Section 11 Using Counting Principles, Permutations, and ......Part II Probability and Statistical Reasoning Section 11 Using Counting Principles, Permutations, and Combinations Permutations

88 Advanced Quantitative Reasoning Copyright © 2015 by Gregory D. Foley, Thomas R. Butts, Stephen W. Phelps, and Daniel A. Showalter

Part II Probability and Statistical Reasoning

Section 11 Using Counting Principles, Permutations, and Combinations

Our world is full of possibilities! If there are eight available toppings at Port Jef-ferson Pizza, and their pizzas come in three sizes, then Port Jefferson offers 84 possible two-topping pizzas. How many ways can you lose a pick-6-numbers-out-of-50 lottery? There are 15,890,699 ways to lose! There are fewer than 2.6 million different 5-card hands from a deck of 52 playing cards, but more than 635 billion 13-card hands. Determining the correct values for such large numbers of possibili-ties requires systematic methods and constitutes the branch of mathematics known as combinatorics.

Quick Question 11.1 So Many Ways

(a) How many ways can three books with different titles be arranged on a shelf?(b) Four books? (c) Five books? (d) Ten books?

Using Problem-Solving Strategies in CombinatoricsCombinatorics is “counting without really counting.” It would simply take too long to list and count all 635,013,559,600 possible hands in the game of contract bridge. So instead we solve related similar problems, make systematic lists, draw diagrams, look for patterns, generalize, and use other problem-solving strategies when asked to determine a large number of possibilities.

In solving Quick Question 11.1, you may have used systematic lists or tree dia-grams to solve parts (a) through (c), noticed the pattern 3 · 2 · 1, 4 · 3 · 2 · 1, 5 · 4 · 3 · 2 · 1, …, generalized it to n · (n – 1) · … · 3 · 2 · 1, and then used this formula with n = 10 to solve part (d). Of course, there are many ways to solve Quick Question 11.1, but in all cases problem-solving strategies are helpful. When solving combinatorics prob-lems, it is often dangerous to try to apply formulas without first using problem-solving methods.

Exploration 11.2 Counting Without Really CountingHow many possible numbers could each student be thinking of? (You may wish to try this exploration as a think–pair–share activity working in groups of four, with each student initially solving one of the four parts.)(a) Tomas thinks of a 3-digit number with at least one digit of 7.(b) Yolanda thinks of a 3-digit number with the second digit less than the

first and the third digit less than or equal to the second.(c) Michael thinks of a 3-digit number with all odd digits.(d) Ashley thinks of a 3-digit number that is greater than 940 and not a

multiple of 5.Working together as a group of four…(e) Could all four students be thinking of the same number? If so, which

number(s) could they all be thinking of?

Section 11Using Counting Principles, Permutations, and CombinationsMain Ideas• Using Problem-Solving Strategies

in Combinatorics• Using the Multiplication Principle

of Counting• Permutations• Combinations• Counting Subsets

Copyright © 2015 by Gregory D. Foley, Thomas R. Butts, Stephen W. Phelps, and Daniel A. Showalter Advanced Quantitative Reasoning 89

Part II Probability and Statistical ReasoningSection 11 Using Counting Principles, Permutations, and Combinations

Using the Multiplication Principle of CountingOne of the authors of this book wanted to get a license plate reading “MATH ED” because he loves mathematics education. But his first choice was already taken, so he had to settle for “MTH ED.” (See Figure 11.1.) Apparently, there is at least one other mathematics education lover in Ohio.

Definition Multiplication Principle of Counting

If a procedure occurs in n stages and the number of ways the stages occur are w1, w2, w3, …, wn, then the number of ways the overall procedure can occur is the product

w1 · w2 · w3 · … · wn.

This is the multiplication principle of counting or the fundamental principle of counting.

ExamplE 1 Permuting LettersHow many ways can the letters M, T, H, E, and D be arranged using all five let-ters one time in each arrangement?Solution There are 5 possible first letters, but once a first letter is picked, there are only 4 choices available for the second letter. Similarly, there are 3 choices for the third letter, 2 for the fourth, and 1 for the fifth. So, by the multiplication prin-ciple of counting, there are

5 · 4 · 3 · 2 · 1 = 120 possible arrangements of these five letters.Looking back. Pólya’s fourth step in the problem-solving process is to look back,

check your work, consider alternative solutions, and generally reflect on what you have done. To solve Example 1 we could have used the brute force approach and arranged all of the possibilities in a systematic list. Table 11.1 shows the list in alpha-betical order:

Figure 11.1 An Ohio license plate consisting of five letters.

more onlineGo to http://regentsprep.org/Regents/math/ALGEBRA/APR1/indexAPR1.htm to learn more about the multiplica-tion principle of counting.

Table 11.1 List of the Possible Arrangements of M, T, H, E, and DDEHMT DEHTM DEMHT DEMTH DETHM DETMHDHEMT DHETM DHMET DHMTE DHTEM DHTMEDMEHT DMETH DMHTE DMHET DMTEH DMTHEDTEHM DTEMH DTHEM DTHME DTMEH DTMHE

EDHMT EDHTM EDMHT EDMTH EDTHM EDTMHEHDMT EHDTM EHMDT EHMTD EHTDM EHTMDEMDHT EMDTH EMHDT EMHTD EMTDH EMTHDETDHM ETDMH ETHDM ETHMD ETMDH ETMHD

HDEMT HDETM HDMET HDMTE HDTEM HDTMEHEDMT HEDTM HEMDT HEMTD HETDM HETMDHMDET HMDTE HMEDT HMETD HMTDE HMTEDHTDEM HTDME HTEDM HTEMD HTMDE HTMED

MDEHT MDETH MDHET MDHTE MDTEH MDTHEMEDHT MEDTH MEHDT MEHTD METDH METHDMHDET MHDTE MHEDT MHETD MHTDE MHTEDMTDEH MTDHE MTEDH MTEHD MTHDE MTHED

TDEHM TDEMH TDHEM TDHME TDMEH TDMHETEDHM TEDMH TEHDM TEHMD TEMDH TEMHDTHDEM THDME THECM THEMD THMDE THMEDTMDEH TMDEH TMEDH TMEHD TMHDE TMHED

http://regentsprep.org/Regents/math/ALGEBRA/APR1/indexAPR1.htm






Organizing such a loooong list takes some planning. Once the layout is planned, making the list is tedious and time consuming. This is not an efficient way to solve Example 1, but the list may give us insight into the fundamental principle of count-ing. Take a few moments to confirm that the list is in alphabetical order. The use of alphabetical order is a systematic way to include all possible arrangements and to avoid repeats—in order to get a complete and accurate list.

Once the list is completed and checked for accuracy, we still need to count the number of entries in the list. Notice that there are five (5) groupings of four (4) rows with six (6) entries per row, for all total of 5 · 4 · 6 = 120 entries. Notice that all six entries in each row start with the same two letters in the same order. In each row, there are 3 · 2 · 1 = 6 ways to arrange the remaining three letters. So, do you see the 5 · 4 · 3 · 2 · 1 pattern embedded in the systematic list?

Arranging three letters is a related simpler problem. It is easy to see that there are six ways to arrange A, B, C:

ABC ACB BAC BCA CAB CBASome people find it easier to understand the multiplication principle of counting

using a tree diagram. Figure 11.2 represents the four stages of creating a 4-letter ar-rangement using each of the letters A, B, C, D one time. Each stage is represented by a set of line segments emanating from a single point. The total number of paths from left to right matches the number of line segments in the final stage and the 24 possible arrangements of the four letters. Why do we multiply 4 · 3 · 2 · 1 to count the paths?

Figure 11.2 A tree diagram showing the 24 ways to arrange four letters.

Tree diagrams can be used to calculate conditional probability. A visual inspection of Figure 11.2, for example, shows that the probability of the second letter being a consonant, assuming that the first letter is a consonant, is 6/9 (or 2/3).

Investigation 11.3 Counting Using the Multiplication PrincipleUse the multiplication principle of counting to determine…(a) How many ways can the letters R, K, H, S be arranged into 5-letter strings

allowing letters to be repeated?(b) How many 3-letter words are theoretically possible using the 26 letters of

the English alphabet and allowing repeated letters?(c) How many 3-letter English words are theoretically possible if each must

contain at least one of the vowels a, e, i, o, or u?(d) How many different license plates can be created using three letters fol-

lowed by four digits? (Examples: XYZ2005, SST7797, IAM8699)

Problem-Solving Strategy Solve an Equivalent Problem

When the notation or context of a problem is unfamiliar or hard to work with, it may be easier to solve another problem that has the same solution.Example In Example 1, to determine how many ways the letters M, T, H, E, D can be arranged, we could have used the letters A, B, C, D, E; the dig-its 1, 2, 3, 4, 5; a set of five books; or any five distinct objects. The letters A, B, C, D, E or the digits 1, 2, 3, 4, 5 are easier to work with than the let-ters M, T, H, E, and D.

Theory Versus Practice

The actual number of 3-letter English words is less than the theoretical val-ues in Investigation 11.3(b) and (c). In the game of Scrabble®, for example, fewer than 1,000 such words are allowed.

Contrary to the theoretical answer in Investigation 11.3(d), U.S. states do not use or allow some letter sequences for various reasons, including those deemed potentially offensive to the public.



PermutationsSeveral of the problems we have investigated so far have involved counting the number of possible arrangements of an n-set (that is, a set of n elements): for exam-ple, 10 books on a shelf or a set of five English letters. Such ordered arrangements of the elements of a set are permutations of the set.

Definition Permutations of an n-Set

A set of n elements has n! = n(n – 1)(n – 2) · … · 3 · 2 · 1 permutations.The notation n! is read “n factorial.” We define 0! = 1 and n! = n(n – 1)! for n = 1, 2, 3, ….Christian Kramp (1760–1826) introduced the exclamation mark as the symbol for factorial because factorials become remarkably large for relatively small numbers n.

Quick Question 11.4 Fascinating FactorialsUsing technology, make a chart of the values 0!, 1!, 2!, …, 20! (These values will take you into the millions, billions, trillions, quadrillions, and quintillions.)Challenge: Try pronouncing the value for 20!

Quick Question 11.5 Using the Permutation Formula

(a) Four of the best-selling musicians of all time—Mariah Carey, Julio Iglesias, Elton John, and Madonna—are going to perform together at a huge out-door summer concert. You have to decide the order in which they should appear on stage. How many permutations are possible?

(b) For being the top AQR student in America, you get to manage the Cincin-nati Reds baseball team for the last day of the season. Regular manager Dusty Baker gets to pick the eight position players and the pitcher, who will bat ninth. You get to pick the batting order for the rest. How many batting orders are possible?

(c) Is it possible to use the same eight position players for all 162 games in a major league season and have a different batting order for every game?

Exploration 11.6 Distinguishable PermutationsTribune Media Services, Inc. publishes the Daily Jumble® in many newspapers. In this game, ordinary words have their letters rearranged and you have to figure out what the original word is. For example, the word BUNDLE could be rearranged as ULBEND.(a) How many distinguishable ways can the letters in the word PUZZLE be

rearranged?(b) How many distinguishable ways can the letters in the word MISSISSIPPI

be rearranged?

If you switch the Zs in PUZZLE, it will still look the same, which cuts the number of distinguishable arrangements in half. So, you should have discovered that there are only 6!/2 – 1 = 359 distinguishable rearrangements of the letters.

Numbers Everywhere Splitting Hairs

Beard shaving is a multibillion dol-lar international industry. Beard hair grows an average of 140 mm/year. Men in the U.S. on average live about 60 years after they begin shaving and have about 11,000 hairs in their beard, each about 0.2 mm in diameter.(a) How long could a single beard

hair grow in a typical man’s life-time if it were never shaved, cut, or otherwise broken?

(b) What is the total volume of beard hair the typical man in the U.S. grows during his lifetime?

Data source: Rimmer (2006).

more onlineGo to http://www.regentsprep.org/regents/math/algebra/APR2/in-dexAPR2.htm to learn more about permutations.

http://www.regentsprep.org/regents/math/algebra/APR2/indexAPR2.htm






The word MISSISSIPPI has 11 letters: 4 Is, 1 M, 2 Ps, and 4 Ss. So the number of distinguishable permutations of MISSISSIPPI is

Thus the number of distinguishable rearrangements is 34,649. Even so, ULBEND might be harder to unscramble than SISIPSIPISM if you are from Mississippi. When humans unscramble words, they look for familiar patterns and do not just check all possible permutations as a computer generally would.

Yet another twist on permutations occurs when selecting elements from a set to form a smaller ordered set.

ExamplE 2 Superlative StudentsAt the end of each 6 weeks of study, a math class with 18 students selects one student to win each of four awards: most creative problem solver, most proficient using the handheld, most diligent, and most improved. No student can receive the same award twice, and any student can be honored in at most one way in any given 6 weeks. At the end of the first 6 weeks, in how many possible ways could the awards be given?Solution There are 18 possible most creative problem solvers, but once that student is chosen, there are only 17 choices available for most proficient using the handheld. Similarly, there are 16 choices for most diligent, and 15 for most improved. So, by the multiplication principle of counting, there are

18 · 17 · 16 · 15 = 73,440 possible ways to make the awards.Example 2 illustrates the idea of the number of permutations of 18 objects taken

4 at a time.

So, the solution to Example 2 could be denoted using nPr notation as 18P4, and solved using the formula

This formula is available on most calculators and handheld computers. See Figure 11.3.

more onlineGo to http://regentsprep.org/Regents/math/algtrig/ATS5/indexATS5.htm to learn more about permutations of n objects taken r at a time.

Figure 11.3 Permutations computed on a handheld.

Definition Distinguishable Permutations of an n-Set

If an n-set consists of n1 elements of one type, n2 elements of a second type, n3 elements of a third type, etc., such as n = n1 + n2 + n3 + … + nk, then the number of distinguishable permutations of the set is

Definition Number of Permutations of n Objects Taken r at a Time

If r elements are chosen from a set of n objects, the number of possible permuta-tions of the r objects is denoted by either nPr or P(n,r) and can be computed using

for integers 0 ≤ r ≤ n.

http://regentsprep.org/Regents/math/algtrig/ATS5/indexATS5.htm




CombinationsIn some counting problems, the arrangements or ordering of objects does not mat-ter. If we were to select a 4-person committee from a group of 18 students, with no ranking or titles, then we may wish to know how many such committees are pos-sible. This situation illustrates the idea of the number of combinations of 18 objects taken 4 at a time.

ExamplE 3 Committee ConundrumThe number of possible 4-student committees from a class of 18 students is

See Figure 11.4.

Quick Question 11.7 First Five to PlayEach National Basketball Association team has a roster of 12 players.(a) How many starting fives can be chosen from these 12 players, ignoring

positions?(b) How many starting fives can be chosen from one of two centers, two of

six forwards, and two of four guards?

Quick Question 11.8 Straight Poker Versus Contract BridgeAt the beginning of the section, it was stated that, “There are fewer than 2.6 million different 5-card hands from a deck of 52 playing cards, but more than 635 billion 13-card hands.” Is this statement true? How do you know?

Counting SubsetsThe number of 13-card hands in a deck of 52 cards is the same as the number of 13-element subsets of any set of 52 elements. So, counting combinations is equiva-lent to counting subsets of an n-set.

ExamplE 4 Golfing GroupsThe golf team at Grandson Mountain High School has only five players. Due to poor grades, illness, injury, or other commitments, sometimes players are not available. How many different subsets of players could be available for the dis-trict tournament?(a) 5C5 = 1. So, there is 1 way all five players could be available.(b) 5C4 = 5. So, there are 5 ways four players could be available.(c) 5C3 = 10. So, there are 10 ways three players could be available.(d) 5C2 = 10. So, there are 10 ways two players could be available.

Figure 11.4 Combinations computed on a handheld.

more onlineGo to http://regentsprep.org/Regents/math/algtrig/ATS5/indexATS5.htm to learn more about combinations and how they are related to permutations.

Definition Number of Combinations of n Objects Taken r at a Time

If r elements are chosen from a set of n objects, the number of possible combi-nations of the r objects is denoted by

or ,

and can be computed using for integers 0 ≤ r ≤ n.






(e) 5C1 = 5. So, there are 5 ways one player could be available.(f) 5C0 = 1. So, there is 1 way no players could be available.See Figure 11.5. There are 32 distinct subsets of a 5-set, so 32 different subsets of players could be available for the district tournament.There are some interesting patterns in the solution to Example 4. For example,

there is symmetry in the numerical results to subproblems (a) through (f). The total is 32 = 25, with the power of 2 matching the number of players on the golf team. Is this just a coincidence? No, it is part of a larger pattern. See Figure 11.6.

Figure 11.6 Pascal’s triangle. A numerical pattern for finding nCr.

To create Pascal’s triangle, (a) write the outer diagonals of ones, and then (b) for each of the remaining entries, write the sum of the two numbers immediately above it. This makes generating the numbers nCr an easy matter if done recursively. We can express this recursive process algebraically using the following formulas.

Exploration 11.9 Pascal Patterns

(a) Create your own Pascal’s triangle using the method just described.(b) To the left, label the single 1 at the top as row 0, the row with two 1s as

row 1, the row with the 1 2 1 pattern as row 2, and so on.(c) What is the sum of each row of Pascal’s triangle for rows 0–10? Describe

the pattern.

If we generalize what we have learned beginning with Example 4, we will realize that the sum of the entries in row n of Pascal’s triangle tells us how many subsets there are for a set of n elements.

Number of Subsets of an n-Set

A set of n elements has 2n subsets.

Figure 11.5 Computations for Example 4.

History Note Whose Triangle Is It?

The numerical array in Figure 11.6 is called Pascal’s triangle, but this triangular pattern was developed long before the birth of Blaise Pascal (1623–1662). Pascal’s triangle was known to, and possibly first developed by, Abu Bakr al-Karaji (c. 953–1019) of Islam’s Baghdad-centered Buyid Dynasty and to Jia Xian (c. 1010–1070) of China’s Song Dynasty, ap-pearing in his work The Key to Math-ematics (Shi suo suan shu).

more onlineGo to http://pages.csam.montclair.edu/~kazimir/index.html and learn more about the history, patterns, and applications of Pascal’s triangle in Investigation 11.9.

Definition Recursive Formulas for the Number of Combinations of n Objects Taken r at a Time

The entries in Pascal’s triangle are nCr = C(n,r) = Cnr = , and can be computed

using

(a) for the integers n = 0, 1, 2, 3, …, and

(b) for all integers n > r > 0.

http://pages.csam.montclair.edu/~kazimir/index.html

http://pages.csam.montclair.edu/~kazimir/index.html



This simple formula can be derived directly from the fundamental principle of counting: When forming a subset of an n-set, each of the n elements in the original set will be either (a) included or (b) excluded—two choices. This then is a procedure with n stages each of which can occur 2 ways. So the overall procedure can occur in

ways.

This formula applies to many counting situations involving including or exclud-ing elements, such as Example 5.

ExamplE 5 That’s The Way … I Like ItThe fast food chain Why So Many Burgers offers hamburgers with or without cheese, lettuce, tomatoes, onions, pickles, salt, black pepper, jalapeños, ketchup, mayonnaise, and mustard. So you can choose any subset of these 11 choices, making for 211 = 2,048 possible ways to order a burger.

Quick Question 11.10 Pizza PossibilitiesAt the beginning of the section, it was stated that, “If there are eight available toppings at Port Jefferson Pizza, and their pizzas come in three sizes, then Port Jefferson offers 84 possible two-topping pizzas.” How many pizzas are possible with any number of toppings? How do you know?

Quick Review for Section 11

Do the calculation in your mind. Write down only the answer. 1. 1 · 2 · 3 · 4 2. 1 · 2 · 3 · 4 · 5 3. 2 · 5 · 6 4. 5 · 4 · 90

5.

6.

Do the calculation using the Probability menu of a calculator. Write down only the answer, inserting commas for answers greater than 1,000. 7. 11! 9. 20P6

8. 17! 10. 20C6

Exercises for Section 11

1. Outfitting. Choosing from three pairs of jeans and five T-shirts, how many outfits are possible?

2. Counting codes. How many 3-digit telephone area codes are possible if the first digit cannot be 0 or 1? (Examples: 888, 477, 910)

3. Breakfast options. The local BLD Restaurant serves break-fast, lunch, and dinner all day. All breakfasts come with scrambled eggs, meat, potatoes, bread, and a drink. • The egg options are one, two, or three. • The meat options are sausage, bacon, country ham, or no

meat.• The potato options are home fries, away fries, or no fries.• The bread options are toast or biscuit.• The drink options are coffee, tea, milk, orange juice, or

grapefruit juice. How many different breakfasts can be ordered at the BLD

Restaurant?

4. Choosing classes. Senior year you must take English, math-ematics, science, and social studies classes.• The English options are Classical World Literature and

Modern World Literature.• The math options are Calculus, Precalculus, and

Advanced Quantitative Reasoning.• The science options are Astronomy, Biology, Chemistry,

Geology, Physics, and Planet Earth.• The social studies options are Civics, Economics, and

Ancient World History. How many ways can you pick classes, choosing exactly one

from each list? 5. Routing. There are three roads from Ausville to Booneville,

six roads from Booneville to Cristabel, and five roads from Cristabel to Deep Notch. How many possible routes are there from Ausville to Deep Notch using only these roads with no backtracking?




6. King, queen, and court. The Festival of the Hills has seven candidates for Festival King and nine candidates for Festival Queen. How many ways can the King, Queen, and two First Runner-Ups be chosen to form the Festival Court?

7. Station to station. How many radio station names are possi-ble that start with K or W and consist of three or four letters? ( Examples: KFMA, WING, WLW, KCC)

8. Socially secure. How many 9-digit Social Security numbers are possible if there are no restrictions on the digits that can be used?

9. Three-letter codes. How many ways can the letters A, C, H, S be arranged into 3-letter strings (a) if letters can be repeated? (b) If letters cannot be repeated?

10. Five-letter codes. How many ways can the letters W, X, Y, Z be arranged into 5-letter strings (a) if letters can be repeated? (b) If letters cannot be repeated?

11. Three chiefs. A local shipping operation is selecting new leadership. There are seven finalists: one each will become the chief executive officer (CEO), the chief operating officer (COO), and the chief financial officer (CFO). In how many ways can the officers be selected from the finalists?

12. Team picture. The 23 members of the school softball team line up for a photograph with their three coaches and one trainer. Ten are in the front row, nine in the middle row, and eight in the back row. How many ways can this group be ar-ranged for the photo?

13. Old-fashioned poker. In the basic game of straight poker, five cards are dealt from a deck of 52 different cards; how many hands are possible?

14. Contract bridge. In the game of contract bridge, 13 cards are dealt from a standard deck of 52 distinct playing cards; how many hands are possible?

15. How many distinct ways can the letters in the word PLAN-CHET be arranged?

16. How many distinct ways can the letters in the word SPRINGHEAD be arranged?

17. How many distinct ways can the letters in the word FILIGREE be arranged?

18. How many distinct ways can the letters in the word COMMANDANT be arranged?

19. How many distinct ways can the letters in the word INTERCHANGEABLE be arranged?

20. How many distinct ways can the letters in the word ENTOMOPHAGOUS be arranged?

Factorial expressions. Calculate using nPr or nCr from the Prob-ability menu of a calculator without using the factorial (!) com-mand. Then check using factorials.

21.

22.

23.

24.

25. Yumyum Pizza. Yumyum Pizza Parlor offers 12 possible toppings on its large thin-crust pizza. You can order a plain cheese pizza or any combination of toppings, including the “Megayum,” which has all 12 toppings. How many types of large thin-crust pizza are possible at Yumyum?

26. Auto choices. Your rich cousin is ordering a brand-new car and has already chosen the make, model, and interior and exterior colors. Now she must decide on the following options: air conditioning, power windows, sun roof, privacy-tint glass, manual transmission, larger engine, dual exhaust, pin stripes, whitewall tires, leather upholstery, and sound system upgrade. In how many ways can your cousin order her new car?

27. Senate committees. The U.S. Senate has 100 members. In how many ways can a 20-member committee be chosen?

28. House committees. The U.S. House of Representatives has 435 voting members. In how many ways can a committee of 32 be chosen from voting members?

Extending What You Have Learned29. Factorial vs. powers of two. Which is greater? (a) 0! or 20 (b) 3! or 23 (c) 6! or 26 (d) 9! or 29 30. Factorial vs. powers of ten. Which is greater? (a) 0! or 100 (b) 3! or 103 (c) 9! or 109 (d) 27! or 1027 31. Case-sensitive passwords. The passwords on the Tg0007

computer system must be two English letters followed by four base-ten digits. Capital and lowercase letters are con-sidered to be distinct. So, for example, AB1234 and ab1234 are different passwords. (Such systems are said to be case-sensitive because upper- and lowercase letters are treated differently.) How many passwords are possible if all such passwords are allowed except for tg0007, Tg0007, tG0007, and TG0007?

32. Senate caucuses. Suppose that the U.S. Senate has 60 mem-bers who are in the Democratic Caucus and 40 in the Republi-can Caucus. (a) In how many ways can a 10-member committee be cho-

sen with 6 members from the Democratic Caucus and 4 from the Republican Caucus?

(b) In how many ways can a 10-member committee be cho-sen without regard to caucus?

33. Poker faces. You deal cards in order to your four friends Benny, Billie, Bobbi, and Bubba, and yourself until you each have five cards. If you use an ordinary deck of 52 cards, how many arrangements of five such 5-card hands are possible?

34. Contract bridge. You deal in order to player Pat, partner Peg, Pat’s partner Pixie, and yourself until you each have 13 cards. If you use an ordinary deck of 52 cards, how many arrange-ments of four such 13-card hands are possible?



Investigations35. Orders of Magnitude for Factorials

(a) Make a table that shows 0!, 10!, 20!, 30!, …, 90!. (b) Make a table that shows log(0!), log(10!), log(20!), log(30!),

…, log(90!). (c) Make a table that shows 100!, 200!, 300!, …. (d) Make a table that shows 400!, 410!, 420!, …. (e) What is the largest factorial for which you can obtain an

exact value on your handheld?(f) What is the largest factorial for which you can obtain an

approximation on your handheld?(g) What are the orders of magnitude for the factorials in

parts (e) and (f)?36. Triangular numbers. Recall that the triangular numbers are

1, 1 + 2, 1 + 2 + 3, 1 + 2 + 3 + 4, …. (a) Make a table for these sums.(b) Why are these numbers called triangular?(c) Where can the triangular numbers be found in Pascal’s

triangle? Explain why.(d) Using the Lists & Spreadsheet application, make a table

of values for n = 1, 2, 3, …, 100.(e) Use your results so far to explain why

(f) Make a table of values for the sequence:

(g) How is the sequence of partial sums of cubes in part (f) related to the triangular numbers?

Looking Ahead37. On a roll. Two 6-sided dice are rolled; each has faces labeled 1

through 6.(a) If one is white and one is blue, how many distinguishable

results are possible?(b) If both are identical green dice, how many distinguishable

results are possible?(c) Which totals are possible?(d) How many ways can each possible total be achieved?(e) Which total do you think is most likely? Why?



Section 12 Reasoning With Probability

Section 12Reasoning With Probability

You likely have encountered some of the concepts of probability in earlier grades or everyday experiences. Probability involves quantifying uncertainty, that is, assign-ing a number to how likely something is to occur. The probability of a phenomenon occurring ranges from impossible (a probability of zero) to certain (a probability of one). See Figure 12.1.

p = 0 = 0% p = 1⁄2 = 50% p = 1 = 100% • • • Impossible Unlikely 50–50 Chance Likely Certain

Figure 12.1 The interval [0, 1] of real numbers includes all possible probability values p.

Quick Question 12.1 Everyday ProbabilitiesWhat are three examples of how probability is used in everyday life?

Understanding and applying probability can be a difficult and tricky business. Professional actuaries who work with and determine probabilities to assess finan-cial risks must know a great deal of mathematics and deeply understand relevant issues. For example, an actuary who sets rates for hurricane insurance must under-stand tropical weather patterns and real estate market fluctuations, and be able to develop and update accurate mathematical models.

In this course, we do not expect you to become an actuary, but we do expect you to gain a deep understanding of the basic concepts of probability and to learn how to apply them to many everyday situations. Research shows that untrained human intuition about probability is often incorrect and can mislead us into mak-ing unsound decisions. In this section and throughout this course, we wish to help you rethink your intuition about probability and rebuild it on a solid mathematical foundation. We urge you to keep an open mind and to adapt your thinking about probability as needed.

What Is Random?We begin our study of probability by exploring the meaning of the word “ random,” first by defining it and then by investigating it within the context of flipping coins.

Definition Random Sample

In statistics, when a sample is drawn from a population, it is a random sample if each combination of members of the population is equally likely to be chosen.

In mathematics the concept of random links statistics with probability because equally likely means has the same probability. The mathematical concept of random is different from its everyday meaning of haphazard. We hear people say, “That’s so random!” In this case, the speaker may be using random to mean unexpected. In mathematics, random phenomena (random happenings, random occurrences) have a predictability and regularity in the long run. In mathematics, random hap-penings are neither haphazard nor unexpected. Random samples are used pre-cisely because they yield predictable, reliable results.

Main Ideas• What Is Random?• Probability Simulations• Theoretical Probability• Independent and Dependent

Events• Empirical Probability


Part II Probability and Statistical ReasoningSection 12 Reasoning With Probability

In the next three investigations, we engage in a series of related probability ex-periments to test our intuition about flipping a fair coin, that is, a coin for which landing with either side up—heads or tails—is equally likely.

Now you can test your intuition about random sequences of Hs and Ts against actual data.

The fundamental principle of counting tells us there are over 1 million ways to fill in a table with 20 boxes using only Hs and Ts. To be exact, there are 220 = 1,048,576 ways to fill such a table. In a true random sample of this population, each of the over 1 million possible sequences would be equally likely to occur. Figure 12.2 is a spreadsheet of Pascal’s triangle from row 0 through row 20, with the rows tilted as

more onlineGo to www.aqrpress.com/sa1201 for an interactive file that can be used with Investigations 12.2, 12.3, and 12.4. The applet will help you get a feel for the randomness of flipping coins.

Investigation 12.2 Imagined Data—Thought Experiments

(a) Using a table like the one below, fill in what you imagine to be a random sequence of Hs and Ts (heads and tails) as if you were flipping a fair coin repeatedly 20 times:

Count the strings of Hs and strings of Ts in your table. [Three Hs in a row is a string (or streak) of 3 Hs.] Overall, what is the longest string of either type?

(b) Share your data with a small group of students. Let everyone in the group look at everyone else’s data. Discuss the similarities and differences, and whether you feel these data are random. Could you change your strategy to make them “more random”?

(c) After completing the small group discussion, repeat the thought experi-ment, and place your results in a table.

Once again, count the strings of Hs and strings of Ts in your table. Over-all, what is the longest string of either type?

Investigation 12.3 Genuine Data—Conducting an Actual Experiment

(a) Using a table like the one below, fill in the results of flipping a fair coin 20 times:

Count the strings of Hs and strings of Ts in your table. Overall, what is the longest string of either type?

(b) Pool the class’s data regarding the longest strings Hs or Ts, and compute an average of the longest string length for the class. Engage in a whole-group discussion of the results. How does the class average for the genu-ine data compare to the thought-experiment data from Investigation 12.2? Which is truly random? Explain.

http://www.aqrpress.com/sa1201




diagonals and row 20 presented in boldface type. Focusing on the ends of row 20, we see that only 1 of the 1,048,576 possible sequences would be all heads, or all tails. Moving toward the middle of row 20, we can see that, 20C9 = 167,960 of the sequences would contain exactly 9 heads (and hence 11 tails). So in a large random sample we would expect about 16% (167,960 ÷ 1,048,576) of the sequences to con-tain 9 Hs and 11 Ts.

Figure 12.2 Pascal’s triangle revealing 20Cr for r = 0, 1, 2, 3, …, 20.

Reflection. Notice that Figure 12.2 does not tell us much about strings of heads or strings of tails. We do know that a sequence containing 9 Hs and 11 Ts could not have a string of length 12 or longer. (Why?) To determine how many among these 167,960 sequences contain strings of length 11, or 10, or 9,… would require further analysis—perhaps on the order of a chapter or a course project.

The point here is to have a problem context that helps us understand the con-cepts of a population, a sample, and a random sample. The population is the set of all possible sequences of Hs and Ts of length 20. A sample would be a collection of these sequences drawn from the population; depending on the circumstances the sequences could be drawn with or without replacement. A random sample in this context is one for which each of the 1,048,576 members of the population are equally likely to be included in the sample.

Probability SimulationsIn Investigation 12.3, you conducted an experiment of flipping a fair coin 20 times. There are several ways to simulate this experiment, that is, to perform a similar experiment with equivalent or nearly equivalent results: • You could draw cards from a well-shuffled deck, with replacement and reshuf-

fling between draws, and record a black card as heads and a red card as tails.• You could roll a single die, counting an even number as heads and an odd as

tails.• You could use computer or handheld software as explained in the Tech Tip.

Figure 12.3 A calculator simulation of flipping a fair coin 20 times, repeated three times.

Tech Tip Seeding and Using a Random Number Generator

If your software or device is brand new, first seed the pseudorandom number generator using the randSeed command followed by the digits of your student ID, telephone number, or other such number. Then, to simu-late flipping a fair coin 20 times, use the randInt command as shown in Figure 12.3. Count each 1 as a heads and each 0 as a tails.



Theoretical ProbabilityThe set of all possible outcomes of a random phenomenon is a sample space, and a subset of the sample space is an event. So in our coin flipping experiment TTHTHHTTTHTTTHHTTHHH would be one of the possible 1,048,576 outcomes that altogether constitute the sample space. The entire set is too large to write down. The set of 167,960 sequences with 9 Hs and 11 Ts would be an event in the context of this experiment, and TTHTHHTTTHTTTHHTTHHH is one of the outcomes of this event. It is important to remember that an event is a set of outcomes.

Recapping the Terminology

• A sample space is the set of all possible outcomes of a random phenomenon.• An event is a subset of the sample space, and is thus a set of outcomes.• An outcome is an element in the sample space of a random phenomenon.

Before electronic computer-based technology (that is, for most of human history), dealing with large sample spaces and events was extremely difficult. Even now, working with a small sample space can be instructive, and is in keeping with the strategy of solving a related simpler problem.

ExamplE 1 Flipping a Fair Coin Four TimesIn the probability experiment of flipping a fair coin four times:(a) The sample space is {HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH,

HTTT, THHH, THHT, THTH, THTT, TTHH, TTHT, TTTH, TTTT}.(b) The outcomes are HHHH, HHHT, … , TTTT, as listed within the sample

space. There are 24 = 16 outcomes in the sample space.(c) The event “a total of 3 heads” is {HHHT, HHTH, HTHH, THHH}, a subset of

the sample space containing 4 outcomes.(d) The event “at least 3 heads” is {HHHH, HHHT, HHTH, HTHH, THHH}, a

subset of the sample space containing 5 outcomes.

more onlineCan you tell a random sequence of coin flips from a nonrandom sequence of coin flips? Go to www.aqrpress.com/sa1202 for an interactive file that will allow you to explore the difference between flip-ping a fair coin and flipping an unfair coin.

Investigation 12.4 Simulating Random Data Collection

(a) Use a nonelectronic method to simulate 20 coin flips. Record the results in a table:

What is the longest string of Hs or Ts?(b) Use an electronic method to simulate 20 coin flips. Record the results in a

table:

What is the longest string of Hs or Ts?(c) Which method is faster? Which method is truly random?(d) What does the word random mean in the context of probability and

statistics?(e) What is meant by the term pseudorandom number generator? How do

you know?





(e) The event “a longest string of 2 in a row” is {HHTH, HHTT, HTHH, HTTH, THHT, THTT, TTHH, TTHT}. This event contains 8 outcomes, half of the sequences in the sample space.

ExamplE 2 Further FlippingIn the probability experiment of flipping a fair coin four times:

(a)

(b)

(c) If the event E is “a longest string of 2 in a row,” then

We can use this same method for any random phenomenon with equally likely outcomes.

ExamplE 3 Tumbling DiceNicholas rolls a pair of fair dice. The 36 possible outcomes are

(a) Because (6, 6) is the only outcome that totals 12,

(b) There are three ways to get a “4,” so

(c) There are six ways to get a “7,” so

(d) When rolling a pair of dice, “doubles” means two of a kind: (1, 1), (2, 2), …, (6, 6).

Thus, p(“doubles”)

The following exploration will help you solidify your understanding of new terminology and notation. Notice that determining a theoretical probability value often boils down to solving a counting problem.

Numbers Everywhere Benford’s Law

Benford’s Law, or the first-digit law, says that numbers when used in everyday ways do not exhibit a uni-form distribution of the possible first digits 0 through 9, but rather a first digit of 1 is the most likely and a first digit of 9 is the least likely. The law states that the probability of a first digit d naturally occurring in actual use is given by the formula

(a) Make a table for the Benford’s Law probabilities of first digits 0 through 9.

(b) Do these probabilities total to approximately 1? Do they total to exactly 1?

(c) How could Benford’s Law be used to catch faked numbers on tax returns and other records?

Theoretical Probability of an Event (Equally Likely Outcomes)

If an event E is a subset of a nonempty, finite sample S containing n(S) equally likely outcomes, and E contains n(E) of these outcomes, then the probability of event E is



Exploration 12.5 Pick ThreeRachel wants to pick a 3-digit number at random. She decides to use the com-mand “randInt(100,999)” on her handheld. (a) If S is the sample space for this random phenomenon, what is the value of

n(S)?(b) If E is the event “an even number,” what are the values for n(E) and p(E)?(c) What is the probability p(an even first digit)?(d) What is the probability p(contains at least one digit of 7)?(e) What is the probability p(a 3-digit result)?(f) What is the probability p(a 2-digit result)?

At the beginning of the section we stated, “The probability of a phenomenon occurring ranges from impossible (a probability of zero) to certain (a probability of one).” We now reinforce and extend that statement.

We now combine these rules with the results from Example 3.

ExamplE 4 Rolling Two Fair DiceJennifer rolls a pair of fair dice.

(a) Because no outcomes total 13, A total of 13 is an impossible event.

(b) By the complement rule,

(c) By the addition rule, p(a total of 4 or “doubles”) =

(d) By the addition rule for disjoint events, p(a total of 7 or “doubles”) =

Independent and Dependent EventsTwo events are independent events if the occurrence of one has no effect on the prob-ability of the occurrence of the other. In our experiment of flipping a fair coin 20 times, the event “heads on the first flip” (or “tails on the first flip”) does not influence what happens on the second flip; there is still a 50–50 chance of heads or tails. So, the events “tails on the first flip” and “tails on the second flip” are independent events—but the events “heads on the first flip” and “tails on the first flip” are not independent. Why?

Figure 12.4 Venn diagram of a sample space S and events E and F within S.

Figure 12.5 Venn diagram of a sample space S and disjoint events E and F within S.

Probability Rules

Let E and F be events of a nonempty, finite sample space S. • • If , then and E is an impossible event.• If E = S, then and E is a certain event.• Addition rule. . (See Figure 12.4).• Addition rule for disjoint events. If , that is, if E and F have

no outcomes in common, then E and F are disjoint (mutually exclusive) events and (See Figure 12.5).

• Complement rule. The complement of an event E is the event that E does not occur, and




Often events that occur in sequence are independent, but not always.

ExamplE 5 Independent and Dependent Events(a) A family’s first child is a girl. The probability that the second child is a girl is

still ½. The events are independent. (b) A standard U.S. roulette wheel has 38 compartments: 18 red, 18 black, 2

green. Landing in any compartment is equally likely. If the first spin lands on green, the probability of the second spin landing on red is still 18/38. The events are independent.

(c) The first card drawn from a standard deck of 52 is a “heart.” Now only 51 cards are left, of which only 12 are hearts. The probability that the second draw is a ”club” is now 13/51 rather than 13/52. In this case, the occurrence of one event does have an effect of the probability of the occurrence of the other event. The events are not independent; the events are dependent.

In part (c) of Example 5, we are dealing with a conditional probability. The prob-ability that a club is the second draw is changed by the condition that a heart was the first card drawn. In such circumstances, a handy notation is , which is read “p of E given F.” This notation means the probability that event E occurs given that event F occurs. In the card example, the probability that a club is the second draw given that a heart was the first draw is 13/51. We can write this in the following way:

Notice that in this case, drawing the first card actually changed the sample space for the second draw. When one event changes the sample space for another event, often the events are dependent—but not always. Recognizing whether events are independent or dependent is not always easy. Truth tables, as shown in the margin, can help determine dependence of events. However, the only way to be sure is to go back to the definition of independent events, as illustrated in Example 6.

ExamplE 6 Proving Independence of EventsIf William uses the randInt(0,9) command to generate one random digit, then the events E “an even digit” and F “a multiple of 5” are independent.proof First, we establish that the sample space S is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, E is {0, 2, 4, 6, 8}, and F is {0, 5}. This allows us to compute and

Next we compute the probability of F given E:

;

and the probability of E given F:

So the events E and F are independent because the occurrence of one has no ef-fect of the probability of the occurrence of the other.Notice that the rule does apply:

Truth Tables

A truth table is a systematic way to evaluate logical claims. The first columns of a truth table list all the combinations possible of the claims being true or false. The rest of the columns can be used to evaluate logical statements that involve the claims. Consider the following two claims after rolling two dice: “The second roll was a 6,” and “The sum was 12.” Refer to the first claim as p and the second as q. Table 12.1 shows how a truth table could be used to classify compound state-ments involving “and” (∧), “or” (∨), and “implies” (⇒). If one claim im-plies another—as in this example—then the events are dependent.

Table 12.1 Truth Tables and Logical Claims

p q p ∧ q p ∨ q p ⇒ qT T T T TT F F T FF T F T TF F F F T

Multiplication Rule for Probabilities of Independent Events

Let E and F be independent events of a nonempty, finite sample space S. Then,



The following formula is valid for both dependent and independent events.

In the case of independent events , so the principle simplifies to the earlier formula . These formulas can be used both to find probabilities and to determine whether two events are independent.

Empirical ProbabilityIt is possible to purchase trick dice that are loaded (internally weighted) so that cer-tain values turn up more often than others. Suppose you buy some trick dice from a magic shop and wish to impress your friends. You decide to test one of them by rolling it 1,000 times to determine how often each number will turn up. The results are shown in Table 12.2.

When we determine the probability of an outcome using repeated trials, the relative frequency of the outcome is its empirical probability. In this instance, the empirical probabilities are

It appears the die has been loaded so that 6 turns up most of the time, which would give an unfair advantage in many games.

Exploration 12.6 Using Empirical Probabilities to Test FairnessIf the die that generated the values in Table 12.2 is rolled twice, (a) What is the probability of two 6s in a row?(b) What is the probability of two 1s in a row?(c) What is the probability of a total of 7 on the two rolls?(d) How could you use a Venn diagram to show the theoretical probability of

getting two 6s in a row?(e) How could you use an area model to show the theoretical probability of

getting a sum of 7? [Hint: See margin note on Area Models.](f) How does the empirical result in part (c) compare with the theoretical

result in part (e)?(g) Use a tree diagram to calculate the probability of getting a sum of 7 in two

rolls given that your first roll is an even number.

more onlineGo to www.aqrpress.com/sa1203 for an interactive file that will allow you to explore the results of rolling a loaded die in Exploration 12.6.

Area Models

One way to represent probability is using an area model. In an area model, the probability of a particular outcome is equal to the portion of the overall area of the shape. For example, the area model in Figure 12.6a shows that the theoretical probability of obtain-ing a heads with a fair coin is .5. We can also use an area model to show compound probability of independent events. Figure 12.6b shows that, if you roll a fair die and flip a fair coin, the theoretical probability of rolling a six (6) and flipping a heads is 1/12.

Figure 12.6 Area models for the probabilities of (a) getting a heads and (b) getting both a 6 and a heads.

Quick Review for Section 12

Do the calculation by an appropriate method. Ensure that your answer is reasonable. 1. 3! 2. 5!

3. 23

4. 27 5. 7. 12P2 6. 8. 15P3 9. 12C2 10. 15C3

Generalized Multiplication Principle for Probability

Let E and F be events of a nonempty, finite sample space S. Then,

Table 12.2 Results From Rolling a Loaded Die 1,000 TimesResult One Two Three Four Five SixFrequency 19 81 84 79 91 646





1. Certain. What is the probability of an event that is certain to happen?

2. Impossible. What is the probability of an impossible event? 3. Random sample. Which of the following are true for a

random sample of a population?(A) Each member of the population is equally likely to be

chosen.(B) The sample tends to be representative of the population

in the long run.(C) You can predict the likelihood that an attribute will occur

within the elements of the sample.(D) Such samples are easy to obtain because any haphazard

method can be used. 4. Random sample. Is it likely that a random sample of 50 U.S.

residents would be drawn from the same county? Is it pos-sible that a random sample of 50 U.S. residents would be drawn from the same county?

Population probabilities. Solve using the state data in Table 12.3 and the 2010 official U.S. census population estimate of 308.7 mil-lion and assuming random selection. 5. What is the probability of a U.S. resident being from California? 6. What is the probability of a U.S. resident being from a state in

the top ten in population? 7. What is the probability of a U.S. resident being from Wyoming? 8. What is the probability of a U.S. resident being from a state in

the bottom ten in population?

Standard Problems 9. What is the probability of flipping three heads in a row using

a fair coin?10. What is the probability of two heads when flipping a fair coin

three times?11. What is the probability of flipping five heads in a row using a

fair coin?

12. What is the probability of four heads when flipping a fair coin five times?

13. What is the probability of at least 3 heads when flipping a fair coin 5 times?

14. What is the probability of exactly 13 heads when flipping a fair coin 15 times?

15. What is the probability of at least 13 heads when flipping a fair coin 15 times?

16. What is the most likely number of heads when flipping a fair coin 15 times?

17. What is the probability of getting a 3 when rolling one fair die?

18. What is the probability of getting more than 2 when rolling one fair die?

19. What is the probability of a sum of 10 when rolling two fair dice?

20. What is the probability of a sum of 6 when rolling two fair dice?

Intermediate Problems21. What is the probability of “7, 11, or doubles” when rolling

two fair dice?22. What is the probability of an odd sum when rolling two fair

dice?23. What is the probability of all hearts in a 5-card hand from a

standard deck of 52?24. What is the probability of all hearts in a 13-card hand from a

standard deck of 52?25. What is the probability of all red cards in a 13-card hand from

a standard deck of 52?26. What is the probability of all red cards in a 5-card hand from

a standard deck of 52?27. If you thoroughly shuffle a deck of 52 cards, and then deal 7

red cards in a row, what is the probability that that 8th card you deal will also be red?

28. If you thoroughly shuffle a deck of 52 cards, and then deal 10 clubs in a row, what is the probability that that 11th card you deal will also be a club?

Challenge Problems29. Yahtzee. In the game of Yahtzee®, you roll five dice, and get

to keep as many of the outcomes as you wish and reroll the remaining dice. After the second roll, again you can keep as many of the original outcomes and as many of the new out-comes as desired, and reroll for a third and final time. If five of a kind is Yahtzee, what is the probability of(a) getting Yahtzee on the first roll?(b) getting three of a kind, which you keep, on the first roll,

and then getting Yahtzee on the second roll?

Exercises for Section 12

Table 12.3 Populations (in Millions) by State, 2010 Census Data in Order by RankCA 37.3 NJ 8.79 MN 5.30 MS 2.97 ME 1.33TX 25.1 VA 8.00 CO 5.03 AR 2.92 NH 1.32NY 19.4 WA 6.72 AL 4.78 KS 2.85 RI 1.05FL 18.8 MA 6.55 SC 4.63 UT 2.76 MT 0.989IL 12.8 IN 6.48 LA 4.53 NV 2.70 DE 0.898PA 12.7 AZ 6.39 KY 4.34 NM 2.06 SD 0.814OH 11.5 TN 6.35 OR 3.83 WV 1.85 AK 0.710MI 9.88 MO 5.99 OK 3.75 NE 1.83 ND 0.673GA 9.69 MD 5.77 CT 3.57 ID 1.57 VT 0.626NC 9.54 WI 5.69 IA 3.05 HI 1.36 WY 0.564Source: U.S. Census Bureau, as reported in The World Almanac and Book of Facts 2012.



30.–34. Medical testing. Modern medicine uses many types of tests; none are 100% accurate. You may wish to use an area model or tree diagram to solve these exercises.30. False positive. Suppose that a recent national study indicates

that 3.7% of high school athletes use steroids and related performance-enhancing drugs. The accuracy of the test for the drugs is 97.2%. So, 2.8% of the time, the test returns an incorrect result (either a false positive or a false negative). What is the probability that a randomly selected high school athlete who tests “positive” is not a user of performance-enhancing drugs?

31. True positive. Suppose that married women who do at-home pregnancy tests have a 30% probability of being pregnant. The average accuracy of such tests is 90% if pregnant, and 80% if not. (That is, 10% of the time, the test returns a false negative for pregnant women, and 20% of the time, the test returns a false positive for women who are not pregnant.) What is the probability that a randomly selected married woman who tests “positive” is actually pregnant?

32. True positive. Suppose that 4.6% of the men tested in Wis-consin who are at least 65 years of age have prostate cancer. If a test is 95.3% accurate among men being tested, what is the probability of a true-positive result? What is the probability of a true-positive result given that the result is positive?

33. False negative. Suppose that 20% of the residents of Illinois have the flu and that 75% of the patients going to a doctor for testing have the flu. If a medical doctor in Illinois uses a test that is 60% accurate, what is the probability that a patient receives a false-negative result? What is the probability of a false-negative result given that the result is negative?

34. True negative. Suppose that 2.7% of the residents of Hawaii have HIV and that 7.3% of the patients being tested have HIV. If a test that is 93% accurate, what is the probability of a true-negative result? What is the probability of a true-negative result given that the result is negative?

Looking Back35. Hot pools. Some natural hot pools can be too hot for human

comfort and safety or too acidic or both. In fact, at Yellow-stone National Park, such dangerous waters have killed more people than have bears. Which combination of temperature and pH would be safe?(A) T = 60°C and pH = 2.7 (C) T = 60°C and pH = 7.7(B) T = 40°C and pH = 2.7 (D) T = 40°C and pH = 7.7

36. pH. How many times more acidic is a pH of 2.7 than a pH of 7.7?

37. °C. What is normal body temperature in °C?38. K. What is normal body temperature on the Kelvin scale?

Investigations39. Blood types. Over 40% of the world’s people live in China

(1.33 billion), India (1.17 billion) or the United States (307 mil-lion). The distribution of blood types varies across these three nations, as is shown in Table 12.4.(a) What is the probability of type A blood in the U.S.?(b) What is the probability of not type O in the U.S.?(c) If you choose at random one Chinese and one Indian,

what is the probability that they both have type AB blood? Use an area model.

(d) If you choose one person at random from each of these three countries, what is the probability that all three have type B blood? Use a tree diagram.

(e) If you choose 4 persons at random each of whom live in one of these three countries, what is the probability that all 4 are Chinese? All 4 are from the U.S.? Use a formula.

(f) If you choose at random one person who lives in one of these three countries, what is the probability that the per-son has type O blood? Use a Venn diagram.

40. Flipping four. Consider the random phenomenon of flipping a fair coin four times.(a) How many outcomes are in the sample space?(b) Use the randInt(0,1,4)command repeatedly. (That

is, press enter repeatedly once the command is placed on a calculator entry line.) How can this be used to simu-late the random phenomenon of flipping a fair coin four times?

(c) How many outcomes are in the events “a total of 4 heads”? “3 heads”? “2 heads”? “1 heads”? “0 heads”? What are the associated probabilities?

(d) Use the sum(randInt(0,1,4)) command repeatedly. What events are being simulated?

(e) How many outcomes are in the events “a longest string of 4 in a row”? A longest string of 3? A longest string of 2? A longest string of 1? What are the associated probabilities?

(f) What is the average length of a string (or streak) for this random phenomenon?

41. Use a truth table to analyze Exercise 30. Let “uses drugs” be p, and “tests positive” be q. In terms of p, q, and logical connectors, what type of student is the exercise referring to? [Note: In logic statements, “¬p” means “not p.”]

42. Use a truth table to analyze Exercise 31. Let “is pregnant” be p, and “tests positive” be q. In terms of p, q, and logical connec-tors, what type of married woman is the exercise referring to?

Table 12.4 Relative Frequency of Blood Types in the World’s Three Most Populous NationsBlood Type China India U.S.

O .35 .37 .44 A .27 .22 .42 B .26 .33 .10 AB .12 .07 .04

Source: http://www.bloodbook.com/ and other sources.

http://www.bloodbook.com/

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