15 Chances, Probabilities, and Odds
15.1 Random Experiments and Sample
Spaces
15.2 Counting Outcomes in Sample
Spaces
15.3 Permutations and Combinations
15.4 Probability Spaces
15.5 Equiprobable Spaces
15.6 Odds
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Excursions in Modern Mathematics, 7e: 15.1 - 1
Random Experiment
• Probability is the quantification of
uncertainty.
• We will use the term random
experiment to describe an activity or a
process whose outcome cannot be
predicted ahead of time.
• Examples of random experiments:
tossing a coin, rolling a pair of dice,
drawing cards out of a deck of cards,
predicting the result of a football game,
and forecasting the path of a hurricane.
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Excursions in Modern Mathematics, 7e: 15.1 - 2
Sample Space
• Associated with every random
experiment is the set of all of its possible
outcomes, called the sample space of
the experiment.
• For the sake of simplicity, we will
concentrate on experiments for which
there is only a finite set of outcomes,
although experiments with infinitely many
outcomes are both possible and
important.
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Excursions in Modern Mathematics, 7e: 15.1 - 3
Sample Space - Set Notation
• We use the letter S to denote a sample
space and the letter N to denote the size
of the sample space S (i.e., the number
of outcomes in S).
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Excursions in Modern Mathematics, 7e: 15.1 - 4
Example 15.1
Tossing a Coin
One simple random experiment is to toss a
quarter and observe whether it lands heads or
tails. The sample space can be described by
S = {H, T}, where H stands for Heads and T
for Tails. Here N = 2.
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Excursions in Modern Mathematics, 7e: 15.1 - 5
Example 15.2
More Coin Tossing
Suppose we toss a coin twice and record the
outcome of each toss (H or T) in the order it
happens. What is the sample space?
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Excursions in Modern Mathematics, 7e: 15.1 - 6
Example 15.2
More Coin Tossing
The sample space now is
S = {HH, HT, TH, TT}, where HT means that
the first toss came up H and the second toss
came up T, which is a different outcome from
TH (first toss T and second toss H). In this
sample space N = 4.
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Excursions in Modern Mathematics, 7e: 15.1 - 7
Example 15.2
More Coin Tossing
Suppose now we toss two distinguishable
coins (say, a nickel and a quarter) at the
same time. What is the sample space?
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Excursions in Modern Mathematics, 7e: 15.1 - 8
Example 15.2
More Coin Tossing
The sample space is still
S = {HH, HT, TH, TT}. (Here we must agree
what the order of the symbols is–for example,
the first symbol describes the quarter and the
second the nickel.)
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Excursions in Modern Mathematics, 7e: 15.1 - 9
Example 15.2
More Coin Tossing
Since they have the same sample space, we
will consider the two previous random
experiments as the same random experiment.
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Excursions in Modern Mathematics, 7e: 15.1 - 10
Example 15.2
More Coin Tossing
Suppose we toss a coin twice, but we only
care now about the number of heads that
come up. What is the sample space?
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Excursions in Modern Mathematics, 7e: 15.1 - 11
Example 15.2
More Coin Tossing
Here there are only three possible outcomes
(no heads, one head, or both heads), and
symbolically we might describe this sample
space as
S = {0, 1, 2}.
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Excursions in Modern Mathematics, 7e: 15.1 - 12
Example 15.5
Dice Rolling
The experiment is to roll a pair of dice. What
is the sample space?
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Excursions in Modern Mathematics, 7e: 15.1 - 13
Example 15.5
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More Dice Rolling
Excursions in Modern Mathematics, 7e: 15.1 - 14
Example 15.5
More Dice Rolling
Here we have a sample space with 36
different outcomes. Notice that the dice are
colored white and red, a symbolic way to
emphasize the fact that we are treating the
dice as distinguishable objects. That is why
the following rolls are distinguishable.
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Excursions in Modern Mathematics, 7e: 15.1 - 15
Example 15.5

More Dice Rolling
The sample space has 36 possible outcomes:
{(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}
where the pairs represent the numbers rolled on
each dice (white, red).
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Excursions in Modern Mathematics, 7e: 15.1 - 16
Example 15.4
Rolling a Pair of Dice
Roll a pair of dice and consider the total of the
two numbers rolled. What is the sample
space?
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Excursions in Modern Mathematics, 7e: 15.1 - 17
Example 15.4
Rolling a Pair of Dice
The possible outcomes in this scenario range
from “rolling a two” to “rolling a twelve,” and
the sample space can be described by
S = {2, 3, 4, 5, 6, 7, 8, 9, 10,11,12}.
Copyright © 2010 Pearson Education, Inc.
Excursions in Modern Mathematics, 7e: 15.1 - 18
Examples
• Page 577, problem 3
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Excursions in Modern Mathematics, 7e: 15.1 - 19
Examples
• Page 577, problem 3
Solution:
• {ABCD, ABDC, ACBD, ACDB, ADBC,
ADCB, BACD, BADC, BCAD, BCDA,
BDAC, BDCA, CABD, CADB, CBAD,
CBDA, CDAB, CDBA, DABC, DACB,
DBAC, DBCA, DCAB, DCBA}
• There are 24 outcomes.
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Excursions in Modern Mathematics, 7e: 15.1 - 20
Not Listing All of the Outcomes
We would like to understand what the
sample space looks like without necessarily
writing all the outcomes down. Our real goal
is to find N, the size of the sample space. If
we can do it without having to list all the
outcomes, then so much the better.
Copyright © 2010 Pearson Education, Inc.
Excursions in Modern Mathematics, 7e: 15.1 - 21
15 Chances, Probabilities, and Odds
15.1 Random Experiments and Sample
Spaces
15.2 Counting Outcomes in Sample
Spaces
15.3 Permutations and Combinations
15.4 Probability Spaces
15.5 Equiprobable Spaces
15.6 Odds
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Excursions in Modern Mathematics, 7e: 15.1 - 22
Example 15.7
Tossing More Coins
• If we toss a coin three times and
separately record the outcome of each
toss, the sample space is given by S =
{HHH, HHT, HTH, HTT, THH, THT, TTH,
TTT}.
• Here we can just count the outcomes and
get N = 8.
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Excursions in Modern Mathematics, 7e: 15.1 - 23
Example 15.7
Tossing More Coins
• Toss a coin 10 times
• In this case the sample space S is too big
to write down
• We can “count” the number of outcomes in
S without having to tally them one by one
using the multiplication rule.
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Excursions in Modern Mathematics, 7e: 15.1 - 24
Multiplication Rule
• Suppose an activity consists of a series of
events in which there are a possible
outcomes for the first event, b possible
outcomes for the second event, c possible
outcomes for the third event, and so on.
• Then the total number of different possible
outcomes for the series of events is:
a·b·c·…
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Excursions in Modern Mathematics, 7e: 15.1 - 25
Example 15.7
Tossing More Coins
• Toss a coin ten times. How many
outcomes are in the sample space?
• There are two outcomes on the first toss
• There are two outcomes on the second
toss, etc.
• The total number of possible outcomes is
found by multiplying ten two’s together.
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Excursions in Modern Mathematics, 7e: 15.1 - 26
Example 15.7
Tossing More Coins
• Total number of outcomes if a coin is
tossed ten times:
N
2
2
22
10 factors
• Thus N = 210 = 1024.
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Excursions in Modern Mathematics, 7e: 15.1 - 27
Example 15.8
The Making of a
Wardrobe
Dolores is a young saleswoman planning her
next business trip. She is thinking about
packing three different pairs of shoes, four
skirts, six blouses, and two jackets. How
many different outfits will she be able to
create by combining these items? (Assume
that an outfit consists of one pair of shoes,
one skirt, one blouse, and one jacket.)
Copyright © 2010 Pearson Education, Inc.
Excursions in Modern Mathematics, 7e: 15.1 - 28
Example 15.8
The Making of a
Wardrobe
Let’s assume that an outfit consists of one
pair of shoes, one skirt, one blouse, and one
jacket. Then to make an outfit Dolores must
choose a pair of shoes (three choices), a skirt
(four choices), a blouse (six choices), and a
jacket (two choices). By the multiplication rule
the total number of possible outfits is
3 4 6 2 = 144.
Copyright © 2010 Pearson Education, Inc.
Excursions in Modern Mathematics, 7e: 15.1 - 29
Example 15.10 Ranking the Candidate
in an Election: Part 2
Five candidates are running in an election,
with the top three vote getters elected (in
order) as President, Vice President, and
Secretary. How many different ways are there
to choose the five candidates to fill these
three positions?
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Excursions in Modern Mathematics, 7e: 15.1 - 30
Example 15.10 Ranking the Candidate
in an Election: Part 2
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Excursions in Modern Mathematics, 7e: 15.1 - 31
Examples
• Page 578, problem 14
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Excursions in Modern Mathematics, 7e: 15.1 - 32
Examples
• Solution to part (a)
8 7 6 5 4 3 2 1 40,320
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Excursions in Modern Mathematics, 7e: 15.1 - 33
Examples
• Solution to part (a)
4 7 6 5 4 3 2 1 20,160
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Excursions in Modern Mathematics, 7e: 15.1 - 34
Examples
• Solution to part (c)
4 4 3 3 2 2 1 1 576
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Excursions in Modern Mathematics, 7e: 15.1 - 35
15 Chances, Probabilities, and Odds
15.1 Random Experiments and Sample
Spaces
15.2 Counting Outcomes in Sample
Spaces
15.3 Permutations and Combinations
(OMIT)
15.4 Probability Spaces
15.5 Equiprobable Spaces
15.6 Odds
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Excursions in Modern Mathematics, 7e: 15.1 - 36
Events
• An event is any subset of the sample
space. That is, an event is any set of
individual outcomes.
• This definition includes the possibility of an
“event” that has no outcomes as well as
events consisting of a single outcome.
• We denote events as E and the outcomes
that make up E are listed inside braces {
and } (that is, set notation)
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Excursions in Modern Mathematics, 7e: 15.1 - 37
Events
• An event that consists of no outcomes is
an impossible event.
– For the impossible event we use the empty
set so that E={ }.
• An event that consists of just one outcome
is a simple event.
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Excursions in Modern Mathematics, 7e: 15.1 - 38
Example 15.16 Coin-Tossing Event
• Suppose we toss a coin three times. The
sample space for this experiment is:
S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT }.
• Here N=8.
• Table 15-2 shows examples of events.
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Excursions in Modern Mathematics, 7e: 15.1 - 39
Example 15.16 Coin-Tossing Event
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Excursions in Modern Mathematics, 7e: 15.1 - 40
Dice Rolling
Recall the example 15.5.
The experiment is to roll a pair of dice, one
red and one white. The sample space is
depicted on the next page.
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Excursions in Modern Mathematics, 7e: 15.1 - 41
Example 15.5
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More Dice Rolling
Excursions in Modern Mathematics, 7e: 15.1 - 42
Events
a) Write the event E of rolling a sum of 7.
b) Write the event E of rolling the same
numbers on both dice.
c) Write the event of rolling at least one even
number.
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Excursions in Modern Mathematics, 7e: 15.1 - 43
Events
(a)
Or
E={(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}
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Excursions in Modern Mathematics, 7e: 15.1 - 44
Events
(b)
E={(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}
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Excursions in Modern Mathematics, 7e: 15.1 - 45
Events
(c)
E = {(1,2), (1,4), (1,6), (2,2), (2,4), (2,6),
(3,2), (3,4), (3,6), (4,2), (4,4), (4,6),
(5,2), (5,4), (5,6), (6,2), (6,4), (6,6),}
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Excursions in Modern Mathematics, 7e: 15.1 - 46
Examples
• Page 580, problem 44
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Excursions in Modern Mathematics, 7e: 15.1 - 47
Examples
• Solution to part (a)
E1 {TTFF , TFTF , TFFT , FTTF , FTFT , FFTT }
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Excursions in Modern Mathematics, 7e: 15.1 - 48
Examples
• Solution to part (b)
E2
{TTFF , TFTF , TFFT , FTTF , FTFT , FFTT ,
TTTF , TTFT , TFTT , FTTT , TTTT }
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Excursions in Modern Mathematics, 7e: 15.1 - 49
Examples
• Solution to part (c)
E3 {TTFF , TFTF , TFFT , FTTF , FTFT , FFTT ,
FFFT , FFTF , FTFF , TFFF , FFFF }
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Excursions in Modern Mathematics, 7e: 15.1 - 50
Examples
• Solution to part (d)
E4
Copyright © 2010 Pearson Education, Inc.
{TTFF , TTFT , TTTF , TTTT }
Excursions in Modern Mathematics, 7e: 15.1 - 51
Probability
• Defined as the long-term proportion of times
the outcome occurs
Building Blocks of Probability
• Experiment - any activity for which the
outcome is uncertain
• Outcome - the result of a single
performance of an experiment
• Sample space (S) - collection of all possible
outcomes
• Event - collection of outcomes from the
sample space
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Excursions in Modern Mathematics, 7e: 15.1 - 52
Probability
• The probability for any event E is
always between 0 and 1.
• If the event is impossible, its
probability is 0.
• If the event is equal to the sample
space, its probability is 1.
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Excursions in Modern Mathematics, 7e: 15.1 - 53
Probability Assignment
• A probability assignment assigns to
each simple event E in the sample space
a number between 0 and 1, which
represents the probability of the event E
and which we denote by Pr(E).
• Pr({ })=0
• Pr(S)=1
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Excursions in Modern Mathematics, 7e: 15.1 - 54
Probability Assignment
Every probability assignment must
obey the Law of Total Probability:
For any experiment, the sum of
all the outcome probabilities in
the sample space must equal 1.
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Excursions in Modern Mathematics, 7e: 15.1 - 55
Example 15.18 Handicapping a
Tennis Tournament
There are six players playing in a tennis
tournament: A (Russian, female), B (Croatian,
male), C (Australian, male), D (Swiss, male),
E (American, female), and F (American,
female). To handicap the winner of the
tournament we need a probability assignment
on the sample space S = {A, B, C, D, E, F }.
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Excursions in Modern Mathematics, 7e: 15.1 - 56
Example 15.18 Handicapping a
Tennis Tournament
With sporting events the probability
assignment is subjective (it reflects an
opinion), but a professional odds-maker
comes up with the following probability
assignment:
Event
A
B
C
Probability
0.08
0.16
0.20 0.25
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D
E
F
0.16
0.15
Excursions in Modern Mathematics, 7e: 15.1 - 57
Example 15.18 Handicapping a
Tennis Tournament
• Each probability is between 0 and 1.
• The sum of all outcome probabilities is 1
Pr(A) Pr(B) Pr(C ) Pr(D) Pr(E ) Pr(F )
0.08 0.16 0.20 0.25 0.16 0.15
1
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Excursions in Modern Mathematics, 7e: 15.1 - 58
Probability Space
Once a specific probability assignment is
made on a sample space, the combination of
the sample space and the probability
assignment is called a probability space.
Example:
S = {A, B, C, D, E, F }
Event
A
B
C
Probability
0.08
0.16
0.20 0.25
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D
E
F
0.16
0.15
Excursions in Modern Mathematics, 7e: 15.1 - 59
Examples
• Page 580, problem 38
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Excursions in Modern Mathematics, 7e: 15.1 - 60
Examples
• Solution part (a)
Pr(o1)=0.15= Pr(o4)
Pr(o2)=0.35= Pr(o3)
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Excursions in Modern Mathematics, 7e: 15.1 - 61
Examples
• Solution part (b)
Pr(o1)=0.15
Pr(o2)=0.35
Pr(o3)=0.22
Pr(o4)=0.28
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Excursions in Modern Mathematics, 7e: 15.1 - 62
15 Chances, Probabilities, and Odds
15.1 Random Experiments and Sample
Spaces
15.2 Counting Outcomes in Sample
Spaces
15.3 Permutations and Combinations
15.4 Probability Spaces
15.5 Equiprobable Spaces
15.6 Odds
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Excursions in Modern Mathematics, 7e: 15.1 - 63
Mutually Exclusive Events
• Events that have no outcomes in common
are called mutually exclusive events.
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Excursions in Modern Mathematics, 7e: 15.1 - 64
Mutually Exclusive Events
• Example of mutually exclusive events:
E1
event of rolling a sum of 7 on two different color dice
E1 {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}
E2
event of rolling the same number on two different color dice
E2
{(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}
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Excursions in Modern Mathematics, 7e: 15.1 - 65
Mutually Exclusive Events
• Example of events which are not mutually
exclusive:
E1
E2
event of rolling a sum of 7 on two different color dice
E1 {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}
event of rolling at least one 6 on two different color dice
E2 {(1,6), (2,6), (3,6), (4,6), (5,6), (6,6),
(6,1),(6,2),(6,3),(6,4),(6,5)}
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Excursions in Modern Mathematics, 7e: 15.1 - 66
Mutually Exclusive Events
• If events A and B are mutually exclusive,
then
Pr(A or B) = Pr(A)+Pr(B)
• If events A, B, and C are mutually
exclusive, then
Pr(A or B or C) = Pr(A)+Pr(B)+Pr(C)
Etc.
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Excursions in Modern Mathematics, 7e: 15.1 - 67
Example 15.18 Handicapping a
Tennis Tournament
(from 15.4) There are six players playing in a
tennis tournament and the events in the
sample space are:
A (Russian, female),
B (Croatian, male),
C (Australian, male),
D (Swiss, male),
E (American, female), and
F (American, female)
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Excursions in Modern Mathematics, 7e: 15.1 - 68
Example 15.18 Handicapping a
Tennis Tournament
The probability space for the tournament
winner is:
S = {A, B, C, D, E, F }
Event
A
B
C
D
Probability
0.08
0.16
0.20 0.25
E
F
0.16
0.15
Use the probability space to find the probability that an
American will win the tournament.
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Excursions in Modern Mathematics, 7e: 15.1 - 69
Example 15.18 Handicapping a
Tennis Tournament
Pr(E or F) = Pr(E) + Pr(F)
= 0.16 + 0.15
= 0.31
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Excursions in Modern Mathematics, 7e: 15.1 - 70
Example 15.18 Handicapping a
Tennis Tournament
What is the probability that a male will win the
tournament?
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Excursions in Modern Mathematics, 7e: 15.1 - 71
Example 15.18 Handicapping a
Tennis Tournament
Pr(B or C or D)
=Pr(B) + Pr(C) + Pr(D)
= 0.16 + 0.20 + 0.25
= 0.61
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Excursions in Modern Mathematics, 7e: 15.1 - 72
Example 15.18 Handicapping a
Tennis Tournament
What is the probability that an American male
will win the tournament?
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Excursions in Modern Mathematics, 7e: 15.1 - 73
Example 15.18 Handicapping a
Tennis Tournament
Pr({ }) = 0
(since this one is an impossible event–there
are no American males in the tournament)
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Excursions in Modern Mathematics, 7e: 15.1 - 74
Calculating Probabilities
• Next we determine how to calculate
probabilities for random experiments.
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Excursions in Modern Mathematics, 7e: 15.1 - 75
What Does Honesty Mean?
• What does honesty mean when applied
to coins, dice, or decks of cards?
• It essentially means that all individual
outcomes in the sample space are
equally probable.
• Thus, an honest coin is one in which H
and T have the same probability of
coming up, and an honest die is one in
which each of the numbers 1 through 6 is
equally likely to be rolled.
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Excursions in Modern Mathematics, 7e: 15.1 - 76
Equiprobable Spaces
• A probability space in which each simple
event has an equal probability of
occurring is called an equiprobable
space.
• Not all probability spaces are
equiprobable as the next example
shows.
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Excursions in Modern Mathematics, 7e: 15.1 - 77
Example 15.20 Rolling a Pair of Honest
Dice
Suppose that you are playing a game that
involves rolling a pair of honest dice, and the
only thing that matters is the total of the two
numbers rolled. The sample space in this
situation is
S = {2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12},
where the outcomes are the possible totals
that could be rolled.
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Excursions in Modern Mathematics, 7e: 15.1 - 78
Example 15.20 Rolling a Pair of Honest
Dice
This sample space would not represent an
equiprobable space. For example, the
likelihood of rolling a 7 is much higher than
that of rolling a 12.
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Excursions in Modern Mathematics, 7e: 15.1 - 79
Equiprobable Spaces
• If we know the probability space is an
equiprobable space, we can determine
the probability of any outcome as follows.
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Excursions in Modern Mathematics, 7e: 15.1 - 80
PROBABILITIES IN
EQUIPROBABLE SPACES
If k denotes the size of an event E and N
denotes the size of the sample space S,
then in an equiprobable space
Pr E
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k
N
Excursions in Modern Mathematics, 7e: 15.1 - 81
Example 15.19 Honest Coin Tossing
Suppose that a coin is tossed three times,
and we have been assured that the coin is an
honest coin. If this is true, then each of the
eight possible outcomes in the sample space:
{HHH HHT HTH THH TTH THT HTT TTT}
has probability 1/8.
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Excursions in Modern Mathematics, 7e: 15.1 - 82
Example 15.19 Honest Coin Tossing
This table shows each of the events with
their respective probabilities.
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Excursions in Modern Mathematics, 7e: 15.1 - 83
Example 15.20
More Dice Rolling
If we roll a pair of honest dice (one red and one white), each
of the 36 outcomes is equally likely:
S = {(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)}
where the pairs represent the numbers rolled on
each dice (white, red).
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Excursions in Modern Mathematics, 7e: 15.1 - 84
Example 15.20 Rolling a Pair of Honest
Dice
Because the dice are honest, each of these
36 possible outcomes is equally likely to
occur, so the probability of each is 1/36.
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Excursions in Modern Mathematics, 7e: 15.1 - 85
Example 15.20 Rolling a Pair of Honest
Dice
Table 15-5 (next slide) shows the probability
of rolling a sum of 2, 3, 4, . . . , 12.
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Excursions in Modern Mathematics, 7e: 15.1 - 86
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Excursions in Modern Mathematics, 7e: 15.1 - 87
Examples
• Page 581, problem 50 (a) and (c)
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Excursions in Modern Mathematics, 7e: 15.1 - 88
Example 15.22 Rolling a Pair of Honest
Dice
There are 16 outcomes in this sample space:
N
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2 2 2 2 2
4
16
Excursions in Modern Mathematics, 7e: 15.1 - 89
Examples
• Solution to part (a)
E1 {TTFF , TFTF , TFFT , FTTF , FTFT , FFTT }
which has k=6 outcomes so that
6
Pr(E1 )
16
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3
8
0.375
Excursions in Modern Mathematics, 7e: 15.1 - 90
Examples
• Solution to part (c)
E3 {TTFF , TFTF , TFFT , FTTF , FTFT , FFTT ,
FFFT , FFTF , FTFF , TFFF , FFFF }
which has k=11 outcomes so that
11
Pr(E3 )
0.6875
16
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Excursions in Modern Mathematics, 7e: 15.1 - 91
Example
Find the probability of drawing an ace when
drawing a single card at random from a
standard deck of 52 cards.
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Excursions in Modern Mathematics, 7e: 15.1 - 92
Example continued
Solution
• The sample space for the experiment:
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Excursions in Modern Mathematics, 7e: 15.1 - 93
Example continued
• If a card is chosen at random, then each card
has the same chance of being drawn.
• There are 52 outcomes in this sample space,
so N = 52.
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Excursions in Modern Mathematics, 7e: 15.1 - 94
Example 5.1 continued
• Let E be the event that an ace is drawn.
• Event E consists of the four aces
{A♥, A♦, A♣, A♠}, so k = 4.
• Therefore, the probability of drawing
an ace is
Pr(E) = 4/52 = 1/13 = 0.0769
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Excursions in Modern Mathematics, 7e: 15.1 - 95
Example

Experiment: A pair of dice are rolled. Define the
following events
A event the sum of the two dice equals 5
B event the sum of the two dice is 2
Find the probability that the sum of the two dice
equals five or the sum of the two dice equals 2.
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Excursions in Modern Mathematics, 7e: 15.1 - 96
Example
The event that the sum of the two dice equals five
or the event that the sum of the two dice equals two
are mutually exclusive and
Pr(A or B)
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Pr(A) Pr(B)
Excursions in Modern Mathematics, 7e: 15.1 - 97
Example
The event that the sum of the two dice equals five.
1,1 1,2 1,3 1,4 1,5 1,6
2,1 2,2 2,3 2,4 2,5 2,6
3,1 3,2 3,3 3,4 3,5 3,6
4,1 4,2 4,3 4,4 4,5 4,6
5,1 5,2 5,3 5,4 5,5 5,6
6,1 6,2 6,3 6,4 6,5 6,6
A {(1,4), (2,3), (3,2), (4,1)}
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Excursions in Modern Mathematics, 7e: 15.1 - 98
Example
The probability that the sum of the two dice equals five.
Pr( A)
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k
N
4
36
1
9
0.111
Excursions in Modern Mathematics, 7e: 15.1 - 99
Example
The event that the sum of the two dice equals two.
1,1 1,2 1,3 1,4 1,5 1,6
2,1 2,2 2,3 2,4 2,5 2,6
3,1 3,2 3,3 3,4 3,5 3,6
4,1 4,2 4,3 4,4 4,5 4,6
5,1 5,2 5,3 5,4 5,5 5,6
6,1 6,2 6,3 6,4 6,5 6,6
B {(1,1)}
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Excursions in Modern Mathematics, 7e: 15.1 - 100
Example
The probability that the sum of the two dice equals two.
Pr(B )
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k
N
1
36
0.0278
Excursions in Modern Mathematics, 7e: 15.1 - 101
Example
Pr(A or B)
Pr(A) Pr(B)
4 1
36 36
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5
36
0.139
Excursions in Modern Mathematics, 7e: 15.1 - 102
Example
• Page 581, problem 55 (d)
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Excursions in Modern Mathematics, 7e: 15.1 - 103
Example
• The outcomes in the sample space consist
of a sequence 10 of correct and incorrect
guesses (possibly all correct and possibly
all incorrect.
• For example, if we denote the event of
getting an answer correct as C and
incorrect as I we have outcomes like:
{ICICICICCI, CIIICCICCC, ETC.}
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Excursions in Modern Mathematics, 7e: 15.1 - 104
Example
• Each sequence has an associated score:
{ICICICICCI, CIIICCICCC, ETC.}
score
5 2.5 2.5
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score
6 2
4
Excursions in Modern Mathematics, 7e: 15.1 - 105
Example
There are 1024 outcomes in this sample
space:
10
N
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2
1024
Excursions in Modern Mathematics, 7e: 15.1 - 106
Example
• To get 8 or more points you can get all 10
correct (score=10 points) or 9 correct and
1 incorrect (score=8.5 points)
Denote:
A = event get all ten correct
B = event you get 9 correct and 1
incorrect
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Excursions in Modern Mathematics, 7e: 15.1 - 107
Example
There is one outcome in the sample space
that corresponds to getting all ten correct:
A={CCCCCCCCCC}
1
Pr(A)
1024
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Excursions in Modern Mathematics, 7e: 15.1 - 108
Example
• There are 10 outcomes in the sample
space to getting 9 correct (C) and 1
incorrect (I):
B={CCCCCCCCCI, CCCCCCCCIC, ETC.}
10
Pr(B)
1024
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Excursions in Modern Mathematics, 7e: 15.1 - 109
Example 15.22 Rolling a Pair of Honest
Dice
Since events A and B are mutually exclusive
we get that
Pr( A or B )
Pr( A) Pr(B )
1
10
1024 1024
11
0.0107
1024
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Excursions in Modern Mathematics, 7e: 15.1 - 110
Complement of E
• The complement of an event E is denoted:
EC
• Collection of outcomes in the sample space
that are not in event E
• Complement comes from the word “to
complete”
• Any event and its complement together
make up the complete sample space
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Excursions in Modern Mathematics, 7e: 15.1 - 111
Example
If a fair coin is tossed three times.
E = event that exactly one heads occurs
(for example, HTT)
What is the complement of E = EC ?
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Excursions in Modern Mathematics, 7e: 15.1 - 112
Example
Sample space:
S {HHH , HHT , HTH , HTT , THH , THT , TTH , TTT }
Event E:
E {HTT , THT , TTH }
Complement of E:
EC
{HHH , HHT , HTH , THH , TTT }
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Excursions in Modern Mathematics, 7e: 15.1 - 113
Example: find the probability of
the complement of an event
E is the event “observing a sum of 4 when
the two fair (honest) dice are rolled,”
Find the probability that you do not roll a 4.
Pr(EC)
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Excursions in Modern Mathematics, 7e: 15.1 - 114
Example continued
Solution
• Which outcomes belong to EC?
• There are the following outcomes in E:
{(3,1)(2,2)(1,3)}.
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Excursions in Modern Mathematics, 7e: 15.1 - 115
Example continued
• All the outcomes except the outcomes from A
in the two-dice sample space:
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Excursions in Modern Mathematics, 7e: 15.1 - 116
Example continued
• There are 33 outcomes in EC and 36
outcomes in the sample space.
• This gives the probability of not rolling a 4 to
be
Pr(EC) = 33/36 = 11/12 = 0.917
• The probability is high that, on this roll at
least, your roommate will not land on
Boardwalk.
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Excursions in Modern Mathematics, 7e: 15.1 - 117
Probabilities for Complements
• For any event E and its complement EC,
Pr(E) + Pr(EC) = 1
OR:
Pr(E) = 1 - Pr(EC)
OR:
Pr(EC ) = 1 - Pr(E)
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Excursions in Modern Mathematics, 7e: 15.1 - 118
Example
Consider the experiment of drawing a
card at random from a shuffled deck of
52 cards. Find the probability of
drawing a card that is not a face card.
NOTE: a face card is a king, queen, or jack
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Excursions in Modern Mathematics, 7e: 15.1 - 119
Example
• The sample space for the experiment where
a subject chooses a single card at random
from a deck of cards is depicted here.
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Excursions in Modern Mathematics, 7e: 15.1 - 120
Example
ANSWER:
There are 52 cards, 12 of which are face
cards.
E event you draw a face card
Pr(E )
EC
k
N
12
52
0.231
event you do not draw a face card
Pr(E C ) 1 0.231 0.769
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Excursions in Modern Mathematics, 7e: 15.1 - 121
Example
Handicapping a
Tennis Tournament
There are six players playing in a tennis
tournament and the events in the sample
space are:
A (Russian, female),
B (Croatian, male),
C (Australian, male),
D (Swiss, male),
E (American, female), and
F (American, female)
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Excursions in Modern Mathematics, 7e: 15.1 - 122
Example
Handicapping a
Tennis Tournament
The probability space for the tournament
winner is:
S = {A, B, C, D, E, F }
Event
A
B
C
D
Probability
0.08
0.16
0.20 0.25
E
F
0.16
0.15
Use the probability space to find the probability that a
Russian will not win the tournament.
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Excursions in Modern Mathematics, 7e: 15.1 - 123
Example
Handicapping a
Tennis Tournament
The probability a Russian will win the
tournament is:
Pr(A) 0.08
The probability a Russian will not win the
tournament is:
C
Pr(A ) 1 0.08 0.92
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Excursions in Modern Mathematics, 7e: 15.1 - 124
Independent Events
Two events are said to be independent
events if the occurrence of one event does
not affect the probability of the occurrence of
the other.
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Excursions in Modern Mathematics, 7e: 15.1 - 125
Independent Events
Examples of independent events:
1.Flip a coin three times; the outcome of any
one flip does not affect the probability of
another flip.
2.Roll a die four times; the outcome of any
one roll does not affect the probability of
another roll.
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Excursions in Modern Mathematics, 7e: 15.1 - 126
Example 15.22 Rolling a Pair of Honest
Dice
Imagine a game in which you roll an honest
die four times. Find the probability that you
roll a number that is not one on all four rolls of
the dice.
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Excursions in Modern Mathematics, 7e: 15.1 - 127
Example 15.22 Rolling a Pair of Honest
Dice
There are 1296 outcomes in this sample
space:
N
6 6 6 6 6
4
1296
k = number of events corresponding to the
event that you roll a number that is not one on
all four rolls of the dice
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Excursions in Modern Mathematics, 7e: 15.1 - 128
Example 15.22 Rolling a Pair of Honest
Dice: Part 3
Let R1, R2, R3, and R4 denote the outcomes
on roll 1, roll 2, roll 3, and roll 4.
Examples of outcomes that you do not roll 1:
R1 = 5, R2 = 3, R3 = 4, R4 = 2
R1 = 4, R2 = 6, R3 = 2, R4 = 2
It is not practical to try and list all of these
events to find k.
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Excursions in Modern Mathematics, 7e: 15.1 - 129
Independent Events
When events E and F are independent, the
probability that both occur is the product of
their respective probabilities; in other words,
Pr(E and F) = Pr(E) • Pr(F)
This is called the multiplication principle
for independent events.
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Excursions in Modern Mathematics, 7e: 15.1 - 130
Example 15.22 Rolling a Pair of Honest
Dice
Let F1, F2, F3, and F4 denote the events “first
roll is not a one,” “second roll is not a one,”
“third roll is not a one,” and “fourth roll is not a
one,” respectively. These events are
independent.
Let
F = F1 and F2 and F3 and F4
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Excursions in Modern Mathematics, 7e: 15.1 - 131
Example 15.22 Rolling a Pair of Honest
Dice
Then
Pr(F1) = 5/6,
Pr(F3) = 5/6,
Pr(F2) = 5/6
Pr(F4) = 5/6
Multiplication principle for independent events
gives the probability that you roll a number
that is not one on all four rolls of the dice.
Pr(F) = 5/6
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5/6
5/6
5/6 = (5/6)4 ≈ 0.482
Excursions in Modern Mathematics, 7e: 15.1 - 132
Example 15.22 Rolling a Pair of Honest
Dice
Imagine a game in which you roll an honest
die four times. If at least one of your rolls
comes up a one, you are a winner. Let E
denote the event “you win”.
Find Pr(E).
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Excursions in Modern Mathematics, 7e: 15.1 - 133
Example 15.22 Rolling a Pair of Honest
Dice
Let R1, R2, R3, and R4 denote the outcomes
on roll 1, roll 2, roll 3, and roll 4.
Examples of rolls where you are a winner:
R1 = 4, R2 = 3, R3 = 1, R4 = 6
R1 = 1, R2 = 5, R3 = 2, R4 = 1
It would be tedious to write out all possible
events in the sample space that correspond
to at least one of the rolls comes up a one.
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Excursions in Modern Mathematics, 7e: 15.1 - 134
Example 15.22 Rolling a Pair of Honest
Dice
The complement of E is the event that none
of the rolls comes up one; that is, the event
that you roll a number that is not one on all
four rolls of the dice.
Therefore we use the previous example and
the complement rule for probabilities to find
the answer.
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Excursions in Modern Mathematics, 7e: 15.1 - 135
Example 15.22 Rolling a Pair of Honest
Dice
Pr(E) = 1 – Pr(F) = 1 – 0.482 = 0.518
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Excursions in Modern Mathematics, 7e: 15.1 - 136
15 Chances, Probabilities, and Odds
15.1 Random Experiments and Sample
Spaces
15.2 Counting Outcomes in Sample
Spaces
15.3 Permutations and Combinations
15.4 Probability Spaces
15.5 Equiprobable Spaces
15.6 Odds
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Excursions in Modern Mathematics, 7e: 15.1 - 137
Example 15.19 Odds of Making
Free Throws
• In Example 15.15 discusses the fact that
Steve Nash (one of the most accurate freethrow shooters in NBA history) shoots free
throws with a probability of p = 0.90.
• We can interpret this to mean that on the
average, out of every 100 free
throws,Nash is going to make 90 and miss
about 10, for a hit/miss ratio of 9/1 or
9 to 1.
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Excursions in Modern Mathematics, 7e: 15.1 - 138
Example 15.19 Odds of Making
Free Throws
• This ratio of hits to misses gives what is
known as the odds of the event (in this
case Nash making the free throw).
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Excursions in Modern Mathematics, 7e: 15.1 - 139
ODDS
Let E be an arbitrary event. If F denotes
the number of ways that event E can
occur (the favorable outcomes or hits)
and U denotes the number of ways that
event E does not occur (the unfavorable
outcomes, or misses), then the odds of
(also called the odds in favor of) the
event E are given by the ratio:
F/U or F to U
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Excursions in Modern Mathematics, 7e: 15.1 - 140
ODDS
the odds against the event E are given
by the ratio
U/F or U to F
Note: “the odds of E” is the same as “the
odds in favor of E”
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Excursions in Modern Mathematics, 7e: 15.1 - 141
Example 15.26 Odds of Rolling a “Natural
Suppose that you are playing a game in
which you roll a pair of dice, presumably
honest. In this game, when you roll a “natural”
(i.e., roll a sum of 7 or sum of 11) you
automatically win. Find the odds of winning.
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Excursions in Modern Mathematics, 7e: 15.1 - 142
Example 15.26 Odds of Rolling a “Natural
If we let E denote the event “roll a natural,”
we can check that out of 36 possible
outcomes 8 are favorable:
6 ways to “roll a 7” plus two ways to “roll an 11”
and the other 28 are unfavorable. It follows
that the odds of rolling a “natural” are
8/28=2/7 or 2 to 7
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Excursions in Modern Mathematics, 7e: 15.1 - 143
Converting Odds to Probability
To convert odds into probabilities: If the
odds of E are F/U or F to U, then
Pr(E) = F/(F + U)
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Excursions in Modern Mathematics, 7e: 15.1 - 144
Example
• Page 582, problem 61
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Excursions in Modern Mathematics, 7e: 15.1 - 145
Example
• Page 582, problem 61
(a)3/8
(b)15/23
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Excursions in Modern Mathematics, 7e: 15.1 - 146
Converting Probability to Odds
To convert probabilities into odds when the
probability is given in the form of a fraction:
If Pr(E) = A/B then the odds of E are:
A
B A
or
A to B A
(When the probability is given in decimal
form, the best thing to do is to first convert
the decimal form into fractional form.)
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Excursions in Modern Mathematics, 7e: 15.1 - 147
Example
• Page 582, problem 60
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Excursions in Modern Mathematics, 7e: 15.1 - 148
Example
• Page 582, problem 60
(a)3/8
(b)3/5
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Excursions in Modern Mathematics, 7e: 15.1 - 149
Casinos and Bookmakers Odds
• There is a difference between odds as
discussed in this section and the payoff
odds posted by casinos or bookmakers in
sports gambling situations.
• Suppose we read in the newspaper, for
example, that the Las Vegas bookmakers
have established that “the odds that the
Boston Celtics will win the NBA
championship are 5 to 2.”
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Excursions in Modern Mathematics, 7e: 15.1 - 150
Casinos and Bookmakers Odds
• What this means is that if you want to bet
in favor of the Celtics, for every $2 that
you bet, you can win $5 if the Celtics win.
• The ratio 5 to 2 may be taken as some
indication of the actual odds in favor of the
Celtics winning, but several other factors
affect payoff odds, and the connection
between payoff odds and actual odds is
tenuous at best.
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Excursions in Modern Mathematics, 7e: 15.1 - 151
Casinos and Bookmakers Odds
This ratio may be taken as some indication of
the actual odds in favor of the Celtics
winning, but several other factors affect
payoff odds, and the connection between
payoff odds and actual odds is tenuous at
best.
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Excursions in Modern Mathematics, 7e: 15.1 - 152