Section 11 Using Counting Principles, Permutations, and Combinations Main Ideas

Part II Probability and Statistical Reasoning
Section 11 Using Counting Principles, Permutations, and Combinations
Section 11
Using Counting Principles, Permutations,
and Combinations
Main Ideas
•Using Problem-Solving Strategies
in Combinatorics
•Using the Multiplication ­Principle
of Counting
•Permutations
•Combinations
•Counting Subsets
Our world is full of possibilities! 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 ways can you lose a pick-6-numbers-outof-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 possibilities 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 Combinatorics
Combinatorics 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 diagrams 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 problems, it is often dangerous to try to apply formulas without first using problemsolving methods.
Exploration 11.2 Counting Without Really Counting
How 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?
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
Using the Multiplication Principle of Counting
Figure 11.1 An Ohio license plate consisting
of five letters.
more online
Go to http://regentsprep.org/Regents/
math/ALGEBRA/APR1/indexAPR1.
htm to learn more about the multiplication principle of counting.
One 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 Letters
How many ways can the letters M, T, H, E, and D be arranged using all five letters 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 principle 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 alphabetical order:
Table 11.1
List of the Possible Arrangements of M, T, H, E, and D
DEHMT DEHTM DEMHT DEMTH DETHM
DHEMT DHETM DHMET DHMTE DHTEM
DMEHT DMETH DMHTE DMHET DMTEH
DTEHM DTEMH DTHEM DTHME DTMEH
DETMH
DHTME
DMTHE
DTMHE
EDHMT
EHDMT
EMDHT
ETDHM
EDHTM
EHDTM
EMDTH
ETDMH
EDMHT
EHMDT
EMHDT
ETHDM
EDMTH
EHMTD
EMHTD
ETHMD
EDTHM
EHTDM
EMTDH
ETMDH
EDTMH
EHTMD
EMTHD
ETMHD
HDEMT
HEDMT
HMDET
HTDEM
HDETM
HEDTM
HMDTE
HTDME
HDMET
HEMDT
HMEDT
HTEDM
HDMTE
HEMTD
HMETD
HTEMD
HDTEM
HETDM
HMTDE
HTMDE
HDTME
HETMD
HMTED
HTMED
MDEHT
MEDHT
MHDET
MTDEH
MDETH
MEDTH
MHDTE
MTDHE
MDHET
MEHDT
MHEDT
MTEDH
MDHTE
MEHTD
MHETD
MTEHD
MDTEH
METDH
MHTDE
MTHDE
MDTHE
METHD
MHTED
MTHED
TDEHM
TEDHM
THDEM
TMDEH
TDEMH
TEDMH
THDME
TMDEH
TDHEM
TEHDM
THECM
TMEDH
TDHME
TEHMD
THEMD
TMEHD
TDMEH
TEMDH
THMDE
TMHDE
TDMHE
TEMHD
THMED
TMHED
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 Reasoning
Section 11 Using Counting Principles, Permutations, and Combinations
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 digits 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 letters M, T, H, E, and D.
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 counting. 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 CBA
Some 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 arrangement 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).
Theory Versus Practice
The actual number of 3-letter English
words is less than the theoretical values 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.
Investigation 11.3 Counting Using the Multiplication Principle
Use 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 followed by four digits? (Examples: XYZ2005, SST7797, IAM8699)
90 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
Permutations
Several 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 example, 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.
more online
Go to http://www.regentsprep.org/
regents/math/algebra/APR2/indexAPR2.htm to learn more about
permutations.
Numbers Everywhere Splitting Hairs
Beard shaving is a multibillion dollar 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 lifetime 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).
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 Factorials
Using 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 outdoor 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 Cincinnati 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 Permutations
Tribune 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.
Copyright © 2015 by Gregory D. Foley, Thomas R. Butts, Stephen W. Phelps, and Daniel A. Showalter
Advanced Quantitative Reasoning 91
Part II Probability and Statistical Reasoning
Section 11 Using Counting Principles, Permutations, and Combinations
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
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 Students
At 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.
more online
Go to http://regentsprep.org/Regents/
math/algtrig/ATS5/indexATS5.htm
to learn more about permutations of n
objects taken r at a time.
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 permutations of the r objects is denoted by either nPr or P(n,r) and can be computed using
for integers 0 ≤ r ≤ n.
So, the solution to Example 2 could be denoted using nPr notation as
solved using the formula
18P4,
and
This formula is available on most calculators and handheld computers. See
Figure 11.3.
Figure 11.3 Permutations computed on a
handheld.
92 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
Combinations
In some counting problems, the arrangements or ordering of objects does not matter. 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 possible. This situation illustrates the idea of the number of combinations of 18 objects
taken 4 at a time.
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 combinations of the r objects is denoted by
more online
Go to http://regentsprep.org/Regents/
math/algtrig/ATS5/indexATS5.htm
to learn more about combinations and
how they are related to permutations.
or
and can be computed using
,
for integers 0 ≤ r ≤ n.
Example 3 Committee Conundrum
The number of possible 4-student committees from a class of 18 students is
See Figure 11.4.
Figure 11.4 Combinations computed on a
handheld.
Quick Question 11.7 First Five to Play
Each 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 Bridge
At 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 Subsets
The 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 equivalent to counting subsets of an n-set.
Example 4 Golfing Groups
The 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 district 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.
Copyright © 2015 by Gregory D. Foley, Thomas R. Butts, Stephen W. Phelps, and Daniel A. Showalter
Advanced Quantitative Reasoning 93
Part II Probability and Statistical Reasoning
Section 11 Using Counting Principles, Permutations, and Combinations
(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 11.5 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, appearing in his work The Key to Mathematics (Shi suo suan shu).
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.
Definition Recursive Formulas for the Number of Combinations of n Objects
Taken r at a Time
n
The entries in Pascal’s triangle are nCr = C(n,r) = C r = using
(a) (b) , and can be computed
for the integers n = 0, 1, 2, 3, …, and
for all integers n > r > 0.
more online
Exploration 11.9 Pascal Patterns
Go 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.
(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.
94 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
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 excluding elements, such as Example 5.
Example 5 That’s The Way … I Like It
The 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 Possibilities
At 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.
Do the calculation using the Probability menu of a calculator.
Write down only the answer, inserting commas for answers
greater than 1,000.
6.
8.17!
7.11!
9. 20P6
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)
4.Choosing classes. Senior year you must take English, mathematics, 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.
3.Breakfast options. The local BLD Restaurant serves breakfast, lunch, and dinner all day. All breakfasts come with
scrambled eggs, meat, potatoes, bread, and a drink.
• The science options are Astronomy, Biology, Chemistry,
Geology, Physics, and Planet Earth.
• 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?
• 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?
Copyright © 2015 by Gregory D. Foley, Thomas R. Butts, Stephen W. Phelps, and Daniel A. Showalter
Advanced Quantitative Reasoning 95
Part II Probability and Statistical Reasoning
Section 11 Using Counting Principles, Permutations, and Combinations
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 possible 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 arranged 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 PLANCHET be arranged?
16. How many distinct ways can the letters in the word
­SPRINGHEAD be arranged?
Factorial expressions. Calculate using nPr or nCr from the Probability menu of a calculator without using the factorial (!) command. Then check using factorials.
22.
24.
96 Advanced Quantitative Reasoning
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 Learned
29. 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 considered to be distinct. So, for example, AB1234 and ab1234
are ­different passwords. (Such systems are said to be casesensitive 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 members who are in the Democratic Caucus and 40 in the Republican 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?
19. How many distinct ways can the letters in the word
­INTERCHANGEABLE be arranged?
23.
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?
(b) In how many ways can a 10-member committee be chosen without regard to caucus?
18. How many distinct ways can the letters in the word
­COMMANDANT be arranged?
21.
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?
(a) In how many ways can a 10-member committee be chosen with 6 members from the Democratic Caucus and 4
from the Republican Caucus?
17. How many distinct ways can the letters in the word
­FILIGREE be arranged?
20. How many distinct ways can the letters in the word
­ENTOMOPHAGOUS be arranged?
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 arrangements of four such 13-card hands are possible?
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
Investigations
Looking Ahead
35. Orders of Magnitude for Factorials
37. On a roll. Two 6-sided dice are rolled; each has faces labeled 1
through 6.
(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?
(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?
(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
values for n = 1, 2, 3, …, 100.
of
(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?
Copyright © 2015 by Gregory D. Foley, Thomas R. Butts, Stephen W. Phelps, and Daniel A. Showalter
Advanced Quantitative Reasoning 97
Part II Probability and Statistical Reasoning
Section 12 Reasoning With Probability
Section 12
Reasoning With Probability
Main Ideas
•What Is Random?
•Probability Simulations
•Theoretical Probability
•Independent and Dependent
Events
•Empirical Probability
You likely have encountered some of the concepts of probability in earlier grades or
everyday experiences. Probability involves quantifying uncertainty, that is, assigning 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 Probabilities
What 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 financial risks must know a great deal of mathematics and deeply understand relevant
issues. For example, an actuary who sets rates for hurricane insurance must understand 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 making 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 happenings are neither haphazard nor unexpected. Random samples are used precisely because they yield predictable, reliable results.
98 Advanced Quantitative Reasoning
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Part II Probability and Statistical Reasoning
Section 12 Reasoning With Probability
In the next three investigations, we engage in a series of related probability experiments 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.
more online
Investigation 12.2 Imagined Data—Thought Experiments
Go 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.
(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 experiment, and place your results in a table.
Once again, count the strings of Hs and strings of Ts in your table. Overall, what is the longest string of either type?
Now you can test your intuition about random sequences of Hs and Ts against
actual data.
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 wholegroup discussion of the results. How does the class average for the genuine data compare to the thought-experiment data from Investigation 12.2?
Which is truly random? Explain.
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
Copyright © 2015 by Gregory D. Foley, Thomas R. Butts, Stephen W. Phelps, and Daniel A. Showalter
Advanced Quantitative Reasoning 99
Part II Probability and Statistical Reasoning
Section 12 Reasoning With Probability
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 contain 9 Hs and 11 Ts.
Figure 12.2 Pascal’s triangle revealing 20Cr for r = 0, 1, 2, 3, …, 20.
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 simulate 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.
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 concepts 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 Simulations
In 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 reshuffling 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.
100 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 12 Reasoning With Probability
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?
Theoretical Probability
The 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 Times
more online
Can 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 flipping a fair coin and flipping an unfair
coin.
In 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.
Copyright © 2015 by Gregory D. Foley, Thomas R. Butts, Stephen W. Phelps, and Daniel A. Showalter
Advanced Quantitative Reasoning 101
Part II Probability and Statistical Reasoning
Section 12 Reasoning With Probability
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 uniform 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?
(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.
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
Example 2 Further Flipping
In 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 Dice
Nicholas 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.
102 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 12 Reasoning With Probability
Exploration 12.5 Pick Three
Rachel wants to pick a 3-digit number at random. She decides to use the command “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.
Figure 12.4 Venn diagram of a sample space S
and 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
We now combine these rules with the results from Example 3.
Example 4 Rolling Two Fair Dice
Jennifer rolls a pair of fair dice.
(a) Because no outcomes total 13,
impossible event.
Figure 12.5 Venn diagram of a sample space S
and disjoint events E and F within S.
A total of 13 is an
(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 Events
Two events are independent events if the occurrence of one has no effect on the probability 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?
Copyright © 2015 by Gregory D. Foley, Thomas R. Butts, Stephen W. Phelps, and Daniel A. Showalter
Advanced Quantitative Reasoning 103
Part II Probability and Statistical Reasoning
Section 12 Reasoning With Probability
Multiplication Rule for Probabilities of Independent Events
Let E and F be independent events of a nonempty, finite sample space S. Then,
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 probability 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:
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 statements involving “and” (∧), “or” (∨),
and “implies” (⇒). If one claim implies 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⇒q
T
T
T
T
T
T
F
F
T
F
F
T
F
T
T
F
F
F
F
T
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 Events
If 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 effect of the probability of the occurrence of the other.
Notice that the rule
does apply:
104 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 12 Reasoning With Probability
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 obtaining 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.
The following formula is valid for both dependent and independent events.
Generalized Multiplication Principle for Probability
Let E and F be events of a nonempty, finite sample space S. Then,
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 Probability
It is possible to purchase trick dice that are loaded (internally weighted) so that certain 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.
Table 12.2 Results From Rolling a Loaded Die 1,000 Times
Result
One
Two
Three Four
Five
Frequency
19
81
84
79
91
Six
646
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.
Figure 12.6 Area models for the probabilities
of (a) getting a heads and (b) getting both a 6
and a heads.
more online
Go 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.
Exploration 12.6 Using Empirical Probabilities to Test Fairness
If 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.
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.
6.
7. 12P2
Copyright © 2015 by Gregory D. Foley, Thomas R. Butts, Stephen W. Phelps, and Daniel A. Showalter
8. 15P3
9. 12C2
10. 15C3
Advanced Quantitative Reasoning 105
Part II Probability and Statistical Reasoning
Section 12 Reasoning With Probability
Exercises for Section 12
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 possible 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 million 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?
Table 12.3
Populations (in Millions) by State, 2010 Census Data in
Order by Rank
CA
TX
NY
FL
IL
PA
OH
MI
GA
NC
37.3
25.1
19.4
18.8
12.8
12.7
11.5
9.88
9.69
9.54
NJ
VA
WA
MA
IN
AZ
TN
MO
MD
WI
8.79
8.00
6.72
6.55
6.48
6.39
6.35
5.99
5.77
5.69
MN
CO
AL
SC
LA
KY
OR
OK
CT
IA
5.30
5.03
4.78
4.63
4.53
4.34
3.83
3.75
3.57
3.05
MS
AR
KS
UT
NV
NM
WV
NE
ID
HI
2.97
2.92
2.85
2.76
2.70
2.06
1.85
1.83
1.57
1.36
ME
NH
RI
MT
DE
SD
AK
ND
VT
WY
1.33
1.32
1.05
0.989
0.898
0.814
0.710
0.673
0.626
0.564
Source: U.S. Census Bureau, as reported in The World Almanac and Book of
Facts 2012.
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?
106 Advanced Quantitative Reasoning
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 Problems
21. 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 Problems
29. 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 outcomes 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?
Copyright © 2015 by Gregory D. Foley, Thomas R. Butts, Stephen W. Phelps, and Daniel A. Showalter
Part II Probability and Statistical Reasoning
Section 12 Reasoning With Probability
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 performanceenhancing 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 Wisconsin 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 truenegative result? What is the probability of a true-negative
result given that the result is negative?
Looking Back
35. Hot pools. Some natural hot pools can be too hot for human
comfort and safety or too acidic or both. In fact, at Yellowstone 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
(B) T = 40°C and pH = 2.7
(C) T = 60°C and pH = 7.7
(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?
Investigations
39. Blood types. Over 40% of the world’s people live in China
(1.33 billion), India (1.17 billion) or the United States (307 million). 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 person has type O blood? Use a Venn diagram.
Table 12.4
Relative Frequency of Blood Types in the
World’s Three Most Populous Nations
Blood 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.
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 simulate 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 connectors, what type of married woman is the exercise referring to?
Copyright © 2015 by Gregory D. Foley, Thomas R. Butts, Stephen W. Phelps, and Daniel A. Showalter
Advanced Quantitative Reasoning 107