a mathematical view of our world

A Mathematical View A Mathematical View of Our Worldof Our World

11stst ed. ed.

Parks, Musser, Trimpe, Parks, Musser, Trimpe, Maurer, and MaurerMaurer, and Maurer

Chapter 10Chapter 10

ProbabilityProbability

Section 10.1Section 10.1Simple ExperimentsSimple Experiments

• GoalsGoals• Study probabilityStudy probability

• Experimental probabilityExperimental probability• Theoretical probabilityTheoretical probability

• Study probability propertiesStudy probability properties• Mutually exclusive eventsMutually exclusive events• Unions and intersections of eventsUnions and intersections of events• Complements of eventsComplements of events

10.1 Initial Problem10.1 Initial Problem

• Three cards were removed from Three cards were removed from wedding presents and then randomly wedding presents and then randomly replaced.replaced.

• What are the chances that at least one What are the chances that at least one of the gifts was paired with the correct of the gifts was paired with the correct card? card? • The solution will be given at the end of the section.The solution will be given at the end of the section.

Interpreting ProbabilityInterpreting Probability• Probability is the mathematics of Probability is the mathematics of

chance.chance.• For example, the statement “The For example, the statement “The

chances of winning the lottery game are chances of winning the lottery game are 1 in 150,000” means that only 1 of 1 in 150,000” means that only 1 of every 150,000 lottery tickets printed is a every 150,000 lottery tickets printed is a winning ticket.winning ticket.

Probability TerminologyProbability Terminology• Making an observation or taking a Making an observation or taking a

measurement is called an measurement is called an experimentexperiment..• An An outcomeoutcome is one of the possible results of is one of the possible results of

an experiment.an experiment.• The set of all possible outcomes is called the The set of all possible outcomes is called the

sample spacesample space..• An An eventevent is any collection of possible is any collection of possible

outcomes.outcomes.

Example 1Example 1

• The experiment consists of rolling a The experiment consists of rolling a standard six-sided die and recording standard six-sided die and recording the number of dots showing on the top the number of dots showing on the top face.face.• List the sample space.List the sample space.• List one possible event.List one possible event.

Example 1, cont’dExample 1, cont’d

• Solution: The sample space contains 6 Solution: The sample space contains 6 possible outcomes and can be written possible outcomes and can be written {1, 2, 3, 4, 5, 6}.{1, 2, 3, 4, 5, 6}.

• One possible event is {2, 4, 6}, which is One possible event is {2, 4, 6}, which is the event of getting an even number of the event of getting an even number of dots.dots.

Example 2Example 2

• The experiment consists of tossing a The experiment consists of tossing a coin 3 times and recording the results in coin 3 times and recording the results in order.order.• List the sample space.List the sample space.• List one possible event.List one possible event.


• Solution: The sample space contains 8 Solution: The sample space contains 8 possible outcomes and can be written possible outcomes and can be written {HHH, HHT, HTH, THH, HTT, THT, {HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.TTH, TTT}.

• One possible event is {HTH, HTT, TTH, One possible event is {HTH, HTT, TTH, TTT}, which is the event of getting a tail TTT}, which is the event of getting a tail on the second coin toss.on the second coin toss.

Example 3Example 3• The experiment The experiment

consists of spinning consists of spinning a spinner twice and a spinner twice and recording the colors recording the colors it lands on.it lands on.• List the sample List the sample

space.space.• List one possible List one possible

event.event.

Example 3, cont’dExample 3, cont’d• Solution: The sample space contains Solution: The sample space contains

16 possible outcomes and can be 16 possible outcomes and can be written {RR, RY, RG, RB, YR, YY, YG, written {RR, RY, RG, RB, YR, YY, YG, YB, GR, GY, GG, GB, BR, BY, BG, YB, GR, GY, GG, GB, BR, BY, BG, BB}.BB}.

• One possible event is {RR, YY, GG, One possible event is {RR, YY, GG, BB}, which is the event of getting the BB}, which is the event of getting the same color on both spins.same color on both spins.

Example 4Example 4

• The experiment consists of rolling 2 The experiment consists of rolling 2 standard dice and recording the standard dice and recording the number appearing on each die.number appearing on each die.• List the sample space.List the sample space.• List one possible event.List one possible event.


• Solution: The sample space contains 36 Solution: The sample space contains 36 possible outcomes and can be written {(1,1), possible outcomes and can be written {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (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), (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), (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), (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)}.(5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}.


• Solution, cont’d: One possible event is Solution, cont’d: One possible event is {(6,1), (5,2), (4,3), (3,4), (2,5), (1,6)}, {(6,1), (5,2), (4,3), (3,4), (2,5), (1,6)}, which is the event of getting a total of 7 which is the event of getting a total of 7 dots on the two dice.dots on the two dice.

Question:Question:

1. An experiment consists of tossing a coin 1. An experiment consists of tossing a coin and then rolling a 4-sided die. List the and then rolling a 4-sided die. List the outcomes in the sample space outcomes in the sample space

a. { H1, H2, H3, H4, T1, T2, T3, T4 }a. { H1, H2, H3, H4, T1, T2, T3, T4 }b. { H, T, 1, 2, 3, 4 }b. { H, T, 1, 2, 3, 4 }c. { H1, H2, H3, H4, H5, H6, T1, T2, T3, c. { H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6 }T4, T5, T6 }d. { H, T, 1, 2, 3, 4, 5, 6 }d. { H, T, 1, 2, 3, 4, 5, 6 }

Probability, cont’dProbability, cont’d• The probability of an event is a number The probability of an event is a number

from 0 to 1, and can be written as a from 0 to 1, and can be written as a fraction, decimal, or percent.fraction, decimal, or percent.• The greater the probability, the more likely The greater the probability, the more likely

the event is to occur.the event is to occur.• An impossible event has probability 0.An impossible event has probability 0.• A certain event has probability 1.A certain event has probability 1.

Experimental ProbabilityExperimental Probability

• One way to find the probability of an One way to find the probability of an event is to conduct a series of event is to conduct a series of experiments. experiments. • The The experimental probabilityexperimental probability is the relative is the relative

frequency with which an event occurs in a frequency with which an event occurs in a particular sequence of trials.particular sequence of trials.

Example 5Example 5• An experiment An experiment

consisted of tossing consisted of tossing 2 coins 500 times 2 coins 500 times and recording the and recording the resultsresults

• Let Let EE be the event of be the event of getting a head on the getting a head on the first coin and find the first coin and find the experimental experimental probability of probability of EE. .

Example 5, cont’dExample 5, cont’d• Solution: The event Solution: The event

EE is {HH, HT}. is {HH, HT}.• Event Event EE occurred a occurred a

total of 137 + 115 = total of 137 + 115 = 252 times out of 252 times out of 500.500.

• The experimental The experimental probability of probability of EE is is252 0.504500

Question:Question:A total of 200 people are given a A total of 200 people are given a taste test of 2 kinds of crackers. The taste test of 2 kinds of crackers. The results are that 148 of them prefer results are that 148 of them prefer Cracker A, 41 of them prefer Cracker Cracker A, 41 of them prefer Cracker B, and 11 have no preference.B, and 11 have no preference.Find the experimental probability of a Find the experimental probability of a randomly selected person preferring randomly selected person preferring Cracker B.Cracker B.

a. 74.0% a. 74.0% b. 5.5%b. 5.5%c. 20.5%c. 20.5% d. 27.7% d. 27.7%

Theoretical ProbabilityTheoretical Probability• Another way to find the probability of an Another way to find the probability of an

event is to use the theory of what event is to use the theory of what “should” happen rather than conducting “should” happen rather than conducting experiments.experiments.• The The theoretical probabilitytheoretical probability is the chance is the chance

an event will occur based on the situation, an event will occur based on the situation, such as tossing a fair coin and knowing such as tossing a fair coin and knowing each side should come up half of the time.each side should come up half of the time.

Theoretical Probability, cont’dTheoretical Probability, cont’d

• If all the outcomes in a sample space If all the outcomes in a sample space are equally likely to occur, then the are equally likely to occur, then the probability of event probability of event EE is equal to the is equal to the number of outcomes in number of outcomes in EE divided by the divided by the number of outcomes in the sample number of outcomes in the sample space space SS..• The probability of event The probability of event EE is written is written P(E).P(E).

Example 6Example 6

• An experiment consists of tossing 2 An experiment consists of tossing 2 fair coins. fair coins.

• Find the theoretical probability of:Find the theoretical probability of:a)a) Each outcome in the sample space.Each outcome in the sample space.

b)b) The event The event EE of getting a head on the first of getting a head on the first coin.coin.

c)c) The event of getting at least one head.The event of getting at least one head.


• Solution:Solution:a)a) There are 4 outcomes in the sample space: There are 4 outcomes in the sample space:

{HH, HT, TH, TT}. Each outcome is equally {HH, HT, TH, TT}. Each outcome is equally likely to occur. likely to occur.


• Solution, cont’d:Solution, cont’d:b)b) The event The event EE is {HH, HT} and the is {HH, HT} and the

theoretical probability of theoretical probability of EE is the number is the number of outcomes in of outcomes in EE divided by the number divided by the number of outcomes in the sample space.of outcomes in the sample space.

• 2 1 0.54 2

P E


• Solution, cont’d:Solution, cont’d:c)c) The event of getting at least one head is The event of getting at least one head is

EE = {HH, HT, TH}. = {HH, HT, TH}.

• 3 0.754

P E

Example 7Example 7

• An experiment consists of rolling 2 fair An experiment consists of rolling 2 fair dice. dice.

• Find the theoretical probability of:Find the theoretical probability of:a)a) Event Event AA: getting 7 dots.: getting 7 dots.

b)b) Event Event BB: getting 8 dots.: getting 8 dots.

c)c) Event Event CC: getting at least 4 dots.: getting at least 4 dots.


• Solution: There are 36 outcomes in the Solution: There are 36 outcomes in the sample space.sample space.

a)a) The event The event AA contains 6 outcomes: {(1,6), (2,5), contains 6 outcomes: {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}. Each outcome is (3,4), (4,3), (5,2), (6,1)}. Each outcome is equally likely to occur.equally likely to occur.

• 6 136 6

P A


• Solution, cont’d:Solution, cont’d:b)b) The event The event BB contains 5 equally likely contains 5 equally likely

outcomes: {(2,6), (3,5), (4,4), (5,3), outcomes: {(2,6), (3,5), (4,4), (5,3), (6,2)}.(6,2)}.

• 536

P B

Example 7, cont’dExample 7, cont’d• Solution, cont’d:Solution, cont’d:

c)c) The event The event CC contains 33 equally likely contains 33 equally likely outcomes: {outcomes: {(1,3), (1,4), (1,5), (1,6), (2,2), (1,3), (1,4), (1,5), (1,6), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (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), (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), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)(6,3), (6,4), (6,5), (6,6)}.}.

• 33 1136 12

P C

Example 8Example 8• A jar contains four A jar contains four

marbles: 1 red, 1 marbles: 1 red, 1 green, 1 yellow, and green, 1 yellow, and 1 white. 1 white.

Example 8, cont’dExample 8, cont’d• If we draw 2 marbles in a row, without If we draw 2 marbles in a row, without

replacing the first one, find the probability replacing the first one, find the probability of:of:

a)a) Event Event AA: One of the marbles is red.: One of the marbles is red.

b)b) Event Event BB: The first marble is red or yellow.: The first marble is red or yellow.

c)c) Event Event CC: The marbles are the same color.: The marbles are the same color.

d)d) Event Event DD: The first marble is not white.: The first marble is not white.

e)e) Event Event EE: Neither marble is blue.: Neither marble is blue.


• Solution: The sample space contains Solution: The sample space contains 12 outcomes: {RG, RY, RW, GR, GY, 12 outcomes: {RG, RY, RW, GR, GY, GW, YR, YG, YW, WR, WG, WY}.GW, YR, YG, YW, WR, WG, WY}.

a)a) Event Event AA: One of the marbles is red.: One of the marbles is red.• AA = {RG, RY, RW, GR, YR, WR}. = {RG, RY, RW, GR, YR, WR}.

• 6 112 2

P A


b)b) Event Event BB: The first marble is red or yellow.: The first marble is red or yellow.• BB = {RG, RY, RW, YR, YG, YW}. = {RG, RY, RW, YR, YG, YW}.

•

c)c) Event Event CC: The marbles are the same color.: The marbles are the same color.• CC = { }. = { }.

• 0 012

P C

6 112 2

P B


d)d) Event Event DD: The first marble is not white.: The first marble is not white.• DD = {RG, RY, RW, GR, GY, GW, YR, YG, = {RG, RY, RW, GR, GY, GW, YR, YG,

YW}.YW}.•

e)e) Event Event EE: Neither marble is blue.: Neither marble is blue.• EE = = SS• 12 1

12P E

9 312 4

P D

Question:Question:

A jar contains 1 red marble, 1 green A jar contains 1 red marble, 1 green marble, and 1 blue marble. You draw marble, and 1 blue marble. You draw 2 marbles in a row, without replacing 2 marbles in a row, without replacing the first one. What is the probability the first one. What is the probability of the event E of the first marble of the event E of the first marble being red or yellow? being red or yellow?

a. P(E) = 0 a. P(E) = 0 b. P(E) = 1/3b. P(E) = 1/3c. P(E) = ½c. P(E) = ½ d. P(E) = 2/3d. P(E) = 2/3

Union and IntersectionUnion and Intersection

• The The unionunion of two events, of two events, AA U U BB, , refers to all outcomes that are in one, refers to all outcomes that are in one, the other, or both events.the other, or both events.

• The The intersectionintersection of two events, of two events, AA ∩ ∩ B,B, refers to outcomes that are in both refers to outcomes that are in both events.events.

Mutually Exclusive EventsMutually Exclusive Events

• Events that have no outcomes in Events that have no outcomes in common are said to be common are said to be mutually mutually exclusiveexclusive..

•

• If A and B are mutually exclusive If A and B are mutually exclusive events, then events, then

A B

P A B P A P B

Example 9Example 9

• A card is drawn from a standard deck of A card is drawn from a standard deck of cards.cards.• Let Let AA be the event the card is a face card. be the event the card is a face card.• Let Let BB be the event the card is a black 5. be the event the card is a black 5.

• Find and interpret P(Find and interpret P(AA U U BB).).

Example 9, cont’dExample 9, cont’d• Solution: The sample space contains Solution: The sample space contains

52 equally likely outcomes. 52 equally likely outcomes.


• Solution, cont’d: Event Solution, cont’d: Event AA has 12 has 12 outcomes, one for each of the 3 face outcomes, one for each of the 3 face cards in each of the 4 suits.cards in each of the 4 suits.• P(P(AA) = 12/52.) = 12/52.

• Event B has 2 outcomes, because Event B has 2 outcomes, because there are 2 black fives.there are 2 black fives.• P(P(BB) = 2/52.) = 2/52.


• Solution, cont’d: Events Solution, cont’d: Events AA and and BB are are mutually exclusive because it is mutually exclusive because it is impossible for a 5 to be a face card. impossible for a 5 to be a face card. • P(P(A A UU B B) = 12/52 + 2/52 = 14/52 = 7/26.) = 12/52 + 2/52 = 14/52 = 7/26.• This is the probability of drawing either a This is the probability of drawing either a

face card or a black 5.face card or a black 5.

Complement of an EventComplement of an Event• The set of outcomes in a sample space The set of outcomes in a sample space SS, ,

but not in an event but not in an event EE, is called the , is called the complement of the event complement of the event EE. . • The complement of The complement of EE is written is written ĒĒ. .

Complement of an Event, cont’dComplement of an Event, cont’d• The relationship between the probability The relationship between the probability

of an event of an event EE and the probability of its and the probability of its complement complement ĒĒ is given by: is given by:

•

•

1P E P E

1P E P E

Example 10Example 10• In a number matching game, In a number matching game,

• First Carolan chooses a whole number from 1 First Carolan chooses a whole number from 1 to 4.to 4.

• Then Mary guesses a number from 1 to 4.Then Mary guesses a number from 1 to 4.

a)a) What is the probability the numbers are What is the probability the numbers are equal?equal?

b)b) What is the probability the numbers are What is the probability the numbers are unequal?unequal?

Example 10, cont’dExample 10, cont’d• Solution: The sample space contains 16 Solution: The sample space contains 16

outcomes: { (1,1), (1,2), (1,3), (1,4), (2,1), outcomes: { (1,1), (1,2), (1,3), (1,4), (2,1), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4), (2,2), (2,3), (2,4), (3,1), (3,2), (3,3), (3,4), (4,1), (4,2), (4,3), (4,4) }.(4,1), (4,2), (4,3), (4,4) }.

a)a) Let Let EE be the event the numbers are equal. be the event the numbers are equal.• P(P(EE) = ¼ ) = ¼

b)b) Then Then ĒĒ is the event the numbers are is the event the numbers are unequal.unequal.

• P(P(ĒĒ) = 1 – ¼ = ¾ ) = 1 – ¼ = ¾

Example 11Example 11

• A diagram of a sample space S for an A diagram of a sample space S for an experiment with equally likely outcomes is experiment with equally likely outcomes is shown.shown.

Example 11, cont’dExample 11, cont’d• Find the probability of each of the events: Find the probability of each of the events:

• SS• • AA• BB• CC• • • •

C

A BA C

A B

Example 11, cont’dExample 11, cont’d• Solution:Solution:

Properties of ProbabilityProperties of Probability• For a sample space For a sample space SS and events and events AA

and and BB::1)1) For any event For any event AA, ,

2)2)

3)3) P(P(SS) = 1) = 1

0 1P A

0P

Properties of Probability, cont’dProperties of Probability, cont’d• For a sample space For a sample space SS and events and events AA and and BB::4)4) If events If events AA and and BB are mutually exclusive, are mutually exclusive,

then then

5)5) If If AA and and BB are any events, then are any events, then

6)6) For any event For any event AA and its complement: and its complement:

P A B P A P B P A B

P A B P A P B

1P A P A


• An experiment An experiment consists of spinning consists of spinning the spinner once the spinner once and recording the and recording the number on which it number on which it lands.lands.

Example 12, cont’dExample 12, cont’d• Define 4 events:Define 4 events:

• A: an even numberA: an even number• B: a number greater than 5B: a number greater than 5• C: a number less than 3C: a number less than 3• D: a number other than 2D: a number other than 2

a)a) Find P(Find P(AA), P(), P(BB), P(), P(CC), and P(), and P(DD).).b)b) Find and interpret P(Find and interpret P(AA U U BB) and P() and P(AA ∩ ∩ BB).).c)c) Find and Find and interpret P(interpret P(BB U U CC) and P() and P(BB ∩ ∩ CC).).

Example 12, cont’dExample 12, cont’d• Solution: The sample space has 8 Solution: The sample space has 8

equally likely outcomes: {1, 2, 3, 4, 5, equally likely outcomes: {1, 2, 3, 4, 5, 6, 7, 8}.6, 7, 8}.

a)a) Find P(Find P(AA),P(),P(BB), P(), P(CC), and P(), and P(DD).).• AA = {2, 4, 6, 8}, so P( = {2, 4, 6, 8}, so P(AA) = 4/8 = 1/2) = 4/8 = 1/2• B B = {6, 7, 8}, so P(= {6, 7, 8}, so P(BB) = 3/8) = 3/8• CC = {1, 2}, so P( = {1, 2}, so P(CC) = 2/8 = 1/4) = 2/8 = 1/4• DD = {1, 3, 4, 5, 6, 7, 8}, so P( = {1, 3, 4, 5, 6, 7, 8}, so P(DD) = 7/8.) = 7/8.

Example12, cont’dExample12, cont’d• Solution, cont’d:Solution, cont’d:b)b) Find and interpret P(Find and interpret P(AA U U BB) and ) and

P(P(AA ∩ ∩ BB).).• AA and and B B are not mutually exclusive, soare not mutually exclusive, so

• AA ∩ ∩ B = B = {6, 8}, so P( {6, 8}, so P(AA ∩ ∩ BB)) = 2/8 = 2/8

•

P A B P A P B P A B

4 3 2 58 8 8 8

P A B

Example 12, cont’dExample 12, cont’d• Solution, cont’d:Solution, cont’d:c)c) Find and Find and interpret P(interpret P(BB U U CC) and ) and

P(P(BB ∩ ∩ CC).). • BB and and C C are mutually exclusive, soare mutually exclusive, so

andand

•

P B C P B P C

3 2 58 8 8

P B C

0P B C

10.1 Initial Problem Solution10.1 Initial Problem Solution

• Three cards were removed from Three cards were removed from wedding presents and then randomly wedding presents and then randomly replaced. What are the chances that at replaced. What are the chances that at least one of the gifts was paired with the least one of the gifts was paired with the correct card? correct card?

Initial Problem Solution, cont’dInitial Problem Solution, cont’d

• Solution: Let Solution: Let E E be the be the event that at least event that at least one gift is paired with one gift is paired with the right card.the right card.• Label the gifts A, B, Label the gifts A, B,

and C and the cards and C and the cards a, b, and c.a, b, and c.

• The sample space The sample space contains 6 outcomes.contains 6 outcomes.


• Solution: Four of the 6 outcomes correspond Solution: Four of the 6 outcomes correspond to at least one gift being paired with the right to at least one gift being paired with the right card.card.

• The probability of this event occurring isThe probability of this event occurring is

4 26 3

P E

Section 10.2Section 10.2Multistage ExperimentsMultistage Experiments

• GoalsGoals• Study tree diagramsStudy tree diagrams

• Study the fundamental counting principleStudy the fundamental counting principle

• Study probability tree diagramsStudy probability tree diagrams• Study the additive propertyStudy the additive property• Study the multiplicative propertyStudy the multiplicative property

10.2 Initial Problem10.2 Initial Problem

• A friend who likes to gamble wagers A friend who likes to gamble wagers that if you toss a coin repeatedly you that if you toss a coin repeatedly you will get 2 tails before you get 3 heads.will get 2 tails before you get 3 heads.

• Should you take the bet?Should you take the bet?• The solution will be given at the end of the The solution will be given at the end of the

section.section.

Tree DiagramsTree Diagrams• A A tree diagramtree diagram is a visual aid that can is a visual aid that can

be used to represent the outcomes of be used to represent the outcomes of an experiment.an experiment.

• You can construct a one-stage tree You can construct a one-stage tree diagram by starting from a single point:diagram by starting from a single point:1)1)Draw one branch for each outcome.Draw one branch for each outcome.

2)2)Place a label at the end of the branch to Place a label at the end of the branch to represent each outcome.represent each outcome.

Example 1Example 1

• A tree diagram is A tree diagram is drawn for the drawn for the experiment of experiment of drawing 1 ball from drawing 1 ball from a box containing 1 a box containing 1 red ball, 1 white ball, red ball, 1 white ball, and 1 blue ball.and 1 blue ball.

Tree Diagrams, cont’dTree Diagrams, cont’d• You can construct a two-stage tree diagram You can construct a two-stage tree diagram

for experiments that consist of a sequence for experiments that consist of a sequence of 2 actions:of 2 actions:1)1) Draw a one-stage tree diagram, made of Draw a one-stage tree diagram, made of

primary branchesprimary branches, for the outcomes of the first , for the outcomes of the first action.action.

2)2) Starting at the end of each branch of the tree Starting at the end of each branch of the tree from Step 1, draw a one-stage tree diagram, from Step 1, draw a one-stage tree diagram, made of made of secondary branchessecondary branches, for each , for each outcome of the second action. outcome of the second action.

Example 2Example 2

• There are There are 12 12 possible possible outcomes.outcomes.

Question:Question:How many outcomes in the How many outcomes in the tree diagram correspond to tree diagram correspond to getting either a red or green getting either a red or green marble first and either a marble first and either a yellow or green marble yellow or green marble second? second?

a. 3a. 3 b. 6b. 6c. 4c. 4 d. 2d. 2

Fundamental Counting PrincipleFundamental Counting Principle• The number of outcomes in an experiment The number of outcomes in an experiment

can also be determined using the can also be determined using the fundamental counting principlefundamental counting principle::• If an event or action If an event or action AA can occur in can occur in rr ways, and , ways, and ,

for each of these for each of these rr ways, an event or action ways, an event or action BB can occur in can occur in ss ways, then the number of ways ways, then the number of ways events or actions events or actions AA and and BB can occur, in can occur, in succession, is succession, is rsrs..

• The principle can be extended to more than two The principle can be extended to more than two events or actions.events or actions.

Example 3Example 3

• The options on a pizza are:The options on a pizza are:• Small, medium, or largeSmall, medium, or large• White or wheat crustWhite or wheat crust• Sausage, pepperoni, bacon, onion, Sausage, pepperoni, bacon, onion,

mushroomsmushrooms• How many different one-topping pizzas How many different one-topping pizzas

are possible?are possible?

Example 3, cont’dExample 3, cont’d• Solution: Solution:

• The first action is choosing 1 of 3 sizes.The first action is choosing 1 of 3 sizes.• The second action is choosing 1 of 2 The second action is choosing 1 of 2

crusts.crusts.• The third action is choosing 1 of 5 The third action is choosing 1 of 5

toppings.toppings.• There are 3(2)(5) = 30 different one-There are 3(2)(5) = 30 different one-

topping pizzas possible.topping pizzas possible.

Question:Question:

In the previous example, what is the In the previous example, what is the probability of someone randomly probability of someone randomly selecting a large, whole wheat pizza selecting a large, whole wheat pizza with any one of the 5 toppings? with any one of the 5 toppings?

a. P(E) = 1a. P(E) = 1 b. P(E) = 1/30b. P(E) = 1/30c. P(E) =1/3c. P(E) =1/3 d. P(E) = 1/6d. P(E) = 1/6

Example 4Example 4

• Find the probability of getting a sum of Find the probability of getting a sum of 11 when tossing a pair of fair dice.11 when tossing a pair of fair dice.

Example 4, cont’dExample 4, cont’d• Solution: There are 36 equally likely Solution: There are 36 equally likely

outcomes in the sample space.outcomes in the sample space.• 6 possible outcomes on the first roll 6 possible outcomes on the first roll • 6 possible outcomes on the second roll6 possible outcomes on the second roll• 6(6) = 366(6) = 36

• There are 2 ways of rolling a sum of There are 2 ways of rolling a sum of 11: (5,6) and (6,5).11: (5,6) and (6,5).• The probability is 2/36 = 1/18. The probability is 2/36 = 1/18.

Example 5Example 5

• Suppose 2 cards are drawn from a Suppose 2 cards are drawn from a standard deck.standard deck.

• Use a two-stage tree diagram to find Use a two-stage tree diagram to find the probability of getting a pair.the probability of getting a pair.

Example 5, cont’dExample 5, cont’d• Solution:Solution: A partial A partial

tree diagram is tree diagram is shown at right.shown at right.• There are 52 primary There are 52 primary

branches.branches.• There are 51 There are 51

secondary branches.secondary branches.• So there are a total So there are a total

of 52(51) = 2652 of 52(51) = 2652 possible outcomes. possible outcomes.

Example 5, cont’dExample 5, cont’d• Solution, cont’d: Solution, cont’d:

Consider only the Consider only the outcomes that result outcomes that result in a pair.in a pair.• There are 52 primary There are 52 primary

branches.branches.• There are 3 There are 3

secondary branches.secondary branches.• So there are a total So there are a total

of 52(3) = 156 pairs. of 52(3) = 156 pairs.


• Solution, cont’d: Solution, cont’d: • The probability of drawing a pair is: The probability of drawing a pair is:

156 12652 17

Probability Tree DiagramsProbability Tree Diagrams

• Tree diagrams can also be used to Tree diagrams can also be used to determine probabilities in multistage determine probabilities in multistage experiments.experiments.

• Tree diagrams that are labeled with the Tree diagrams that are labeled with the probabilities of events are called probabilities of events are called probability tree diagramsprobability tree diagrams..

Probability Tree Diagrams, cont’dProbability Tree Diagrams, cont’d

Example 6Example 6

• Draw a probability tree diagram to Draw a probability tree diagram to represent the experiment of drawing represent the experiment of drawing one ball from a container holding 2 red one ball from a container holding 2 red balls and 3 white balls.balls and 3 white balls.

Example 6, cont’dExample 6, cont’d• Solution: The Solution: The

first tree has one first tree has one branch for each branch for each ball. ball.

• The second tree The second tree was simplified was simplified by combining by combining branches.branches.

Probability Tree Diagrams, cont’dProbability Tree Diagrams, cont’d

• If an event If an event EE is the union of events is the union of events EE11, , EE22, …, , …, EEnn, where each pair of events is , where each pair of events is mutually exclusive, then:mutually exclusive, then:

• This is called the This is called the additive propertyadditive property of of probability tree diagrams.probability tree diagrams.

1 2

1 2

n

n

P E P E E E

P E P E P E

Example 7Example 7• Draw a probability Draw a probability

tree diagram to tree diagram to represent the represent the experiment of experiment of spinning the spinner spinning the spinner once.once.

• Find the probability Find the probability of landing on white of landing on white or on green.or on green.

Example 7, cont’dExample 7, cont’d• Solution: There Solution: There

are 4 outcomes in are 4 outcomes in the sample space.the sample space.• They are not all They are not all

equally likely.equally likely.• They are all They are all

mutually mutually exclusive.exclusive.


• Solution, cont’d: Use Solution, cont’d: Use the central angles to the central angles to find the probability of find the probability of each outcome and each outcome and draw the probability draw the probability tree diagram. tree diagram.

Example 7, cont’dExample 7, cont’d• Solution, cont’d: Solution, cont’d:

The probability of The probability of white or green is:white or green is:

1 1 73 4 12

P W G

P W P G

Example 8Example 8

• A jar contains 3 marbles, 2 A jar contains 3 marbles, 2 black and 1 red. black and 1 red.

• A marble is draw and A marble is draw and replaced, and then a replaced, and then a second marble is drawn. second marble is drawn. What is the probability What is the probability both marbles are black?both marbles are black?

Example 8, cont’dExample 8, cont’d• Solution: Draw a probability tree diagram to Solution: Draw a probability tree diagram to

represent the experiment.represent the experiment.

Example 8, cont’dExample 8, cont’d• Solution, cont’d: Assign a probability to the end of Solution, cont’d: Assign a probability to the end of

each secondary branch.each secondary branch.


• Solution, cont’d: Either tree diagram can be Solution, cont’d: Either tree diagram can be used to find that P(used to find that P(BBBB) = 4/9.) = 4/9.

Probability Tree Diagrams, cont’dProbability Tree Diagrams, cont’d• The previous example illustrates another The previous example illustrates another

property of probability tree diagrams.property of probability tree diagrams.• If an experiment consists of a sequence of If an experiment consists of a sequence of

actions represented by branches of a tree actions represented by branches of a tree diagram, the probability of the sequence of diagram, the probability of the sequence of actions is the product of all the probabilities actions is the product of all the probabilities on those branches.on those branches.• This is called the This is called the multiplicative propertymultiplicative property of of

probability tree diagrams. probability tree diagrams.

Example 9Example 9

• A jar contains 3 red balls and 2 green balls.A jar contains 3 red balls and 2 green balls.

Example 9, cont’dExample 9, cont’d• First a coin is tossed. First a coin is tossed.

• If the coin lands heads, a red ball is added to If the coin lands heads, a red ball is added to the jar. the jar.

• If the coin lands tails, a green ball is added to If the coin lands tails, a green ball is added to the jar.the jar.

• Second a ball is selected from the jar.Second a ball is selected from the jar.• What is the probability a red ball is chosen?What is the probability a red ball is chosen?

Example 9, cont’dExample 9, cont’d• Solution: A probability tree diagram is Solution: A probability tree diagram is

created.created.

Example 9, cont’dExample 9, cont’d• Solution, cont’d: The Solution, cont’d: The

probability of probability of choosing a red ball choosing a red ball is found by adding is found by adding the probabilities at the probabilities at the end of the the end of the branches labeled branches labeled RR..

• 4 3 7

12 12 12P R

Example 10Example 10• A jar contains 3 marbles, 2 A jar contains 3 marbles, 2

black and 1 red.black and 1 red.• A marble is drawn and not A marble is drawn and not

replaced before a second replaced before a second marble is drawn.marble is drawn.

• What is the probability that What is the probability that both marbles were black?both marbles were black?

Example 10, cont’dExample 10, cont’d• Solution: Create a probability tree diagram Solution: Create a probability tree diagram

to represent the experiment.to represent the experiment.


• Solution, cont’d: The probability of choosing Solution, cont’d: The probability of choosing 2 black marbles is: 2 black marbles is: 2 1 1

3 2 3P BB

Question:Question:

What is the probability of drawing a What is the probability of drawing a black marble second?black marble second?

a. P(E) = ½a. P(E) = ½b. P(E) = 2/3b. P(E) = 2/3c. P(E) = 1/6c. P(E) = 1/6d. P(E) = 1/3 d. P(E) = 1/3

Example 11Example 11• A jar contains 2 red gumballs and 2 A jar contains 2 red gumballs and 2

green gumballs.green gumballs.• An experiment consists of drawing An experiment consists of drawing

gumballs one at a time from a jar gumballs one at a time from a jar until a red one is chosen.until a red one is chosen.

• Find the probability of: Find the probability of: • A: only 1 draw is neededA: only 1 draw is needed• B: exactly 2 draws are neededB: exactly 2 draws are needed• C: exactly 3 draws are neededC: exactly 3 draws are needed

Example 11, cont’dExample 11, cont’d• Solution: Create a Solution: Create a

probability tree diagram.probability tree diagram.• The probabilities are: The probabilities are:

•

•

• 2 1 115 4 10

P C

2 3 35 4 10

P B

35

P A

Example 12Example 12• Both spinners are spun.Both spinners are spun.• Find the probability both spinners stop Find the probability both spinners stop

on the same color.on the same color.

Example 12, cont’dExample 12, cont’d• Solution: Create a probability tree diagram.Solution: Create a probability tree diagram.

• The desired event is {The desired event is {WW, RR, GGWW, RR, GG}.}.

Example 12, cont’dExample 12, cont’d• Solution, cont’d: The 3 outcomes are Solution, cont’d: The 3 outcomes are

mutually exclusive.mutually exclusive.•

•

• •

1 1 13 2 6

P WW

1 1 14 3 12

P GG

1 1 14 6 24

P RR

1 1 1 76 24 12 24

P WW RR GG

10.2 Initial Problem Solution10.2 Initial Problem Solution

• A friend who likes to gamble wagers A friend who likes to gamble wagers that if you toss a coin repeatedly you that if you toss a coin repeatedly you will get 2 tails before you get 3 heads. will get 2 tails before you get 3 heads. Should you take the bet?Should you take the bet?

• You can figure out what you should do You can figure out what you should do by creating a probability tree diagram.by creating a probability tree diagram.


• Add the probabilities of the outcomes Add the probabilities of the outcomes that result in a win for youthat result in a win for you..•

• You are much more likely to lose than to You are much more likely to lose than to win, so you probably should not take the win, so you probably should not take the bet. bet.

1 1 1 1 516 16 16 8 16

P Win

Section 10.3Section 10.3Conditional Probability, Expected Conditional Probability, Expected

Value, and OddsValue, and Odds

• GoalsGoals• Study conditional probabilityStudy conditional probability

• Study independent eventsStudy independent events

• Study odds Study odds

10.3 Initial Problem10.3 Initial Problem• If you bet $100 on If you bet $100 on

one number on one number on the roulette wheel, the roulette wheel, what is your what is your expected gain or expected gain or loss? loss? • The solution will be given The solution will be given

at the end of the section.at the end of the section.

Conditional ProbabilityConditional Probability

• Sometimes an imposed condition forces Sometimes an imposed condition forces us to focus on a portion of the sample us to focus on a portion of the sample space called the space called the conditional sample conditional sample spacespace..

Conditional Probability, cont’dConditional Probability, cont’d

• For example, in the experiment of For example, in the experiment of tossing 3 fair coins suppose you know tossing 3 fair coins suppose you know the first coin came up heads.the first coin came up heads.• The sample space is {HHH, HHT, HTH, The sample space is {HHH, HHT, HTH,

THH, HTT, THT, TTH, TTT}.THH, HTT, THT, TTH, TTT}.• The conditional sample space is {HHH, The conditional sample space is {HHH,

HHT, HTH, HTT}. HHT, HTH, HTT}.


• The probability of The probability of AA given given BB is called a is called a conditional probabilityconditional probability..

• The probability of The probability of AA given given BB means the means the probability of event probability of event AA occurring within occurring within the conditional sample space of event the conditional sample space of event BB..• The probability of The probability of AA given given BB is written is written

P(P(AA | | BB),),


• Suppose Suppose AA and and BB are events in a are events in a sample space sample space S S and that the probability and that the probability of of BB is not zero. is not zero.

• The formula for conditional probability is The formula for conditional probability is

P A BP A B

P B

Example 1Example 1

• One jar of marbles contains 2 white marbles and 1 One jar of marbles contains 2 white marbles and 1 black marble.black marble.

• Another jar of marbles contains 1 white marble and Another jar of marbles contains 1 white marble and 2 black marbles.2 black marbles.

Example 1, cont’dExample 1, cont’d• A coin is tossed.A coin is tossed.

• Heads: a marble is selected from the first Heads: a marble is selected from the first jar.jar.

• Tails: a marble is selected from the second Tails: a marble is selected from the second jar.jar.

• Find the probability that the coin landed Find the probability that the coin landed heads up, given that a black marble was heads up, given that a black marble was drawn.drawn.


• Solution: Create a Solution: Create a probability tree probability tree diagram.diagram.• The sample space is The sample space is

{HW, HB, TW, TB}.{HW, HB, TW, TB}.• The probabilities are The probabilities are

labeled on the labeled on the diagram.diagram.


• Solution, cont’d:Solution, cont’d:

•

•

•

1 2 36 6 6

P B

16

P H B

1 163 36

P H BP H B

P B

Question:Question:

For the experiment in the previous For the experiment in the previous example, find the probability the example, find the probability the coin landed tails given that a white coin landed tails given that a white marble was drawn.marble was drawn.

a. P(T|W) = ½a. P(T|W) = ½b. P(T|W) = 1/6b. P(T|W) = 1/6c. P(T|W) = 2/3c. P(T|W) = 2/3d. P(T|W) = 1/3d. P(T|W) = 1/3

Example 2Example 2• Suppose a test for a viral infection is not Suppose a test for a viral infection is not

100% accurate.100% accurate.• Of the population, ¼ is infected and ¾ is not.Of the population, ¼ is infected and ¾ is not.• Of those infected, 90% test positive.Of those infected, 90% test positive.• Of those not infected, 80% test negative.Of those not infected, 80% test negative.

a)a) What is the probability the test is correct?What is the probability the test is correct?b)b) Given that a person’s test is positive, what Given that a person’s test is positive, what

is the probability the person is infected? is the probability the person is infected?

Example 2, cont’dExample 2, cont’d• Solution: Create a Solution: Create a

probability tree probability tree diagram.diagram.

• Of the population, ¼ Of the population, ¼ is infected and ¾ is is infected and ¾ is not.not.

• Of those infected, Of those infected, 90% test positive.90% test positive.

• Of those not infected, Of those not infected, 80% test negative.80% test negative.

Example 2, cont’dExample 2, cont’d• Solution, cont’d:Solution, cont’d:a)a) The probability of a The probability of a

correct test is the correct test is the probability of a probability of a positive test for an positive test for an infected person or a infected person or a negative test for an negative test for an uninfected person.uninfected person.

• 1 9 3 4 33 82.5%4 10 4 5 40

Example 2, cont’dExample 2, cont’d• Solution, cont’d:Solution, cont’d:b)b) Find the Find the

conditional conditional probability:probability:

• P(positive) P(positive)

= = 9 3 1540 20 40

Example 2, cont’dExample 2, cont’db)b) Solution, cont’d:Solution, cont’d:

• P(infected and P(infected and positive) = positive) =

• P(infected|positive) P(infected|positive)

= =

940

9 940 60%15 1540

Example 3Example 3

• Results from an inspection of a candy company’s 2 Results from an inspection of a candy company’s 2 production lines are shown in the table. production lines are shown in the table.

• If a customer find a sub-standard piece of candy, If a customer find a sub-standard piece of candy, what is the probability it came from the Bay City what is the probability it came from the Bay City factory?factory?


• Solution: Use the numbers in the table Solution: Use the numbers in the table to solve:to solve:• P(Bay City and sub-standard) = P(Bay City and sub-standard) =

4 4212 4 137 7 360


• Solution, cont’d:Solution, cont’d:• P(sub-standard) = P(sub-standard) = 4 7 11

360 360


• Solution, cont’d:Solution, cont’d:• P(Bay City|sub-standard) = P(Bay City|sub-standard) =

4 436011 11360

Independent EventsIndependent Events

• Two events are called Two events are called independentindependent if if one event does not influence the other.one event does not influence the other.

• When 2 events are independent, their When 2 events are independent, their probabilities follow the rule given below:probabilities follow the rule given below:

• P A B P A P B

Example 4Example 4

• Draw 2 marbles without replacement Draw 2 marbles without replacement from a jar containing 8 white marbles from a jar containing 8 white marbles and 2 red marbles.and 2 red marbles.

• Find the probability that:Find the probability that:a)a)The first marble is red.The first marble is red.

b)b)The second marble is red.The second marble is red.

c)c) Both marbles are red.Both marbles are red.

Example 4, cont’dExample 4, cont’d• Solution: Create a probability tree Solution: Create a probability tree

diagram.diagram.


• Solution, cont’d:Solution, cont’d:a)a)P(red first) = P(RW) + P(RR) P(red first) = P(RW) + P(RR)

16 2 18 190 90 90 5


b)b) P(red second) = P(red second) = P(RR) + P(WR) P(RR) + P(WR)

c)c) P(both red) = P(both red) = P(RR) P(RR)

16 2 18 190 90 90 5

2 190 45


• Solution, cont’d: Notice that the two Solution, cont’d: Notice that the two events are not independent becauseevents are not independent because1 1 15 5 45

Question:Question:

Draw 2 marbles with replacement Draw 2 marbles with replacement from a jar containing 8 while marbles from a jar containing 8 while marbles and 2 red marbles. Are the events and 2 red marbles. Are the events “The first marble is red” and “The “The first marble is red” and “The second marble is red” independent? second marble is red” independent?

a. yesa. yes b. nob. no

Example 5Example 5• A student’s name is chosen at random from the A student’s name is chosen at random from the

college enrollment list and the student is college enrollment list and the student is interviewed.interviewed.

• Let A be the event the student regularly eats breakfast.Let A be the event the student regularly eats breakfast.• Let B be the event the student has a 10:00 AM class.Let B be the event the student has a 10:00 AM class.

• Explain in words what is meant by:Explain in words what is meant by:a)a)

b)b)

c)c)

P A B

P A

P A B

Example 5, cont’dExample 5, cont’d• Solution: Solution:

a)a) : the probability the student : the probability the student regularly eats breakfast and has a 10:00 regularly eats breakfast and has a 10:00 AM class.AM class.

b)b) : the probability the student : the probability the student regularly eats breakfast given that the regularly eats breakfast given that the student has a 10:00 AM classstudent has a 10:00 AM class

c)c) :the probability the student :the probability the student does not regularly eat breakfast.does not regularly eat breakfast.

P A B

P A

P A B

Expected ValueExpected Value• The average numerical outcome for many The average numerical outcome for many

repetitions of an experiment is called the repetitions of an experiment is called the expected valueexpected value..

• If the outcomes of an experiment are If the outcomes of an experiment are vv11, , vv22, …, v, …, vnn and the outcomes have and the outcomes have probabilities of probabilities of pp11, p, p22, …, p, …, p33, respectively, , respectively, the expected value is the expected value is

E = vE = v11(p(p11) + v) + v22(p(p22) + … + v) + … + vnn(p(pnn).).• If If EE = 0, the game is said to be fair = 0, the game is said to be fair..

Example 6Example 6

• An experiment consists of rolling a fair An experiment consists of rolling a fair die and noting the number on top of die and noting the number on top of the die.the die.

• Compute the expected value of one Compute the expected value of one roll of the die. roll of the die.


• Solution: The calculations are shown Solution: The calculations are shown below:below:

Example 7Example 7

• How much should an insurance company How much should an insurance company charge as its average premium in order to charge as its average premium in order to break even? break even?


• Solution: The calculations are shown Solution: The calculations are shown above. above.

• The average premium should cost $760.The average premium should cost $760.

OddsOdds

• The The odds in favorodds in favor of an event compare of an event compare the number of favorable outcomes to the number of favorable outcomes to the number of unfavorable outcomes.the number of unfavorable outcomes.

• If the odds in favor are If the odds in favor are aa::bb, then , then

aP Ea b

Odds, cont’dOdds, cont’d• The The odds againstodds against of an event compare the of an event compare the

number of unfavorable outcomes to the number of unfavorable outcomes to the number of favorable outcomes.number of favorable outcomes.• The odds in favor of an event E are The odds in favor of an event E are

• The odds against an event E are The odds against an event E are

:P E P E

:P E P E

Example 8Example 8

• Suppose a card is randomly drawn from Suppose a card is randomly drawn from a standard deck.a standard deck.

• What are the odds in favor of drawing a What are the odds in favor of drawing a face card? face card?


• Solution: There are 12 face cards in the Solution: There are 12 face cards in the deck, and there are 40 other cards.deck, and there are 40 other cards.

• The odds in favor are 12:40, which The odds in favor are 12:40, which simplifies to 3:10. simplifies to 3:10.

Question: Question:

A card is randomly selected from a A card is randomly selected from a standard deck. What are the odds standard deck. What are the odds against drawing a diamond? against drawing a diamond?

a. 4:3a. 4:3b. 3:1b. 3:1c. 3:4c. 3:4d. 1:3d. 1:3

Example 9Example 9

• Find P(Find P(EE) given the following odds:) given the following odds:a)a) The odds in favor of The odds in favor of EE are 3:7. are 3:7.

b)b) The odds against The odds against EE are 5:13. are 5:13.


• Solution:Solution:a)a) The odds in favor of The odds in favor of EE are 3:7. are 3:7.

b)b) The odds against The odds against EE are 5:13. are 5:13.

3 33 7 10

P E

13 135 13 18

P E


• Find the odds in favor of event Find the odds in favor of event EE, , given the following probabilities:given the following probabilities:

a)a) P(P(EE) = 1/4. ) = 1/4.

b)b) P(P(EE) = 3/5. ) = 3/5.


• Solution:Solution:a)a) The odds in favor The odds in favor

of of EE are 1:3. are 1:3.

b)b) The odds in favor The odds in favor of of EE are 3:2. are 3:2.

1 1 14 41 3 31 4 4

3 3 35 53 2 21 55

10.3 Initial Problem Solution10.3 Initial Problem Solution• A roulette wheel has 38 slots numbered A roulette wheel has 38 slots numbered

00, 0 and 1 through 36. You place a bet 00, 0 and 1 through 36. You place a bet on a number or combination of numbers. on a number or combination of numbers. If you bet on the winning number, you win If you bet on the winning number, you win your bet plus 35 times your bet. If you your bet plus 35 times your bet. If you lose, you lose the money you bet. lose, you lose the money you bet.

• If you bet $100 on one number, what is If you bet $100 on one number, what is your expected gain or loss?your expected gain or loss?


• Solution:Solution:• The probability of winning is 1/38 The probability of winning is 1/38

and the amount won would be and the amount won would be $3500.$3500.

• The probability of losing is 37/38 The probability of losing is 37/38 and the amount lost would be $100. and the amount lost would be $100.


• The expected value is approximately The expected value is approximately -$5.26. You should expect to lose this -$5.26. You should expect to lose this amount, on average, for every $100 amount, on average, for every $100 you bet.you bet.

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