15 Chances, Probabilities, and Odds

Home , Outcome (probability), Sample space

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

Copyright © 2010 Pearson Education, Inc. 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.

Copyright © 2010 Pearson Education, Inc. 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.

Copyright © 2010 Pearson Education, Inc. 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).

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.

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?

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}.

Copyright © 2010 Pearson Education, Inc. 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.

Copyright © 2010 Pearson Education, Inc. 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).

Roll a pair of dice and consider the total of the two numbers rolled. What is the sample space?

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 - 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.

Copyright © 2010 Pearson Education, Inc. 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.

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

Copyright © 2010 Pearson Education, Inc. 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.

Copyright © 2010 Pearson Education, Inc. 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.

• 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 · …

• 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: N 2222 10 factors • Thus N = 210 = 1024.

Copyright © 2010 Pearson Education, Inc. 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?

8 7 6 5 4 3 2 1 40,320

4 7 6 5 4 3 2 1 20,160

4 4 3 3 2 2 1 1 576

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 Copyright © 2010 Pearson Education, Inc. 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)

Copyright © 2010 Pearson Education, Inc. 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.

• 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.

The experiment is to roll a pair of dice, one red and one white. The sample space is depicted on the next page.

Copyright © 2010 Pearson Education, Inc. 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.

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

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

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),}

E1 {TTFF ,TFTF ,TFFT , FTTF , FTFT , FFTT}

E2 {TTFF,TFTF,TFFT , FTTF, FTFT , FFTT , TTTF ,TTFT ,TFTT , FTTT ,TTTT}

E3 {TTFF,TFTF,TFFT, FTTF, FTFT, FFTT, FFFT, FFTF, FTFF,TFFF, FFFF}

E4 {TTFF ,TTFT ,TTTF ,TTTT}

• 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

• 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.

Copyright © 2010 Pearson Education, Inc. 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

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.

Copyright © 2010 Pearson Education, Inc. 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 }.

Copyright © 2010 Pearson Education, Inc. 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 D E F Probability 0.08 0.16 0.20 0.25 0.16 0.15

Copyright © 2010 Pearson Education, Inc. 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

Copyright © 2010 Pearson Education, Inc. 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 D E F Probability 0.08 0.16 0.20 0.25 0.16 0.15

Pr(o1)=0.15= Pr(o4)

Pr(o2)=0.35= Pr(o3)

Pr(o1)=0.15

Pr(o2)=0.35

Pr(o3)=0.22

Pr(o4)=0.28

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

• 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)}

• Example of events which are not mutually exclusive:

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 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)}

• If events A, B, and C are mutually exclusive, then Pr(A or B or C) = Pr(A)+Pr(B)+Pr(C)

Etc.

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)

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

Event A B C D E F Probability 0.08 0.16 0.20 0.25 0.16 0.15

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

Pr(B or C or D) =Pr(B) + Pr(C) + Pr(D) = 0.16 + 0.20 + 0.25 = 0.61

Copyright © 2010 Pearson Education, Inc. 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?

Pr({ }) = 0

(since this one is an impossible event–there are no American males in the tournament)

Copyright © 2010 Pearson Education, Inc. 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.

Copyright © 2010 Pearson Education, Inc. 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.

Copyright © 2010 Pearson Education, Inc. 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.

Copyright © 2010 Pearson Education, Inc. 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.

Copyright © 2010 Pearson Education, Inc. 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.

If k denotes the size of an event E and N denotes the size of the sample space S, then in an equiprobable space

k Pr E 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.

This table shows each of the events with their respective probabilities.

Copyright © 2010 Pearson Education, Inc. 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).

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.

Copyright © 2010 Pearson Education, Inc. 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.

Copyright © 2010 Pearson Education, Inc. Excursions in Modern Mathematics, 7e: 15.1 - 86 Copyright © 2010 Pearson Education, Inc. Excursions in Modern Mathematics, 7e: 15.1 - 87 Examples • Page 581, problem 50 (a) and (c)

N 2 2 2 2 24 16

E {TTFF ,TFTF ,TFFT , FTTF , FTFT , FFTT} 1 which has k=6 outcomes so that

6 3 Pr(E ) 0.375 1 16 8

E3 {TTFF,TFTF,TFFT, FTTF, FTFT, FFTT, FFFT, FFTF, FTFF,TFFF, FFFF} which has k=11 outcomes so that

11 Pr(E ) 0.6875 3 16

Find the probability of drawing an ace when drawing a single card at random from a standard deck of 52 cards.

Solution • The sample space for the experiment:

• 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.

• 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

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.

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) Pr(A) Pr(B)

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)}

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

k 4 1 Pr(A) 0.111 N 36 9

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)}

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

k 1 Pr(B) 0.0278 N 36

Pr(A or B) Pr(A) Pr(B)

4 1 5 0.139 36 36 36

• 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.}

score 5 2.5 2.5 score 6 2 4

There are 1024 outcomes in this sample space: 10 N 2 1024

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

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

A={CCCCCCCCCC}

1 Pr(A) 1024

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

Copyright © 2010 Pearson Education, Inc. 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

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

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:

S {HHH, HHT, HTH, HTT,THH,THT,TTH,TTT}

Event E: E {HTT,THT,TTH}

Complement of E:

EC {HHH, HHT, HTH,THH,TTT}

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)

Solution

• Which outcomes belong to EC?

• There are the following outcomes in E: {(3,1)(2,2)(1,3)}.

• 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.

• 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)

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

Example

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

Copyright © 2010 Pearson Education, Inc. 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 k 12 Pr(E) 0.231 N 52 EC event you do not draw a face card

Pr(EC ) 1 0.231 0.769

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)

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

Event A B C D E F Probability 0.08 0.16 0.20 0.25 0.16 0.15

Use the probability space to find the probability that a Russian will not win the tournament.

The probability a Russian will not win the tournament is:

Pr(AC ) 1 0.08 0.92

Copyright © 2010 Pearson Education, Inc. 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.

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.

Copyright © 2010 Pearson Education, Inc. 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.

4 N 6 6 6 6 6 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

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.

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.

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

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

Copyright © 2010 Pearson Education, Inc. 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).

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.

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.

Pr(E) = 1 – Pr(F) = 1 – 0.482 = 0.518

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

Copyright © 2010 Pearson Education, Inc. 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 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.

Copyright © 2010 Pearson Education, Inc. 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).

Copyright © 2010 Pearson Education, Inc. 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

Note: “the odds of E” is the same as “the odds in favor of E”

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.

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

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

(a)3/8 (b)15/23

Copyright © 2010 Pearson Education, Inc. 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 or A to B A 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.)

(a)3/8 (b)3/5

Copyright © 2010 Pearson Education, Inc. 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.”

Copyright © 2010 Pearson Education, Inc. 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.

Copyright © 2010 Pearson Education, Inc. 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.