unit six combinatorics and probability math 621b 12 hours · unit six combinatorics and probability...

132

UNIT SIX

COMBINATORICS AND PROBABILITY

MATH 621B

12 HOURS

Revised April 9, 02

133

Previous Knowledge

Grade 7

- identify situations for which the probability would be near 0 14

12

34

1, , , , and

- solve probability problems, using simulations and by conducting experiments

- identify all possible outcomes of two independent events, using tree diagrams and area models

- create and solve problems, using the numerical definition of probability

- compare experimental results with theoretical results

- use fractions, decimals and percents as numerical expressions to describe probability

Grade 8

- conduct experiments and simulations to find probabilities of single and complementary events

- determine theoretical probabilities of single and complementary events

- compare experimental and theoretical probabilities

- demonstrate an understanding of how data is used to establish broad probability patterns

Grade 9

- make predictions of probabilities involving dependent and independent events by designing and conducting experiments and simulations

- determine theoretical probabilities of independent and dependent events

- demonstrate an understanding of how experimental and theoretical probabilities are related

- recognize and explain why decisions based on probabilities may be combinations of theoretical calculations, experimental results and subjective judgements

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Combinatorics Introductory Lesson

Combinatorics (from NCTM Standards) is the mathematics of systematic counting.

G20 demonstrate an understanding of the Fundamental Counting Principle

1. Count the number of ways that customers at Kent’s Deli can construct a sandwich. They have a choiceof 3 types of meat - ham, chicken or beef. They also have a choice of 2 types of bread - white or rye.a) construct a tree diagram by choosing bread first then meat second.b) construct a tree diagram by choosing meat first then bread second.c) how do the answers from parts (a) and (b) compare?

2. At Bonnie’s Deli customers have a wider selection. They can choose from 4 types of meat - ham,chicken, beef or salami and 3 choices of bread - white, rye or brown. Use a tree diagram to illustrate allpossible types of sandwiches.

3. From the above problems make a conjecture as to the patten developing. Write a conjecture thatexplains how to solve counting problems where there are different levels of choices to be made.

This conjecture is called the Fundamental Counting Principle.

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SCO: By the end of grade12, students will beexpected to:

G20 demonstrate an understanding of the FCP

Elaborations - Instructional Strategies/ SuggestionsFundamental Counting Principle (7.1)

Invite student groups to do the Explore and Inquire on p.334.

Student groups should read and discuss p.335-6. Students shouldunderstand how tree diagrams can be streamlined into the FCP greatlysimplifies the solving of problems.

Ex. How many license plates can be generated containing 4 letters of thealphabet?a) where repetitions are not allowed.b) where repetitions are allowed.

a) 26 × 25 × 24 × 23 = 358,800in the first blank any of the 26 letters can be placedin the second blank any of the remaining 25 letters can be placed in the third blank any of the remaining 24 letters can be placedin the fourth blank any of the remaining 23 letters can be placed

Note to Teachers: This is a permutation problem 26 P4 .On the TI-83 press 26 math < < < PRB 2: nPr 4.

b) 26 × 26 × 26 × 26 = 456,976in the first blank any of the 26 letters can be placedin the second blank any of the 26 letters can be placedin the third blank any of the 26 letters can be placedin the fourth blank any of the 26 letters can be placed

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Worthwhile Tasks for Instruction and/or Assessment Suggested Resources

Fundamental Counting Principle (7.1)

Group DiscussionA series of recipes is written on index cards. The recipes areeither for appetizer, entree or dessert. There are 5 appetizercards, 7 for entrees and 9 for desserts. In how many ways cana three course meal be selected from the menu?

JournalWhat is the major advantage of using the FCP instead of treediagrams to solve counting problems?

ActivityA store carries 8 different sizes of shirts, each size is availablein men’s and women’s designs. Each is made with eithercotton, silk or micro-fiber. Each shirt is either a solid colour ora print design. How many different kinds of shirts does thestore carry?

Group ActivityThe block diagram shown represents nine city blocks, with thelines indicating streets. A fire truck at point A wishes to travelto point B by travelling only North or East. Use a tree diagramto find the number of different routes this truck could take. (See tree diagram at the end of the unit)

Fundamental Counting Principle(7.1)

Math Power 12 p.336 #1-7 odd, 8-14 even

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A19 describe what a permutation is, how to evaluate a permutation and how to use permutations to solve problems

Elaborations - Instructional Strategies/ SuggestionsPermutations (7.2)Definition: The process or result of making or becoming different.Synonyms: change, variation, modification, mutation.Factorial notation should be introduced to students and its calculation onscientific calculators (and the TI-83) demonstrated.5 math < < < PRB 4: ! enter then press enter again to evaluate 5 !

Invite students to read p.338-341 and do the Explore and Inquire.A key concept is that in permutations “order is important”. An examplewill illustrate what this means:Five people are to get their picture taken. If the people are moved todifferent positions then these are considered to be mutations orpermutations. There are considered to be 5! or 120 different ways( 5P5)(permutations) of arranging these 5 people for the photo. In the next section on combinations, “order is not important” and nomatter how the people are re-positioned the same 5 people are in thepicture and this situation is said to have only 1 combination.There are two mutations of permutations:1) no repetitions allowed:a) n distinct objects taken n at a time (essentially this is a FCP problem)

n! or nPn

b) n distinct objects taken r at a time.

n rP nn r

=−

!( ) !

Note to Teachers: The familiar permutation formula

only works for no repetitions.n rP nn r

=−

!( ) !

2) repetitions allowed n distinct objects taken n at a time, where a objects are alike, b objectsare alike...

n

a b!

! !. . .

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Permutations (7.2)ActivityExpress 7 × 6 × 5 × 4 × 3 × 2 × 1 in factorial notation.

Pencil/Paper/TechnologyDetermine the value of:a) 4!b) 9P6c) 8P3

ActivityA baseball team has 9 players on the field. In how many wayscan the 9 players on the field be re-positioned?

Group DiscussionThe block diagram shown represents nine city blocks, with thelines indicating streets. A fire truck at point A wishes to travelto point B by travelling only North or East. Use permutationtheory to find the number of different routes this person couldtake.

Note to Teachers: This is the same problem as in section 7.1on the FCP. It is a case 2 permutation problem (repetitionsare allowed). A person must travel 3 blocks East and 3 blocksNorth ( for example EENENN). So one must make a total of 6moves where there are 3 East moves (3 repetitions of East)and 3 North moves (3 repetitions of North).

Thus there are 6

3 320!

! != ways to go from A to B

Permutations (7.2)

Do Permutation Worksheet at end ofunit.

Math Power 12 p.342 #1-19 odd

Applications p.342 # 21,30

Note to Teachers: Why is 0! = 1?

The FCP predicts that if 6 objects arechosen from a list of 6 objects thenthey can be chosen in 6 × 5 × 4 × 3× 2 × 1 = 6! Ways.

This is also a permutations problemand can be solved using:

6 6

66 6

60

P nn r

=−

=−

=

!( ) !

!( ) !

!!

Which means that the denominator 0!must be equivalent to 1.

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B24 analyze and count the number of permutations and combinations possible in various situations

G14 describe what a combination is and how to evaluate a combination and use them to solve problems

G15 distinguish between combinations and permutations

Elaborations - Instructional Strategies/ SuggestionsCombinations (7.3)

Challenge student groups to do the Explore and Inquire on p.345.

Invite student groups to read and discuss p.345-348.

In the Explore on p.345 the tree diagram looks like this:

From 5 numbers we pick 2 numbers with no repetitions and order is not important.

Therefore we can do 5C2 = 10

A combination is a selection of objects whereorder is not important.

n rn rC Pr

nn r r

= =−!

!( ) ! !

For a real world example illustrating the difference betweenpermutations and combinations see the explanation sheet at the end ofthe unit.

Ex. A five person committee is to be selected from a class with 10women and 8 men. In how many ways can a committee be formed if itmust contain 3 women?

Solution: 10C3 × 8C2 = 3360 ways to form the committee

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Combinations (7.3)Group DiscussionThere are 2000 CD players in a company warehouse. Theshipper is going to select 40 of these using their serialnumbers to check for defects. In how many ways can theshipper choose the 40 players?

Group ActivityA salesperson handles 25 brands of golf clubs. At a tradeshow, he has, however, space to display only 18 of the brands. In how many ways can the display be set up?

Group ActivityIn a class of 35 students, in how many ways can 8 people bechosen to work on the decorations committee for thegraduation dance? If 13 people say that they don’t want to beon that committee, how many ways can the 8 people bechosen?

JournalMary says “A permutation is the same as a combination.”Write to explain that her statement is incorrect. Group DiscussionThe block diagram shown below represents nine city blocks,with the lines indicating streets. A fire truck at point A wishesto travel to point B by going only North or East. How manydifferent ways can the truck go from A to B?

Solution: The answer is 6C3 = 20. There are 6 blanks to fill in . Put an E in any 3 of these blanks. Nowthe question is: In how many ways can one select the 3 blanksfor an N?

Combinations (7.3)

Math Power 12 p.348 #1,3,5,20,19

Applications p.348 #15-18,21,22,23

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G17 connect Pascal’s Triangle with combinational coefficients

Elaborations - Instructional Strategies/ SuggestionsPascal’s Triangle (7.4)

Invite student groups to read and discuss p.350-352 and do the Exploreand Inquire.See the connections between Combinations and Pascal’s Triangle sheetat the end of the unit.

The block problems we have been looking at using FCP, Permutationsand Combinations can be done as well using Pascal’s Triangle.Consider the nine city block problem that we having been looking at:

Looking at this block diagram superimposed on Pascal’s Triangle weimmediately see the solution.

Encourage student groups torealize that Pascal’s Triangle is generated from combinations as shownon the top of p.351.

The pathway problems like Ex. 1 and 2 on p.351 can be solved usingPascal’s Triangle. The solution is the sum of the numbers in the correctrow of Pascal’s Triangle.

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Pascal’s Triangle (7.4)

Group PresentationDetermine the number of ways to go from A to B bytravelling only East or North if the street pattern is a 4 × 4square pattern using:a) FCPb) permutationsc) combinationsd) Pascal’s Triangle

Activity/DiscussionJulie has invited 8 people to a birthday party at her house. Onthe invitation sent out she did not ask for a RSVP, so she isnot sure of how many will come to the party. How manycombinations of guests could come? How does the answer of(a) relate to Pascal’s Triangle?

JournalExplain how Pascal’s Triangle could have been used to solveJulie’s problem.

ResearchWrite a short essay on the life and contributions of BlaisePascal.

Pascal’s Triangle (7.4)

Math Power 12 p.352 # 1-3,5-8

Pascal’s Triangle Worksheet (7.4)

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G16 demonstrate an understanding of binomial expansion and its connection to combinations

Elaborations - Instructional Strategies/ SuggestionsThe Binomial Theorem (7.5)Student groups should do the Explore and Inquire on p.354. Challengestudent groups to read and discuss p.355-356. Students shouldunderstand that combinations are used to determine the coefficients ofan expanded binomial.

The Binomial Theorem states:

( )a b C a b

where r is the number of the term

nn

r

n

rn r r+ =

=

−∑0

The general term can be obtained using: t C a bn n rn r r

+−=1

Ex. Find the 7th term of (x ! 3)11.

Solution:

To find the 7th term; r = 6, a = x, b = !3, n = 11

n rn r rC a b

C x

x

x

−

− −

×11 6

11 6 6

5

5

3

462 729

336798

( )

Note to Teachers: Most texts stop at very simple and abstract examplesof the Binomial Theorem. At the end of the unit are worksheets on basicprobability and applied Binomial Theorem problems. It assumesstudents have a basic knowledge of probability; which students shouldhave obtained in Junior High and Grade 10 Data Management

For binomial events: ie. either a success or a failure occurs.The formula for binomial application problems is:

let F = the probability of failurelet S = the probability of success

( )F S C F Snn s

n s s+ = −

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The Binomial Theorem (7.5)

Pencil/PaperUse Pascal’s Triangle and the pattern of decreasing powers ofa and increasing powers of b to expand:a) (x ! 3)6

b) (2x + 5y)5

Group ActivityUse the Binomial Theorem to expand:a) (x + 2)6

b) (3x ! 2y)4

Group ActivityUse the general term of the Binomial Theorem to determine:a) the 4th term of (x + 5)8

b) the 9th term of (x ! 2y)13

Group PresentationA true-false test has 12 questions. Suppose all questions areguessed. Determine the probability of guessing 6 correctanswers.

Group ActivityAbout 9% of women will contract breast cancer sometime intheir lifetimes. There are 100 women at a sales meeting. Findthe probability that exactly four of them will contract breastcancer in their lifetimes.

A company manufactures computer chips. About 2% of thechips are defective. A quality control inspector pulls 50 chipsat random off an assembly line. Find the probability thatexactly 10 of them are defective.

The Binomial Theorem (7.5)

Math Power 12 p.356#1-21odd, 25,27,30

Binomial Theorem Worksheet

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G11 apply area and tree diagrams to interpret and determine probabilities

G12 find probabilities for sequences of 2 or more events

G21 apply the FCP to probability

Elaborations - Instructional Strategies/ SuggestionsProbability & Sample Space (8.1)As an introduction to probability, students should do the ProbabilityWorksheet at the end of the unit on probability of single events.Challenge student groups to do the Explore and Inquire on p.368.

Invite student groups to read and discuss p.369-371. Students should beable to express probabilities as common fractions, decimals andpercents.Students should be able to combine the concepts of probability and treediagrams to form a probability tree diagram.

Invite students to discuss the concepts of: < outcomes - things that may happen in an experiment < event - a collection of outcomes (a subset of the sample space) < sample space - the set of all possible outcomes of an experiment < complementary events - two events whose sum is 1.Note to the Teachers: Some of the problems in this section are involvedwith making more than one selection. These are called compoundevents and will be dealt with in more depth in section 8.2.For a simple example, see Ex. 1 , p.369. A second example is below:Ex. 10 p.372

The probability that both trades are with the US is 0.494.

The probability that neither trade is with the US is 0.084.

The probability that the export is to the US but the import is not from the US is 0.266.Note that the sum of all probabilities in the sample space must equal 1.

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Probability & Sample Space (8.1)

Pencil/PaperUse a probability tree to determine the probability of thefollowing events:a) obtaining a tail and a queen of hearts. (1/2 × 1/52 = 1/104)b) rolling a 5 and obtaining a head. (1/6 × ½ = 1/12)c) picking a day of the week and an hour of the day. (1/7 ×1/24 = 1/168)

ResearchDo research to determine the percentage of Islanders that areof Acadian descent. Statistics Canada claims that about 12%of Canadian adults have a university degree. What is theprobability that:a) an Island Acadian has a university degree.b) an Island non-Acadian has a university degree.c) an Island Acadian hasn’t a university degree.d) an Island non-Acadian hasn’t a university degree.e) What is the sum of the probabilities for parts (a) to (d)? Is this the entire sample space?

JournalWrite to explain the differences between an outcome, an eventand a sample space.

Group ActivityIn your class, what is the probability that:a) A person selected is a female and has brown hair?b) A person selected is male and has blue eyes?

Probability & Sample Space (8.1)

Probability Worksheet

Math Power 12 p.371 #1-7,9,10

147


G8 demonstrate an understanding of “and” “or” and “complementary” as they relate to probability

G9 demonstrate an understanding and apply complementary probability


Elaborations - Instructional Strategies/ SuggestionsClassifying Events (8.2)

Note to Teachers: This section has to do with making more than oneselection (compound events). They are classified below. At the end of theunit is a more detailed explanation of the classifications. This topicshould be dealt with carefully and kept to an introductory level.

Invite student groups to read and discuss p.375-379 and do the Exploreand Inquire on p.375.

Classifications for discussion are:

< independent vs dependent events

< mutually exclusive and non-mutually exclusive events

< complementary events

both selections must be made

For independent: P(A & B) = P(A) × P(B)For dependent: P(A & B) = P(A) × P(B *A)

Note: P(B*A) deals with conditional probability and will be dealt withlater in the unit.

one selection or the othermust be made

For mutually exclusive: P(A or B) = P(A) + P(B)

For non-mutually exclusive: P(A or B) = P(A) + P(B) ! P(A & B)

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Classifying Events (8.2)

Group Activity/PresentationClassify the events as either dependent or independent:a) selecting 2 blue socks, one at a time, from a drawer thatcontains 6 black socks and 4 blue socks.b) selecting 2 drinks from a cooler, one at a time, that contains2 bottles of pepsi and 4 bottles of juice.c) select a card from a deck of cards, replace the card, then make a second selection.d) roll two dice and get a sum of 7, then roll a second time and get a sum of 4.

Group Activity/PresentationClassify the events as either mutually exclusive or not mutually exclusive:a) selecting a penny or a dime from 4 pennies, 3 nickels and 6 dimes.b) selecting a boy or a brown-haired person from 12 girls, 5 of whom have brown hair , and 15 boys, 6 of whom have brown hair.c) selecting a king or queen from a deck of cards.d) selecting a card from a deck and it being either red or a face card.

Classifying Events (8.2)

Note to Teachers: At the end of theunit is a more detailed explanation of section 8.2 and accompanyingexercises to help point out thedifferences.

Math Power 12 p.380 # 1-13

149



Elaborations - Instructional Strategies/ SuggestionsProbability & Combinatorics (8.3)Student groups should do the Explore and Inquire on p.382. Invite student groups to read and discuss p.382-385. When treediagrams become too complicated, combinatorics can be used todetermine the size of the sample space and the size of the event set. Theproblems in section 8.2 that were dependent employed conditionalprobability. These problems can be solved using combinatorics as well.

Students should do the Combinatorics Worksheet at the end of the unit.

Ex.1 Find the probability that 2 cards drawn, without replacement, areboth face cards.Solution:The phrase: “without replacement” makes this a dependent situation.Method 1: P(A) = 12/52 = 3/13; P(B*A) = 11/51

P(A & B) = P(A) × P(B*A) = .0498

Method 2: Event set: From the 12 face cards we need to choose, in anyorder, 2 face cards: 12C2.

Sample space: From the total 52 cards we choose 2 cards: 52C2

P face cards CC

( ) .2 049812 2

52 2= =

Ex. 2 Four door prizes are awarded at a party. 26 people hold one ticketeach. a) What is the probability that Sara, Jane and Mary win the first,second and third prizes in that order? b) What is the probability that thesame 3 people win the prizes but not necessarily in that order?

a) P Sara Jane MaryP

( , , ) = =1 1

1560026 3

b) P in any orderP

( ) !3 3 61560026 3

= =

There are 3! = 6 ways of arranging Sara, Jane and Mary and therefore 6times as much of a chance of these 3 people winning if they don’t haveto win in a specific order.

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Probability & Combinatorics (8.3)Pencil/PaperA committee of 5 people is to be selected from your class. Determine the number of women and men in your class andfind the probability that there are exactly 3 women on thecommittee.DiscussionThere are 7 women and 5 men in a room. A committee of 4 isto be randomly selected. What is the probability of therebeing:a) 4 women on the committee.b) 3 women on the committee.c) 2 women on the committee.d) 4 men on the committee.e) 3 men on the committee.f) 2 men on the committee.

Group ActivityThree balls are randomly drawn from a bag containing 6 whiteballs and 4 black balls. What is the probability of drawing:a) 3 white balls.b) 2 white balls and 1 black ball.c) 1 white ball and 2 black ballsd) 3 black balls

Group DiscussionFour people are to be randomly selected from a group of 9boys and 7 girls. What is the probability of each event?a) exactly 3 being girls.b) all 4 being boys.c) 2 being girls.Group ActivityFour balls are randomly drawn from a bag containing 3 whiteballs and 5 black balls. What is the probability that there areexactly 2 white balls?

PresentationFive cards are dealt from a deck of 52 cards. What is theprobability of each event?a) a hand containing one pair of aces and 3 kings.b) a hand containing 2 face cards and 3 10's.c) a hand containing 3 hearts and 2 diamonds.

Probability & Combinatorics (8.3)

Math Power 12 p.385 #1-3, 10-12

Combinatorics Worksheet at end ofunit

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Elaborations - Instructional Strategies/ SuggestionsApplications of Conditional Probability (8.4)

Invite student groups to read and discuss p.388-391.

Conditional Probability is the probability that an event will occur giventhat another event has occurred.

For two events A and B:a) if they are independent, then the joint probability, P(A & B), of themboth happening is: P(A & B) = P(A) × P(B) Note that if the events are independent, the second event in unaffectedby the first event happening. The expression on the left above, P(A & B)could have been written as P(B & A). A particular event does not haveto happen before the other. Hence conditional probability does not comeinto play here.

b) if they are dependent, then the joint probability, P(A & B), of themboth happening is: P(A & B) = P(A) × P(B*A) or P(B & A) = P(B) × P(A*B) The joint probabilities P(A & B) and P(B & A) are equal. Thus from theabove two equations we can get:

P B A P A BP A

P B P A BP A

( ) ( & )( )

( ) ( )( )

= =×

or P A B P A BP B

P A P B AP B

( ) ( & )( )

( ) ( )( )

= =×

This is known as BAYES’ Law or Theorem.

Note to Teachers: This may be rather abstract for students. Conditionalprobability problems can be solved using a joint probability table. Theconditional probability worksheet has solutions done both ways. Students may very well find that using the table is a simpler and moreunderstandable procedure.

152


Applications of Conditional Probability (8.4)

ActivityA household is considered prosperous if its income exceeds$75,000. A household is considered educated if thehouseholder has completed college. A household is selectedat random and event A is: the selected household isprosperous, and event B is: the household is educated. According to Statistics Canada P(A) = 0.15, P(B) = 0.25 andP(A & B) = 0.09. What is the probability of each event?a) a household is educated, given that it is prosperous.b) a householder is prosperous, given that it is educated.

Group DiscussionIn the table below are the numbers (in hundreds) of earneddegrees in Canada in a recent year, classified by level and bygender of the degree recipient.

a) If a degree recipient is chosen at random, what is the probability that the person chosen is a women?b) What is the probability that a woman was chosen, given that the person chosen had received a professional degree?c) What is the probability that the person selected had received a Bachelor’s degree, given that the person chosen was male?

Applications of ConditionalProbability (8.4)

Math Power 12 p.391 #1-3

Applications of ConditionalProbability Worksheet at the end ofthe unit.

153

Block diagram for FCP Worthwhile Task (7.1)

This represents only the top half (10 paths) in the tree diagram. Students should quickly see that moreefficient methods must be found to solve problems like this. This problem will be revisited inPermutations (7.2) where students will write the solution as:

# !! !

paths =6

3 3

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Permutation Worksheet (7.2)

no repetitions allowed

1. In how many ways can 6 people be arranged in a row for a photo shoot? 6P6 = 720

2. In how many ways can 5 students be selected from a class containing 30 students? 30P5 = 17,100,720

repetitions allowed

3. In how many ways can 6 people be arranged in a row for a photo shoot if there are 2 identical twins in the group? 6!/2! = 360

4. How many distinctly different arrangements can be made from the following words:a) selected 8!/3! = 6720b) license 7!/2! = 2520c) initials 8!/3! = 6720d) letters 7!/2!2! = 1260e) arrangements 12!/2!2!2!2! = 29,937,600f) mississippi 11!/4!4!2! = 34650

5. Rearrange the letters to form as many words as possiblea) mslieb) ptoc) iterd) opst

Note to Teachers: One method is to write all possible arrangements(permutations) and see which onesmake words. Ex: pto 6 pto, opt, top, otp, tpo, pot

6. For a four letter word (no repetitions) there are 4! or 24 different arrangements. Write out all 24 arrangements for the letters opst.

7. Write to explain what is meant by “ order is important”.

155

Permutations and Combinations Explanation Sheet (7.3)

For a group of 10 people, a 3 person committee is to be chosen having a president, a vice president and asecretary. In how many different ways can the committee be chosen if:a) we will pick the 3 people to be on committee now and at some later time decide which will be president and so on.b) the first person chosen is to be president, the second is to be VP and the third is to be secretary.

a) Here order is not important (combinations) because if from the 10 people we pick Jane, Bob and Maryit doesn’t make any difference who is chosen first, second or third. We don’t have the added restrictionthat the first has to be president and so on. Usually adding extra conditions reduces the number of answersbut just the opposite happens here. Adding restrictions makes different groupings of the same threepeople distinguishably different creating more possible answers. Thus the number of ways (combinations)is:

10C3 = 120

b) Here order is important (permutations) and the number of ways to pick the committee is :

From the FCP we get : 10 × 9 × 8 = 720

or from permutations we get 10P3 = 720

In this scenario if we had picked:

1st pick < Jane < she would be president2nd pick < Mary < she would be VP3rd pick < Bob < he would be secretary

on the other hand if we had picked:

1st pick < Mary < she would now be president2nd pick < Bob < he would be VP3rd pick < Jane < she would be secretary

The same three people were picked from the 10 to choose from but each of the two scenarios yieldsdifferent committees. Thus the order in which the people are picked is important.

How many different ways can 3 people be arranged for a picture or a committee:FCP = 3 × 2 × 1 = 6 or 3P3 = 6

So in part (a) there is only one combination that has Jane, Mary and Bob but this one combination yields6 different permutations. Hence the answer for (b) is 6 times the answer for part (a).

Pascal’s Triangle for block problems (7.4)

156

The 20 inPascal’s Triangle is found using 6C3,

157

Connections between Combinations and Pascal’s Triangle (7.4)

158

Pascal’s Triangle Worksheet 7.4

Join the following series of points with as many different line segments as possible (including diagonals).Complete the accompanying table.

Predict the number of line segments that can be drawn for an octagon. Verify your answer by drawing anoctagon and all possible line segments.

Predict the number of line segments that can be drawn for a dodecagon (12 sided figure).

Predict the number of diagonals that can be drawn for an octagon. Verify your answer by drawing anoctagon and all possible diagonals.

Predict the number of diagonals that can be drawn for a dodecagon. Verify your answer by drawing adodecagon and all possible diagonals.

Probability of Single Events (8.1)

159

1. Find the probability of drawing:a) a face card. 12/52b) a black, non-face card. 20/52c) an ace 4/52

2. A box contains 5 yellow, 6 blue and 4 black marbles. What is the probability that the marble selected will be:

a) black 4/15b) not be blue 9/15c) green 0d) yellow 5/15

3. One flower is selected at random from a vase containing 5 red flowers, 2 white flowers and 3 orange flowers. Find the probability of each:

a) P(red) 5/10b) P(not orange) 7/10c) P(white) 2/10d) P(not white) 8/10

160

Binomial Theorem Applications Worksheet (7.5)

1. A pitcher throw 4 strikes for every 7 pitches. What is the probability that 3 of the next four pitches will be strikes?

2. A baseball player’s batting average is .200 . Find the probability that for the next 5 times at bat: a) he has 1 hit. b) he has 3 hits.

3. A true-false test has 12 questions. Suppose all questions are guessed. Determine the probability of guessing 5 correct answers.

4. About 9% of women will contract breast cancer sometime in their lifetimes. There are 100 women at a sales meeting. Find the probability that exactly four of them will contract breast cancer in their lifetimes.

5. A company manufactures computer chips. About 2% of the chips are defective. A quality control inspector pulls 50 chips at random off an assembly line. Find the probability that exactly 10 of them are defective.

6. A multiple-choice test has 12 questions. Each question has 5 choices, only one of which is correct. All questions are guessed at. What is the probability of getting 3 correct answers? 7 correct answers?

7. An archer can hit the bullseye 80% of the time. What is the probability that the archer can hit the bullseye 9 times out of 10.

8. The Twins and the Orioles are rival baseball teams. In the past few years, the Twins have win 60% of the games between the two teams. What is the probability that the Twins will win 5 of the next 7 games?

9. The Flames and the Avalanche are in the Stanley Cup Finals. Each team hopes to win the best-of- seven series. The probability of an Avalanche win is 70%. What is the probability that the Avalanche will win in 5 games?

10. Eight out of every 10 persons who contract a certain viral infection can recover. If a group of 7 people become infected, what is the probability that 3 will recover?

11. During the Gulf War in 1990-1991, SCUD missiles hit 20% of their targets. In one attack six missiles were fired at a fuel storage installation. What is the probability that 4 of them hit the target?

12. Ten percent of African-Americans are carriers of the genetic disease sickle-cell anemia. Find the probability that in a random sample of 20 African-Americans that 8 of them carry the disease.

161

Solutions:1. P(F) = 4/7; P(S) = 3/7;

( ) ( )

.

F S

C

n+ = +

= FHGIKJFHGIKJ=

37

47

37

47

32

4

4 3

1 3

2. P(F) = .8; P(S) = .2;a)

(. . )

( ) (. ) (. ).

8 2

1 8 24096

5

5 14 1

+

==

P hit C

b)

(. . )

( ) (. ) (. ).

8 2

3 8 20512

5

5 32 3

+

==

P hits C

3. P(F) = .5; P(S) = .5

(. . )

( ) (. ) (. ).

5 5

5 5 51934

12

12 57 5

+

==

P correct C

4. P(F) = .91; P(S) = .09

(. . )

( ) (. ) (. ).

91 09

4 91 0903

100

100 496 4

+

==

P will get cancer C

162

5. P(F) = .98; P(S) = .02

(. . )

( ) (. ) (. )

.

98 02

10 98 02

4 688 10

50

50 1040 10

8

+

=

= × −

P defective C

6. P(F) = .8; P(S) = .2

P correct C( ) (. ) (. ).

3 8 2236212 3

9 3==

P correct C( ) (. ) (. ).

7 8 2003312 7

5 7==

This problem an all the others can be solved with the TI-83.

The formula for the general term of a binomial expression is n rn r rC a b−

For this problem; n = 12; r = x; a = .8 and b = .2 . The window must have the horizontal dimension be94 divided by some integral value (I used 10 here). From the table the answer can be see to be .00332 andsimilarly from the graph we can see the same answer.

Note to Teachers: The graph shows the shape of this particular binomial distribution. If you are alsoteaching Math 621A these, along with normal distributions, will be done in more detail in Unit 6.

163

7. P(F) = .2; P(S) = .8

(. . )

( ) (. ) (. ).

2 8

9 2 82684

10

10 91 9

+

==

P bullseyes C

8. P(F) = .4; P(S)== .6

(. . )

( ) (. ) (. ).

4 6

5 4 62613

7

7 52 5

+

==

P wins C

9. P(F) = .3; P(S) = .7

(. . )

( ) (. ) (. ).

3 7

5 3 73177

7

7 52 5

+

==

P wins C

10. P(F) = .2; P(S) = .8

(. . )

( ) (. ) (. ).

2 8

3 2 80287

7

7 34 3

+

==

P live C

11. P(F) = .8; P(S) = .2

(. . )

( ) (. ) (. ).

8 2

4 8 20154

6

6 42 4

+

==

P hits C

12. P(F) = .9; P(S) = .1

(. . )

( ) (. ) (. )

.

9 1

8 9 1

3558 10

20

20 812 8

4

+

=

= × −

P with disease C

164

Classifying Events (8.2)Independent vs Dependent EventsIndependent events are when the first event does not affect the second event.

Dependent events are when the first event does affect the second event.

Ex. Find the probability of selecting a face card and then selecting an ace:a) when the first card is replaced in the deck.b) when the first card is not replaced in the deck.

Solution:a) Event A does not affect event B and these events are therefore independent. P(A & B) = P(A) × P(B).

For event A: P(A) = 12/52 = 3/13 For event B: P(B) = 4/52 = 1/13

Therefore P(A & B) = 3/13 × 1/13 = 3/169 = .0178

b) Event A does affect event B and these events are thus dependent. P(A & B) = P(A) × P(B*A).

P(B*A) is known as the conditional probability of event B happening, given that event A has alreadyhappened. Here we will deal with only simple conditional probability problems. This topic will be dealtwith in more detail in section 8.4.

For event A: P(A) = 12/52 = 3/13 For event B: P(B*A) = 4/51

Therefore P(A & B) = 3/13 × 4/51 = 4/221 = .0181

Independent/Dependent exercises

Determine if each event is independent or dependent. Then determine the probability.

1. Statistics collected in a particular area show that the probability that a person will develop diabetes is 5/11.As well, the probability that a person in that same area will develop arthritis is 1/5. If one health problem does not affect the other, what is the probability that a randomly selected person from that area of Canada will not develop diabetes but will develop arthritis.

2. Determine the probability of rolling a sum of 7 on the first toss of two dice and a sum of 4 on the second toss.

3. Tasha has 3 rock, 4 country and 2 rap CD’s in her car. She pulls 2 CD’s, one at a time without replacement, from the CD case without looking. What is the probability that both CD’s are rock?

4. Determine the probability of randomly selecting 2 green socks, one at a time without replacement, from a drawer that contains 7 black socks and 5 green socks.

5. There are 2 bottles of fruit juice and 4 bottles of pop in a cooler. Without looking, Joan chose a bottle

165

for herself and then one for her friend. What is the probability of choosing 2 bottles of pop?

6. Determine the probability of selecting a blue marble, without replacement, then a yellow marble from a box that contains 5 blue marbles and 4 yellow marbles.

7. Determine the probability of randomly selecting 2 oranges, one at a time without replacement, from a bowl of 5 oranges and 4 tangerines, if the first selection is replaced.

8. A green number cube and a red number cube is tossed. What is the probability that a 4 is shown on the green cube and a 5 is shown on the red cube?

9. Find the probability of randomly taking 2 blue notebooks, one at a time without replacement, from a shelf which has 4 blue and 3 black notebooks.

10. A bag contains 4 nickels, 4 dimes and 7 quarters. Three coins are removed in sequence without replacement. What is the probability of selecting a nickel, a dime and a quarter in that order?

Note to teachers: If the phrase “in that order” is not there, then there are 3! = 6 different orders ofchoosing a nickel, a dime and a quarter. The probability would then be 6 × as great. This idea appliesonce you start working with different objects/people.

11. What is the probability of removing 13 cards from a deck of cards, one at a time without replacement, and have all of them red?

12. What is the probability of randomly selecting a knife, a fork and a spoon in that order, without replacement, from a kitchen drawer containing 8 spoons, 8 forks and 12 knives? (Note that the order of selecting is given).

13. Determine the probability of selecting 3 different coloured crayons from a box containing 5 red, 4 black and 7 blue crayons, assuming each crayon is replaced?

14. Find the probability of drawing a black marble, replacing it, then drawing a yellow marble from a bag with 3 yellow and 5 black marbles.

15. Determine the probability of rolling a sum of 2 and the first toss of two dice and a sum of 6 on the second toss. (Note order is given).

16. Determine the probability of randomly selecting 2 yellow markers from a box, one at a time without replacement, that contains 4 yellow and 6 pink markers.

17. A box contains 3 black and 9 white balls. A ball is drawn and not replaced. A second ball is drawn from the box. What is the probability that both balls drawn are black?

18. A box contains 3 black and 9 white balls. A ball is drawn and replaced. A second ball is drawn from the box. What is the probability that both balls drawn are black?

166

Solutions:1. independent. P(A) = 1 ! 5/11; P(B) = 1/5; P(A & B) = 6/55

2. independent. P(A) = 6/36; P(B) = 3/36; P(A & B) = 1/72

3. dependent. P(A) = 3/9; P(B*A) = 2/8; P(A & B) = 1/12







10. dependent. P(A) = 4/15; P(B*A) = 4/14; P(C*A&B) = 7/13; P(A & B & C) = 8/195

11. dependent. P = 19/1,160,054

12. dependent. P(A) = 12/28; P(B*A) = 8/27; P(C*A&B) = 8/26; P(A & B & C) = 32/819

13. independent. P = 35/1024





18. independent. P(a) = 3/12; P(B) = 3/12; P(A & B) = 1/16

167

Classifying Events (8.2) cont’d.

Mutually Exclusive vs Non-Mutually Exclusive Events

Mutually Exclusive events are two events that share no common outcomes. That is they can’t happen atthe same time.

P(A or B) = P(A) + P(B)

Non-Mutually Exclusive events are two events that may share common outcomes. That is the two sets ofoutcomes may have an intersection set.

P(A or B) = P(A) + P(B) ! P(A & B)

Example of Mutually Exclusive:In a contest, Jim can win a car if he selects a blue ball or a red ball. He must select the ball from a box

containing 2 blue, 3 red, 9 yellow and 11 green balls. What is the probability of his winning the car?

Solution:Theses are mutually exclusive events. He is allowed to make only one draw and he is permitted someflexibility in what he can draw to win.Event A: P(A) = 2/25; Event B: P(B) = 3/25

P(A or B) = 2/25 + 3/15 = 1/5.

The probability of winning the car is 1/5.

Example of Non-Mutually Exclusive:What is the probability of rolling a 2 dice where the first die shows a 2 or the sum of the two dice is 6

or 7?

Solution:It is possible to have a 2 on the first die and have the sum of the two dice being 6 or 7. Therefore theseevents are non-mutually exclusive.

Event A: P(A) = 6/36 First die showing a 2

Event B: P(B) = 11/36 Sum of the dice is 6 or 7

Event A & B: 2/36 First die showing a 2 and the sum is 6 or 7

P(A) + P(B) ! P(A & B) = 6/36 + 11/36 ! 2/36 = 15/36

The probability of the first die showing a 2 or the sum of the dice being 6 or 7 is 15/36 = 5/12.

168

Event A: Event B

Event A & B

P(A or B) = 15/36

169

Mutually Exclusive/ Non-Mutually Exclusive Worksheet

Determine if the events are mutually exclusive or non-mutually exclusive. Then determine theprobability of each.

1. Find the probability of choosing a penny or a dime from 4 pennies, 3 nickels and 6 dimes.

2. Find the probability of selecting a boy or a blond-haired person from 12 girls, 5 of whom have blond hair, and 15 boys, 6 of whom have blond hair.

3. Find the probability of drawing a king or queen from a standard deck of cards.

4. The probability for a driver’s license applicant to pass the road test the first time is 5/6. The probability of passing the written test on the first attempt is 9/10. The probability of passing both test the first time is 4/5. Are the events mutually exclusive? What is the probability of passing either test on the first attempt?

5.Find the probability of tossing two dice and either on showing a 4.

6. Find the probability of selecting an ace or a red card from a deck of cards.

7. Determine the probability that a card drawn from a deck is red or a face card.

8. Find the probability of two dice being tossed and showing a sum of 6 or a sum of 9.

9. A weather forecaster states that the probability of rain is 3/5, the probability of lightning is 2/5, and the probability of both is 1/5. What is the probability of a sporting event being cancelled due to rain or lightning?

10. A bag contains cards numbered from 1 to 14. One card is drawn at random. Find the probability of: a) selecting a prime number or a multiple of four. b) selecting a multiple of two or a multiple of three. c) selecting a 3 or a 4. d) selecting an 8 or a number less than 8.

170

Solutions:1. mutually exclusive; P(A) = 4/13; P(B) = 6/13; P(A or B) = 4/13 + 6/13 = 10/13

2. non- mutually exclusive; P(A) = 15/27; P(B) 11/27; P(A & B) = 6/27P(A or B) = 15/27 + 11/27 ! 6/27 = 20/27

3. mutually exclusive; P(A) = 4/52; P(B) = 4/52; P(A or B) = 8/52 = 2/13

4. non mutually exclusive; P(A) = 5/6; P(B) = 9/10; P(A & B) = 4/5P(A or B) = 5/6 + 9/10 ! 4/5 = 14/15

5. non -mutually exclusive; P(A) = 1/6; P(B) = 1/6; P(A & B) = 1/36P(A or B) = 1/6 + 1/6 ! 1/36 = 11/36

6. non-mutually exclusive; P(A) = 4/52; P(B) = 26/52; P(A & B) = 2/52P(A or B) = 4/52 + 26/52 ! 2/52 = 28/52 = 7/13

7. non-mutually exclusive; P(A) = 26/52; P(B) = 12/52; P(A & B) = 6/52P(A or B) = 26/52 + 12/52 ! 6/52 = 32/52 = 8/13

8. mutually exclusive; P(A) = 5/36; P(B) = 4/36P(A or B) = 9/36 = 1/4

9. non- mutually exclusive; P(A) = 3/5; P(B) = 2/5; P(A & B) = 1/5P(A or B) = 3/5 + 2/5 ! 1/5 = 4/5

10. a) mutually exclusive; P(A) = 6/14; P(B) = 3/14P(A or B) = 9/14 b) non-mutually exclusive; P(A) = 7/14; P(B) = 4/14; P(A & B) = 2/14P(A or B) = 7/14 + 4/14 ! 2/14 = 9/14

c) mutually exclusive; P(A) = 1/14; P(B) = 1/14P(A or B) = 2/14 = 1/7

d) mutually exclusive; P(A) = 1/14; P(B) = 7/14P(A or B) = 8/14 = 4/7

Combinatorics Worksheet (8.3)

171

1. A committee of 5 people is to be selected from your class. Determine the number of women and men in your class and find the probability that there are exactly 3 women on the committee.

2. There are 7 women and 5 men in a room. A committee of 4 is to be randomly selected. What is the probability of there being: a) 4 women on the committee. 7C4 5C0 / 12C4 = 35/495 b) 3 women on the committee. 7C3 5C1 / 12C4 = 175/495 c) 2 women on the committee. 7C2 5C2 / 12C4 = 210/495 d) 4 men on the committee. 7C0 5C4 / 12C4 = 5/495 e) 3 men on the committee. 7C1 5C3 / 12C4 = 70/495 f) 2 men on the committee. 7C2 5C2 / 12C4 = 210/495

3. Three balls are randomly drawn from a bag containing 6 white balls and 4 black balls. What is the probability of drawing: a) 3 white balls. 6C3 4C0 / 10C3 = 20/120 b) 2 white balls and 1 black ball. 6C2 4C1 / 10C3 = 60/120 c) 1 white ball and 2 black balls 6C1 4C2 / 10C3 = 36/120 d) 3 black balls 6C0 4C3 / 10C3 = 4/120

4. Four people are to be randomly selected from a group of 9 boys and 7 girls. What is the probability of each event? a) exactly 3 being girls. 7C3 9C1 / 16C4 = 315/1820 b) all 4 being boys. 7C0 9C4 / 16C4 = 126/1820 c) 2 being girls. 7C2 9C2 / 16C4 = 756/1820

5. Four balls are randomly drawn from a bag containing 3 white balls and 5 black balls. What is the probability that there are exactly 2 white balls? 3C2 5C1 / 8C4 = 15/70

6. Five cards are dealt from a deck of 52 cards. What is the probability of each event? a) a hand containing one pair of aces and 3 kings. 4C2 4C3 / 52C5 = 24/2598960 b) a hand containing 2 face cards and 3 10's. 12C2 4C3 / 52C5 = 264/2598960 c) a hand containing 3 hearts and 2 diamonds. 13C3 13C2 / 52C5 = 22308/2598960

172

Applications of Conditional Probability Worksheet (8.4)

1. A household is considered prosperous if its income exceeds $75,000. A household is considered educated if the householder has completed college. A household is selected at random and event A is: the selected household is prosperous, and event B is: the household is educated. According to Statistics Canada P(A) = 0.15, P(B) = 0.25 and P(A & B) = 0.09. What is the probability of each event? a) a household is educated, given that it is prosperous. b) a householder is prosperous, given that it is educated.

2. In the table below are the numbers (in hundreds) of earned degrees in Canada in a recent year, classified by level and by gender of the degree recipient.

a) If a degree recipient is chosen at random, what is the probability that the person chosen is a women? b) What is the probability that a woman was chosen, given that the person chosen had received a professional degree? c) What is the probability that the person selected had received a Bachelor’s degree, given that the person chosen was male?

3. A new medical test for glaucoma is 95% accurate. Suppose 0.8% of the population have glaucoma. What is the probability of each event? a) A randomly selected person will test negative. b) A person has glaucoma, given that they test negative. c) A person does not have glaucoma, given that they test positive.

4. In a recent survey, 60% of the respondents were women. Studies have shown that women report their

173

heights accurately 10% of the time while men report their heights accurately 5% of the time. What is the probability of each event? a) a respondent has reported accurately on their height. b) a respondent is male, given that they did not report their height accurately?

5. 60% of the population are women. 30% of men smoke while 35% of women smoke? What is the probability of each event? a) a person selected is a man. b) a person selected is a smoker. c) a person selected is a man who smokes. d) a person selected is a smoker, given that they a man. e) a person selected is a man, given that they are a smoker.

6. A medical test is 97% accurate. Suppose 0.25% of the population has this disease. What is the probability of each event? a) a randomly selected person tests negative for the disease. b) a person tests positive, given that they have the disease.

7. Polygraph tests are reportedly only 70% accurate. Assuming that a person tells the truth 90% of the time, determine the probability that a person is telling the truth, given that the polygraph indicates that a truth is being told.

8. The test for diabetes is 85% accurate. Statistics have shown that 20% of the people over the age of 85 have diabetes. What is the probability of each event? a) a person over 85 does not have diabetes, given that the test was positive. b) When the situation in part (a) happens, the doctor usually orders the test repeated. What is the probability that a person over 85 does not have diabetes, given that the test showed positive twice? three times? ( Hint different applications of the test are independent events.)

9. A breathalyzer test is 95% accurate. Assume that 80% of the drivers that are asked to take the test have consumed too much alcohol. What is the probability of each event? a) A driver is not impaired, given that the test was positive. b) A driver is impaired, given that the test was positive.

10. Three manufacturing companies make a certain product. Company A has a 50% market share, company B has a 30% market share and company C has a 20% market share. 5% of company A’s products are defective, 7% of company B’s products are defective and 10% of company C’s products are defective. What is the probability of each event? a) a product is defective. b) a product comes from company B, given that it is defective. c) a product is defective, given that it comes from company C.

Solutions:1.

174

a) b) P B A P A BP A

( ) ( & )( )

.

..= = =

0915

6 P A B P A BP B( ) .( & )

( ).09.25= = = 36

2. a) P(woman) = 856/1626

b) P(W*P) = 30/74 This is the probability that a woman was chosen, given that the person chosen hadreceived a professional degree. That means you are restricted to looking at the professional column andin that column find the fraction of women to the total number that received that degree.

c) P(B*M) = 529/770 You must look in the row that relates to men, what fraction of the total menreceived a Bachelor’s Degree.

175

For the following problems we will use two methods; Bayes Theorem and a joint probability tablelike the one given in # 2 above.

3. Bayes’ Theorem

a) P(test is !) = .0004 + .9424 = .9428b) P(G*test is !) = P(G & !)/P(!) = .0004/(.0004 + .9424) = .0004c) P(no G* +) = P(no G & +)/ P(+) = .0496/(.0076 + .0496) = .8671

3. Joint Probability TableAssume 1000 people where 992 don’t have glaucoma (99.2%) and 8 (.8%) do have glaucoma.

Test + Test ! Total

Glaucoma .95 × 8 = 7.6 .05 × 8 = .4 8

no Glaucoma .05 × 992 = 49.6 .95 × 992 = 942.4 992

Total 57.2 942.8 1000

The table is generated with the following logic:the first row talks about the 8 people with glaucoma; since the test is only 95% accurate then .08 × .95 =7.6 of them will test + as it should. However 5% of the time the test is inaccurate so that even though thepeople have glaucoma .05 × 8 = .4 of them will test ! indicating that they don’t have glaucoma. Eventhough we have fractions of a person, I felt it best to limit the sample size to 1000 rather than get thenumbers too large.The second row talks about the 992 people that don’t have glaucoma. Again because of the fallibility ofthe test 49.6 of these people will be stressed when the test indicates that they have glaucoma when in factthey don’t. As well, 942.4 will test negative confirming that they in fact don’t have glaucoma.

4. Bayes’ Theorem

176

a) P(A) = .06 + .02 = .08

b) P(M*NA) = P(M & NA)/P(NA) = .38/(.54 + .38) = .413

4. Joint Probability Table Assume 100 people where 60 are women and 40 are men.

Accurate Not Accurate Total

Men .05 × 40 = 2 .95 × 40 = 38 40

Women .10 × 60 = 6 .90 × 60 = 54 60

Total 8 92 100

P(A) = 8/100 = .08

P(M*NA) = 38/92 = .413 What fraction of the respondents that didn’t report their heights accuratelywere men? We are restricted to looking at the Not Accurate column; what fraction of that total weremen?

177

5. Bayes’ Theorem

a) P(M) = .12 + .28 = .40b) P(S) = .21 + .12 = .33c) P(M & S) = .12d) P(S*M) = P(S & M)/P(M) = .12/(.12 + .28) = .3e) P(M*S) = P(M & S)/P(S) = .12/(.12 + .21) = .36

Joint Probability TableAssume 100 people where 60 are women and 40 are men.

Smoker Non-Smoker Total

Men .30 × 40 = 12 .70 × 40 = 28 40

Women .35 × 60 = 21 .65 × 60 = 39 60

Total 33 67 100 a) P(M) = 40/100 = .40b) P(S) = 33/100 = .33c) P(M & S) = 12/100 = .12d) P(S*M) = 12/40 = .3e) P(M*S) = 12/33 = .36

178

6. Bayes’ Theorem

a) P(!) = .000075 + .967575 = .96765

b) P(D*+) = P(D & +)/P(+) = .002425/(.002425 + .029925) = .07496

Joint Probability TableAssume 10,000 people where 25 have the disease and 9975 don’t have the disease.

Test + Test ! Total

Disease .97 × 25 = 24.25 .03 × 25 = .75 25

no Disease .03 × 9975 = 299.25 .97 × 9975 = 9675.75 0075

Total 323.5 9676.5 10,000

a) P(!) = 9676.5/10,000 = .96765

b) P(D*+) = 24.25/323.5 = .07496

179

7. Bayes’ Theorem

P(T*t) = P(T & t)/P(t) = .63/(.63 + .03) = .95

Joint Probability TableAssume 100 people where 90 are telling the truth and 10 are not.

Test indicates personis telling the truth

(t)

Test indicates personis not telling the truth

(nt)

Total

Person Telling theTruth (T)

.70 × 90 = 63 .30 × 90 = 27 90

Person Not Tellingthe Truth (NT)

.30 × 10 = 3 .70 × 10 = 7 10

Total 66 34 100

P(T*t) = 63/66 = .95

180

8. Bayes’ Theorem

a) P(no D*+) = P(no D & +)/P(+) = .12/(.12 +.17) = .4138b) P(no D*+ twice) = (.4138)2 = .171 P(no D *+ 3 times) = (.4138)3 = .071

Note to Teachers: At this stage a doctor can be about 93% sure that the patient doesn’t have diabetesbecause a person without diabetes will test positive 3 times only 7.1% of the time.

Joint Probability TableAssume 100 people over 85 where 20 have diabetes and 80 don’t have diabetes.

Tests + Tests ! Total

Diabetes .85 × 20 = 17 .15 × 20 = 3 20

no Diabetes .15 × 80 = 12 .85 × 80 = 68 80

Total 29 71 100

P(no D*+) = 12/29 = .4138

181

9. Bayes’ Theorem

a) P(not D*+) = P(not D & +)/P(+) = .01/(.01 + .76) = .0130

b) P(D*+) = P(D & +)/P(+) = .76/(.76 + .01) = .9870

Joint Probabilities TableAssume 100 drivers where 80 are drunk and 20 are not drunk.

Tests + Tests ! Total

Drunk .95 × 80 = 76 .05 × 80 = 4 80

not Drunk .05 × 20 = 1 .95 × 20 = 19 20

Total 77 23 100

a) P(not D*+) = 1/77 = .0130

b) P(D*+) = 76/77 = .9870

182

10. Bayes’ Theorem

a) P(D) = .025 + .021 + .02 = .066b) P(B*D) = P(B & D)/P(D) = .021/.066 = .3182c) P(D*C) = P(D & C)/P(C) = .02/.20 = .10

Joint Probability TableAssume 100 items; 50 made by company A, 30 by company B and 20 by company C.

Defective Not Defective Total

A .05 × 50 = 2.5 .95 × 50 = 47.5 50

B .07 × 30 = 2.1 .93 × 30 = 27.9 30

C .10 × 20 = 2 .90 × 20 = 18 20

Total 6.6 93.4 100

a) P(D) = 6.6/100 = .066b) P(B*D) = 2.1/6.6 = .3182c) P(D*C) = 2/20 = .10

unit six combinatorics and probability math 621b 12 hours · unit six combinatorics and probability...

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