15.1 Random Experiments and Sample Spacesjga001/chapter 15 slides.pdf• For the sake of simplicity, we will concentrate on experiments for which there is only a finite set of outcomes,

Excursions in Modern Mathematics, 7e: 15.1 - 1 Copyright © 2010 Pearson Education, Inc.

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


• 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.

Random Experiment


• 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.

Sample Space


• 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).

Sample Space - Set Notation


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.

Example 15.1 Tossing a Coin


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?

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.



Suppose now we toss two distinguishable

coins (say, a nickel and a quarter) at the

same time. What is the sample space?



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.)



Since they have the same sample space, we

will consider the two previous random

experiments as the same random experiment.



Suppose we toss a coin twice, but we only

care now about the number of heads that

come up. What is the sample space?



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}.



The experiment is to roll a pair of dice. What is the sample space?

Example 15.5 Dice Rolling


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.



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).


Roll a pair of dice and consider the total of the

two numbers rolled. What is the sample

space?

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}.

Example 15.4 Rolling a Pair of Dice


• Page 577, problem 3

Examples



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.

Examples


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.

Not Listing All of the Outcomes




Spaces


Spaces




15.6 Odds


• 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.

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.



• 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 · …

Multiplication Rule


• 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.



• Total number of outcomes if a coin is

tossed ten times:

• Thus N = 210 = 1024.


factors 10

2222N


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.)

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.

Example 15.8 The Making of a

Wardrobe


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?

Example 15.10 Ranking the Candidate

in an Election: Part 2


Example 15.10 Ranking the Candidate

in an Election: Part 2



Examples


• Solution to part (a)

Examples

320,4012345678



Examples

160,2012345674


• Solution to part (c)

Examples

57611223344




Spaces


Spaces


(OMIT)



15.6 Odds


• 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)

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.

Events


• 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.

Example 15.16 Coin-Tossing Event


Example 15.16 Coin-Tossing Event


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.

Dice Rolling




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.

Events


(a)

Or

E={(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}

Events


(b)

E={(1,1), (2,2), (3,3), (4,4), (5,5), (6,6)}

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),}

Events



Examples



Examples

},,,,,{1 FFTTFTFTFTTFTFFTTFTFTTFFE


• Solution to part (b)

Examples

},,,,

,,,,,,{2

TTTTFTTTTFTTTTFTTTTF

FFTTFTFTFTTFTFFTTFTFTTFFE



Examples

},,,,

,,,,,,{3

FFFFTFFFFTFFFFTFFFFT



• Solution to part (d)

Examples

},,,{4 TTTTTTTFTTFTTTFFE


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


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.


• 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

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.

Probability Assignment


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 }.

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:


Tennis Tournament

Event A B C D E F

Probability 0.08 0.16 0.20 0.25 0.16 0.15


• Each probability is between 0 and 1.

• The sum of all outcome probabilities is 1


Tennis Tournament

1

15.016.025.020.016.008.0

)Pr()Pr()Pr()Pr()Pr()Pr( FEDCBA


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:

Probability Space

Event A B C D E F

Probability 0.08 0.16 0.20 0.25 0.16 0.15

S = {A, B, C, D, E, F }



Examples


• Solution part (a)

Examples

Pr(o1)=0.15= Pr(o4)

Pr(o2)=0.35= Pr(o3)


• Solution part (b)

Examples

Pr(o1)=0.15

Pr(o2)=0.35

Pr(o3)=0.22

Pr(o4)=0.28




Spaces


Spaces




15.6 Odds


• Events that have no outcomes in common

are called mutually exclusive events.

Mutually Exclusive Events


• Example of mutually exclusive events:

dicecolor different on two 7 of sum a rolling ofevent 1E

dicecolor different on twonumber same therolling ofevent 2E

)}1,6(),2,5(),3,4(),4,3(),5,2(),6,1{(1E

)}6,6(),5,5(),4,4(),3,3(),2,2(),1,1{(2E



• Example of events which are not mutually

exclusive:

dicecolor different on two 7 of sum a rolling ofevent 1E

dicecolor different on two 6 oneleast at rolling ofevent 2E

)}1,6(),2,5(),3,4(),4,3(),5,2(),6,1{(1E

}(6,5)(6,4),(6,3),(6,2),(6,1),

),6,6(),6,5(),6,4(),6,3(),6,2(),6,1{(2E



• 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.



(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)


Tennis Tournament


The probability space for the tournament

winner is:


Tennis Tournament

Event A B C D E F

Probability 0.08 0.16 0.20 0.25 0.16 0.15

S = {A, B, C, D, E, F }

Use the probability space to find the probability that an

American will win the tournament.


Pr(E or F) = Pr(E) + Pr(F)

= 0.16 + 0.15

= 0.31


Tennis Tournament


What is the probability that a male will win the

tournament?


Tennis Tournament


Pr(B or C or D)

=Pr(B) + Pr(C) + Pr(D)

= 0.16 + 0.20 + 0.25

= 0.61


Tennis Tournament


What is the probability that an American male

will win the tournament?


Tennis Tournament


Pr({ }) = 0

(since this one is an impossible event–there

are no American males in the tournament)


Tennis Tournament


• Next we determine how to calculate

probabilities for random experiments.

Calculating Probabilities


• 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.

What Does Honesty Mean?


• 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.

Equiprobable Spaces


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.

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.


Dice


• If we know the probability space is an

equiprobable space, we can determine

the probability of any outcome as follows.

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

PROBABILITIES IN

EQUIPROBABLE SPACES

Pr E

k

N


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.

Example 15.19 Honest Coin Tossing


This table shows each of the events with

their respective probabilities.

Example 15.19 Honest Coin Tossing


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).


Because the dice are honest, each of these

36 possible outcomes is equally likely to

occur, so the probability of each is 1/36.


Dice


Table 15-5 (next slide) shows the probability

of rolling a sum of 2, 3, 4, . . . , 12.


Dice



• Page 581, problem 50 (a) and (c)

Examples


There are 16 outcomes in this sample space:


Dice

1622222 4N



which has k=6 outcomes so that

Examples

},,,,,{1 FFTTFTFTFTTFTFFTTFTFTTFFE

375.08

3

16

6)Pr( 1E



which has k=11 outcomes so that

Examples

},,,,

,,,,,,{3

FFFFTFFFFTFFFFTFFFFT


6875.016

11)Pr( 3E


Example

Find the probability of drawing an ace when

drawing a single card at random from a

standard deck of 52 cards.


Example continued

Solution

• The sample space for the experiment:


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.


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


Experiment: A pair of dice are rolled. Define the

following events

Find the probability that the sum of the two dice

equals five or the sum of the two dice equals 2.

5 equals dice two theof sum event the A

2 is dice two theof sum event the B

Example


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()Pr()or Pr( BABA


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

)}1,4(),2,3(),3,2(),4,1{(A


Example

The probability that the sum of the two dice equals five.

111.09

1

36

4)Pr(

N

kA


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

)}1,1{(B


Example

The probability that the sum of the two dice equals two.

0278.036

1)Pr(

N

kB


Example


139.036

5

36

1

36

4


• Page 581, problem 55 (d)

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.}

Example


• Each sequence has an associated score:

{ICICICICCI, CIIICCICCC, ETC.}

Example

5.25.25score 426score



1024210N

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

Example


There is one outcome in the sample space that corresponds to getting all ten correct:

A={CCCCCCCCCC}

1024

1)Pr(A

Example


• There are 10 outcomes in the sample

space to getting 9 correct (C) and 1

incorrect (I):

B={CCCCCCCCCI, CCCCCCCCIC, ETC.}

Example

1024

10)Pr(B



Dice

0107.01024

11

1024

10

1024

1


Since events A and B are mutually exclusive we get that


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


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 ?


Example

Sample space:

Event E:

Complement of E:

},,,,,,,{ TTTTTHTHTTHHHTTHTHHHTHHHS

},,,,{ TTTTHHHTHHHTHHHEC

},,{ TTHTHTHTTE


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)


Example continued

Solution

• Which outcomes belong to EC?

• There are the following outcomes in E:

{(3,1)(2,2)(1,3)}.


Example continued

• All the outcomes except the outcomes from A in the two-dice sample space:


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

• The probability is high that, on this roll at least, your roommate will not land on Boardwalk.

Pr(EC) = 33/36 = 11/12 = 0.917


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)


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


• The sample space for the experiment where a subject chooses a single card at random from a deck of cards is depicted here.

Example


Example

ANSWER:

There are 52 cards, 12 of which are face

cards.

769.0231.01)Pr( CE

card face a drawyou event E

card face a drawnot doyou event CE

231.052

12)Pr(

N

kE


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)

Example Handicapping a

Tennis Tournament


The probability space for the tournament

winner is:


Tennis Tournament

Event A B C D E F

Probability 0.08 0.16 0.20 0.25 0.16 0.15

S = {A, B, C, D, E, F }

Use the probability space to find the probability that a

Russian will not win the tournament.


The probability a Russian will win the

tournament is:

The probability a Russian will not win the

tournament is:


Tennis Tournament

08.0)Pr(A

92.008.01)Pr( CA


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.

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.

Independent Events


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.


Dice



k = number of events corresponding to the event that you roll a number that is not one on all four rolls of the dice


Dice

129666666 4N


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.


Dice: Part 3


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.

Independent Events


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


Dice


Then

Pr(F1) = 5/6, Pr(F2) = 5/6

Pr(F3) = 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 5/6 5/6 5/6 = (5/6)4 ≈ 0.482


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).


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.


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.


Dice


Pr(E) = 1 – Pr(F) = 1 – 0.482 = 0.518


Dice




Spaces


Spaces




15.6 Odds


• In Example 15.15 discusses the fact that

Steve Nash (one of the most accurate free-

throw 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.

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).

Example 15.19 Odds of Making

Free Throws


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

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”

ODDS


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.

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

Example 15.26 Odds of Rolling a “Natural


To convert odds into probabilities: If the

odds of E are F/U or F to U, then

Pr(E) = F/(F + U)

Converting Odds to Probability



Example



(a)3/8

(b)15/23

Example


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:

(When the probability is given in decimal

form, the best thing to do is to first convert

the decimal form into fractional form.)

Converting Probability to Odds

ABAAB

A toor



Example



(a)3/8

(b)3/5

Example


• 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.”

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.



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.


Documents

15.1 Random Experiments and Sample Spacesjga001/chapter 15 slides.pdf• For the sake of simplicity, we will concentrate on experiments for which there is only a finite set of outcomes,