nossi ch 10

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

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

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

• List one possible event.List one possible event.

Example 1Example 1

• The sample space contains 6 possible The sample space contains 6 possible outcomes: {1, 2, 3, 4, 5, 6}.outcomes: {1, 2, 3, 4, 5, 6}.

• One possible event: The event of rolling One possible event: The event of rolling an even number: {2,4,6}an even number: {2,4,6}

Example 2Example 2

• An experiment consists of tossing a An 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.

Example 2, cont’dExample 2, cont’d

• 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

• An experiment An 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 3Example 3

• 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 The experiment

consists of rolling 2 consists of rolling 2 standard dice and standard dice and recording the number recording the number appearing on each appearing on each die.die.• List the sample space.List the sample space.

• List one possible List one possible event.event.

Example 4Example 4

• 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)}.

Example 4Example 4

• 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.

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.

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 Find the first coin Find the experimental experimental probability of probability of EE. .

Example 5Example 5 The even of getting a head The even of getting a head

on the first coin: on the first coin:

{HH, HT}.{HH, HT}.

occurred a total of occurred a total of

137 + 115 = 252 times 137 + 115 = 252 times out of 500.out of 500.• The experimental The experimental

probability of probability of EE is is

2520.504

500=

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.

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.

Example 6, cont’dExample 6, cont’d

• 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.

Example 6Example 6

The event The event EE is {HH, HT} and the theoretical probability of is {HH, HT} and the theoretical probability of EE is the is the number of outcomes in number of outcomes in EE divided by the number of outcomes divided by the number of outcomes

in the sample space.in the sample space.

( ) 2 10.5

4 2P E = = =

Example 6Example 6

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}.

( ) 30.75

4P 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.

Example 7Example 7• Solution: There are 36 outcomes in the Solution: There are 36 outcomes in the

sample space. (see pg 635 for a list of the sample space. (see pg 635 for a list of the sample space.)sample space.)

a)a) Event Event AA: getting 7 dots contains 6 outcomes:: getting 7 dots contains 6 outcomes:

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

Each outcome is equally likely to occur.Each outcome is equally likely to occur.

( ) 6 1

36 6P A = =

Example 7Example 7

• Solution, cont’d:Solution, cont’d:

b)b) Event Event BB: getting 8 dots contains 5 : getting 8 dots contains 5 equally likely outcomes: equally likely outcomes:

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

• ( ) 5

36P B =

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

a)a) Event Event CC: getting at least 4 dots (or a total of 4 : getting at least 4 dots (or a total of 4 or more) contains 33 equally likely outcomes: or more) contains 33 equally likely outcomes:

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

•

( ) 33 11

36 12P C = =

Example 8Example 8

• A jar contains four A jar contains four marbles:marbles:

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

Example 8Example 8

• 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.

Example 8Example 8• If we draw 2 marbles in a row, If we draw 2 marbles in a row,

without replacing the first one, without replacing the first one, find the probability of:find the probability of:

• Event Event AA: One of the marbles is : One of the marbles is red.red.

AA = {RG, RY, RW, GR, YR, WR}. = {RG, RY, RW, GR, YR, WR}.

( ) 6 1

12 2P A = =

Example 8Example 8• 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 of:replacing the first one, find the probability of:

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

BB = {RG, RY, RW, YR, YG, YW}. = {RG, RY, RW, YR, YG, YW}.

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

( ) 00

12P C = =

( ) 6 1

12 2P B = =

Example 8Example 8If we draw 2 marbles in a row, without If we draw 2 marbles in a row, without

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

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

DD = {RG, RY, RW, GR, GY, GW, = {RG, RY, RW, GR, GY, GW, YR, YG, YW}.YR, YG, YW}.

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

( ) 121

12P E = =

( ) 9 3

12 4P D = =

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 9Example 9• Solution: The sample space contains 52 equally Solution: The sample space contains 52 equally

likely outcomes. (this chart is on pg 639)likely outcomes. (this chart is on pg 639)

Example 9Example 9

• Event Event AA has 12 outcomes, one for each of the 3 face cards has 12 outcomes, one for each of the 3 face cards in each of the 4 suits.in each of the 4 suits.

• P(P(AA) = 12/52.) = 12/52.• Event B has 2 outcomes, because there are 2 black fives.Event B has 2 outcomes, because there are 2 black fives.

• P(P(BB) = 2/52.) = 2/52.• Events Events AA and and BB are mutually exclusive because it is are 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 face card or a This is the probability of drawing either a face card or a black 5.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 , but not

in an event in an event EE, is called the , is called the complement of the complement of the event event EE. . • The complement of The complement of EE is written is written ĒĒ. .

• For example: The weather man says there is a 45% For example: The weather man says there is a 45% chance of rain. The complement of this event is the chance of rain. The complement of this event is the chance that it will not rain which is 55%.chance that it will not rain which is 55%.

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 10Example 10

• 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 (complement of E) (complement of E) is the event the is the event the numbers are unequal.numbers are unequal.

• P(P(ĒĒ) = 1 – ¼ = ¾ ) = 1 – ¼ = ¾

Example 12Example 12

• 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 12Example 12• 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 12Example 12• 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).).

a)a) AA = {2, 4, 6, 8}, so P( = {2, 4, 6, 8}, so P(AA) = 4/8 = 1/2) = 4/8 = 1/2

b)b) B B = {6, 7, 8}, so P(= {6, 7, 8}, so P(BB) = 3/8) = 3/8

c)c) CC = {1, 2}, so P( = {1, 2}, so P(CC) = 2/8 = 1/4) = 2/8 = 1/4

d)d) DD = {1, 3, 4, 5, 6, 7, 8}, so P( = {1, 3, 4, 5, 6, 7, 8}, so P(DD) = 7/8.) = 7/8.

Example12Example12• 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 5

8 8 8 8P A B∪ = + − =

Example 12Example 12• 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).).

c)c) BB and and C C are mutually exclusive, soare mutually exclusive, so

andand

c)c)

( ) ( ) ( )P B C P B P C∪ = +

( ) 3 2 5

8 8 8P B C∪ = + =

( ) 0P B C∩ =

Section 10.2Section 10.2

Multistage 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.

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.

Example 2Example 2• There are There are

12 possible 12 possible outcomes.outcomes.

• 12 is the 12 is the product of 4 product of 4 branches branches times 3 times 3 branches.branches.

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 r times sr times s..

• 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 3Example 3

• 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.

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 4Example 4• 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.

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 DiagramsProbability Tree Diagrams

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.

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 7Example 7

• Solution: There are 4 outcomes Solution: There are 4 outcomes in the sample space.in the sample space.

• They are not all equally They are not all equally likely.likely.

• They are all mutually They are all mutually exclusive. (no overlapping)exclusive. (no overlapping)

• Use the central angles to Use the central angles to find the probability of each find the probability of each outcome and draw the outcome and draw the probability tree diagram. probability tree diagram.

Example 7Example 7• The probability of The probability of

white or green is:white or green is:

( )( ) ( )

1 1 7

3 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 8Example 8• Solution: Draw a probability tree diagram to Solution: Draw a probability tree diagram to

represent the experiment.represent the experiment.

Example 8Example 8• Assign a probability to the end of each secondary branch. Assign a probability to the end of each secondary branch.

(multiply fractions as you move down branches, add (multiply fractions as you move down branches, add factions as you move across the end of branches.)factions as you move across the end of branches.)

Example 8Example 8

• Either tree diagram can be used to find that Either tree diagram can be used to find that P(P(BBBB) = 4/9.) = 4/9.

Example 9Example 9• A jar contains 3 red balls and 2 A jar contains 3 red balls and 2

green balls.green balls.• First a coin is tossed. First a coin is tossed.

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

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

• Second a ball is selected from the Second a ball is selected from the jar.jar.

• What is the probability a red ball is What is the probability a red ball is chosen?chosen?

Example 9Example 9• Solution: A probability tree diagram is Solution: A probability tree diagram is

created.created.

Example 9Example 9

• The probability of The 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 10Example 10• Solution: Create a probability tree diagram Solution: Create a probability tree diagram

to represent the experiment.to represent the experiment.

Example 10Example 10

• The probability of choosing 2 black marbles The probability of choosing 2 black marbles is: is: ( ) 2 1 1

3 2 3P BB = ⋅ =

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 11Example 11• Solution: Create a Solution: Create a

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

•

•

• ( ) 2 1 11

5 4 10P C = ⋅ ⋅ =

( ) 2 3 3

5 4 10P B = ⋅ =

( ) 3

5P A =

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.

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

Initial Problem SolutionInitial Problem Solution

• 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 5

16 16 16 8 16P Win = + + + =

Section 10.3Section 10.3

Conditional 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

• In the experiment of tossing 3 fair coins In the experiment of tossing 3 fair coins suppose you know the first coin came suppose you know the first coin came up heads.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}.

• Then the conditional sample space is Then the conditional sample space is {HHH, HHT, HTH, HTT} because the first {HHH, HHT, HTH, HTT} because the first coin was Head.coin was Head.

Conditional ProbabilityConditional Probability

• 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),),

Conditional ProbabilityConditional Probability

• 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 1Example 1• 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.

Example 1Example 1• A coin is tossed.A coin is tossed.

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

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

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

• Solution: Create a probability Solution: Create a probability tree diagram.tree diagram.

• The sample space is {HW, The sample space is {HW, HB, TW, TB}.HB, TW, TB}.

• The probabilities are labeled The probabilities are labeled on the diagram.on the diagram.

Example 1Example 1

• Solution, cont’d:Solution, cont’d:

•

•

•

( ) 1 2 3

6 6 6P B = + =

( ) 1

6P H B∩ =

( ) ( )( )

1 163 3

6

P H BP H B

P B

∩= = =

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 2Example 2• 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 2Example 2

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 2Example 2b)b) Given that a person’s test Given that a person’s test

is positive, what is the is positive, what is the probability the person is probability the person is infected? Find the infected? Find the conditional probability:conditional probability:

• P(positive) = P(positive) =

9 3 15

40 20 40+ =

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

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

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

= =

9

40

9 940 60%15 15

40

= =

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?

Example 3Example 3

• 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 4

212 4 137 7 360=

+ + +

Example 3Example 3

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

4 7 11

360 360

+=

Example 3Example 3

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

4 436011 11

360

=

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 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, Example 5, • 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 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.

Example 6Example 6

• Solution: The calculations are shown below: Solution: The calculations are shown below: Multiply 1(1/6) and 2(1/6) and 3(1/6) etc. Multiply 1(1/6) and 2(1/6) and 3(1/6) etc.

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?

Example 7Example 7

• Solution: The calculations are shown above. The Solution: The calculations are shown above. The product row is the product of the 2 numbers product row is the product of the 2 numbers directly above each value in the product row.directly above each value in the product row.

• 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 E

a b=

+

OddsOdds• The The odds againstodds against an event compare the 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?

Example 8Example 8

• 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.

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.

Example 9, cont’dExample 9, cont’d

• 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 3

3 7 10P E = =

+

( ) 13 13

5 13 18P E = =

+

Example 10Example 10

• 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.

Example 10, cont’dExample 10, cont’d

• 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?

Initial Problem SolutionInitial Problem Solution

• 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.

Initial Problem SolutionInitial Problem Solution

• 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.

Chapter 10 Assignments

• Section 10.1 pg 645 (7,11,15,17,31,35)

• Section 10.2 pg 664 (1,3,11,13,19,25)

• Section 10.3 pg 686 (5a,6a,15,27,3339,41)