likelihood that the
underfilled bottle came
from Plant A
has been reduced from
55% to 47.83%
5 - 51
Counting
Rules
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
actorials
!
… this is just a shorthand notation
that is sometimes used to save time!
Examples:
5! … Means 5*4*3*2*1 = 120
4! … Means 4*3*2*1 = 24
By definition, 1! =1 and 0! =1
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
5 - 52
5 - 53
ermutation
…is a counting technique
that is used when order is
n!
n
Pr =
important!
(n – r)!
ombination
…is a counting technique
that is used when order
n
Cr =
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
is NOT important!
n!
r!(n – r)!
5 - 54
ermutation
Pr =
n!
(n – r)!
…How many ways can you arrange n things,
taking r at a time, when order is important?
n
Example:
You are assigned the task of choosing 2 of your 6 classmates to
serve on a task force. One will act as the
Chair of the task force, and the other will be the Secretary.
In how many ways can you make this assignment?
6P2
= 6! / (6-2)! = 6! / 4! = 6*5 = 30
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5 - 55
ombination
…is a counting technique
that is used when order is NOT important!
n
Cr =
n!
r(n – r)!
Example:
You are assigned the task of choosing 2 of your 6 classmates to
serve on a task force. Responsibilities are evenly shared.
In how many ways can you make this assignment?
6C2
= 6! / (2!(6-2)!) = 6! /2!4! = (6*5)/2 = 15
Using…
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.
Using…
i Texas Instruments
ombination
5 - 56
BAII PLUS
30
15
ermutation
6
6
nCr
nPr
2
2
15
30
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Test your learning…
www.mcgrawhill.ca/college/lind
Online Learning Centre
for quizzes
extra content
data sets
searchable glossary
access to Statistics Canada’s E-Stat data
…and much more!
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5 - 57
5 - 58
This completes Chapter 5
Copyright © 2004 by The McGraw-Hill Companies, Inc. All rights reserved.