Download Determining the `reasonability of media`s statistics`

TWO
Probability
- If one learns by memory, and does not think, all remains dark.
Confucius
In the mid-1980’s, the ten states with the greatest longevity were Hawaii, Montana, Iowa, Utah,
North Dakota, Nebraska, Wisconsin, Kansas, Colorado, and Idaho. During the same time period,
these same states except for Hawaii, were the states that consumed the most animal fat per
human diet. Does this mean eating large quantities or even ridiculous amounts of animal fat
prolongs life? An inference such as this from the data seems ridiculous and for a dairy company
to propagate such lies to the American public would have all the indecency of two priests
swapping stories they heard from confessionals. So what would be a more appropriate
interpretation of the data? It also turned out that there was another occurrence in these same
states, again except for Hawaii; these same states had the highest rate of graduation from high
school. Thus, one could infer, the population’s demographics with regard to education may have
played a significant role in longevity. A greater percentage of people who had more advantages,
maybe better health insurance, maybe better living conditions on the whole contributed to greater
longevity in these states.
There is a fundamental difference between correlating a relationship, which just means drawing
a parallel between the states where people lived the longest and states the consumed the most
animal fat, versus inferring causation, deducing from the correlation that eating animal fat makes
one live longer.
It is said the ordinary chess player’s eyes roam over the entire board while the great grand
masters scan the few squares that really matter. Let’s examine our own vision. Again, in the
mid-1980’s there was a correlation between the thirteen states that were voted to be the most
scenic, WY, NV, CO, NM, MT, AZ, FL, OR, OK, CA, VT, AL, and WA according to the
Automobile Association of America (AAA) and those states whose residents commit the most
suicides. Should we conclude that beautiful scenery surrounding a neighborhood is a risky
venture because it increases the rate of suicide? Is there perhaps a more reasonable explanation?
Unlike the above data, the insight we need to see is not in the form of another set of hidden data.
Rather, it is in the interpretation of the data it self. Do you see it? If not, we will tell you at the
end of this opening section. Think about it. Scan the few squares that really matter.
Statistics you may not have heard.
 While 79 out of 1000 people are injured at home, it is a surprise that nearly 58 out of
1000 are injured at work.
 In this country, there are roughly two suicides for every murder.
Surprising, only because we tend to focus on injuries at home, or worry about our loved ones
being in mortal danger, but rarely do we fixate on injuries at work or the possibility that suicide
can take a loved one.
And why do we talk about so many people out of 1000? Because, raw numbers, when shown in
a comparative light, give you a perspective of the world around you. For example, we know
there are roughly 6.4 billion people living around the world, but since this number is too big to
comprehend we use proportions to visualize a smaller number to represent the whole, say 1000
people. These 1000 people are assumed to provide an average slice of the greater population,
preserving proportions to the world’s true demographic profile. This method is no different than
shrinking the earth’s age to week’s time or shrinking the sun down to the size of a soccer ball to
visualize the age of the earth or the vastness of space. Now, let’s examine the world around us
through these 1000 representatives of the earth’s human population.
By continent, there are
 570 Asians
 210 Europeans
 and just 60 North Americans
By religion, there are
 300 Christians
 175 Moslems, the fastest growing
religion on earth
 55 Buddhists
 33 Jews
Literacy,
 700 would be illiterate
By income,
 60 people would own half the
world’s wealth
Poverty,
 167 would earn less than 1 dollar per
day
Hunger,
 500 would go to sleep hungry.
- 40 -
On the surface of these numbers, things may look really awful. Well, not really. A slight
perspective change could show that as bad as things seem on a global level, it was far worse in
the past. For example, in the 1700’s half of the world’s population died before they were eight
years old.
You may be thinking, well, yes, while this is interesting in a trivial sort of way, maybe even eye
opening if I was to stretch it, but to be honest, so what? How can a ‘feel’ for numbers, basic
numeracy in a literate society be of help to me. Well, we all walk around in darkness on many
subjects, but with a smidge of numeric awareness, maybe every now and then a little light may
shine through.
For a moment, let’s just examine one of those conversations parents of teenagers must be
prepared to convincingly win. Their adorable child who they have protected from the dangers of
the cruel world says to them, “Why should I work hard in school, pops? I’m gonna be a
professional athlete after high school. My future is in the NBA (or NFL or Major League
Baseball or on the LPGA Tour… ).” Well, let’s reply in a way that would make even the
Huxtable’s proud by simply examining the dream through a numerical argument. The parent’s
reply, “Junior, what is the greatest number of high school athletes you believe could actually
make such a leap? Is it reasonable or even generous to assume that at most there are 50 players
per high school graduating class in the United States that can conceivably make the NBA by way
of the draft? With the influx of European players, non-senior NCAA athletes and some high
school ballers tossed into the NBA draft, this 50 is actually quite optimistic considering each
draft has 32 teams and only 3 rounds. Moreover, did you know only 1 out of every 50,000 high
school athletes go on to play a professional sport and this includes bowling. The odds are not in
your favor. Try altering your dream. Why don’t you aim for a Ph.D? Last year, there were
roughly 44,000 doctoral degrees conferred in this country. Do you think the odds against each of
those newly appointed Dr.’s were bordering on 1 in 50,000? That would mean for their high
school class, if we multiplied 44,000 by 50,000, 2.2 billion students in that high school class
must have set their sites on a Ph.D too. Does this seem reasonable? Of course not, this number
is absurd, it is 7 ½ times the size of our country, for goodness sakes. Set the Ph.D. in your sights.
It’s a more realistic dream. Pick a book. Study … “
Now, to formulate that argument, you would have to know some of the figures quoted. We
easily researched ours and found the “1 out of 50,000” portion of the argument was formed by
Art Young, the director of Urban Youth Sports at Northeastern University Center for the Study
of Sports in Society. And the 44,000 doctorals degrees was actually a modest adjustment of the
true figure, 44,1660, according to the National Center for Education Statistics (US Department of
Education). The point is, mathematics and the ability to be comfortable with numbers and the
use of logic with the ability to draw proper inferences, does not exist independently from us.
Rather, mathematical reasoning can enhance your life and with it sharpened, you’ll be a better
decision maker. The statements, symbols and methods we will study are not meaningless
statements governed by rules, but rather a way of viewing life around us. The trick here is to
understand how to work with numbers, and at the same time to not have the personality of an
accountant.
- 41 -
Back to the correlation between the thirteen most scenic states and the same thirteen state’s with
the highest rate of suicide. Is it reasonable to infer that nature’s beauty causes one to ponder
whether of not to end their own life? Of course not. But, it does seem reasonable to conclude
that a larger number of transient people move to those same scenic states seeking to start over,
picking scenic states as a place because the state offers nature’s comfort. Many are looking for
a new job, new life. Unfortunately, for a percentage of these people, their dreams don’t come
true.
Exercise Set
Ratios designed to help you understand the world around you and allow you to
make forecasts
1. Internet Project: The World Population
compressed to 100
If we could shrink the Earth's population to
a 100 people, with all the human ratios
remaining the same; what would this
collection of people look like? How many
would be …



















Asians
Europeans
From the western hemisphere (north
and south)
Africans
Female
Male
Non-white
White
Non-Christian
Christian
60 % of the entire world's wealth
would be in the hands of this many
people
from the United States
live in substandard housing
unable to read
suffer from malnutrition
be near death
be near birth
have a college education
own a computer
For problems 2 – 9, use the following data
from Table 1.
The following data can be found at
http://www.census.gov/ at the US Census Bureau web
site.
Table 1. Population of Selected States in the
years 1990, 2000 and 2003.
Fill in the remaining information for your
home state.
State
Florida
New York
California
Arizona
Your home
state
1990
12,941,197
17,987,163
29,816,592
3,665,228
2000
15,982,378
18,976,457
33,871,648
5,130,632
2003
17,019,068
19,190,115
35,484,453
5,580,811
2. Fill in the table below by finding the
numeric population growth (actual
difference in population) for each state for
the indicated time period.
State
Numerical
Growth
from ’90 – ‘00
Numerical
Growth
from ’00 – ‘03
Florida
New York
California
Arizona
Your home
state
- 42 -
6. Population Forecast for each state.
Predict the each state’s population for the
year 2010 based on the percent growth for
the indicated time period.
3. Fill in the table below by finding the
average annual population growth
(population growth per year) for each state
for the indicated time period.
State
Average Annual
Numeric
Growth from ’90
– ‘00
Average Annual
Numeric
Growth from 00 –
’03
Florida
New York
California
Arizona
Your home
state
4. Fill in the table below by finding the
percent growth (ratio of 2000 to the 1990
population or ratio of 2003 to the 2000
population respectively) for each state for
the indicated time period.
State
Percent Growth
from ’90 – ‘00
Percent Growth
from ’00 – ‘03
State
Predicted 2010
population
based on
Percentage of
Growth from ’90 –
‘00
Predicted 2010
population
based on
Percentage of
Growth from ’00 –
‘03
Florida
New York
California
Arizona
Your home
state
7. Why are the population projections for
the year 2010 larger when based on the
percent change?
8. Which of the four predictions, Annual
Numeric Growth from ’90 – ’00, Annual
Numeric Growth from ’90 – ’00, Percent
Growth from ’00 – ‘03 or Percent Growth
from ’00 – ’03, do you feel is the most
accurate? Why?
9. For which of the states is the projected
predictions the largest and why?
Florida
New York
California
Arizona
Your home state
5. Population Forecast for each state.
Predict the each state’s population for the
year 2010 based on the annual numeric
growth for the indicated time period.
State
Predicted 2010
population
based on Average
Annual
Numeric Growth
from ’90 – ‘00
Predicted 2010
population
based on Average
Annual
Numeric Growth
from ’00 – ‘03
Florida
New York
California
Arizona
Your home
state
- 43 -
Introduction to Probability, Coincidences
- I don’t know what’s the matter with people; they don’t learn by understanding; they learn by
some other way – by rote, or something. Their knowledge is so fragile! Richard P. Feyman, winner of
Noble Prize in Physics. 1985. “Surely You’re Joking, Mr. Feyman!”, WW Norton & Company, Inc.
I had a friend in college who had the world’s worst pick-up line. He would set his eyes on some
gorgeous co-ed and amble over to her. In a dead pan voice he would ask, “have you ever been to
Africa?” The co-ed, with eyes rolling, would reply, “no” and he would say, “what a coincidence,
never have I. See, we already have something in common.”
The fact that two events coincide means nothing. Situations where two events coincide gives us
the root of the word coincidence, right? Consider one’s parent who routinely has said over the
years “you and I were talking about so and so yesterday, and guess who I ran into today? So and
so. What’s the odds of that?” When one considers all the names spoken within a family each
and every day, all the tens or even hundreds of conversations they have each week, all the
hundreds or even thousands of conversations they have each month, sooner or later, one of the
people discussed, on one of the many days, weeks, months, and years they have been together,
would show up somewhere in the presence one of the family members. No one should be
stunned by this occurrence. Life can sometimes seem like photographs in our minds and we crop
away the edges of anything we do not want to see. What is construed as a coincidence with some
special meaning assigned to it is nothing more than basic probability presenting itself. As was
the college pick up line. “Well, what are the probability she has been to Africa,” he would say
after his personalized probability lecture with respect to being so cheesy. “At least I have a can’t
miss opening.” Can’t miss, no. But, it is more likely than not that the unknowing woman will
have that ‘something uncommon in common’.
We are all familiar with the bizarre coincidences between the Kennedy and Lincoln presidencies,
and these Kennedy/Lincoln coincidences certainly make interesting reading. They captivate
many filling chat rooms and web pages. Let’s pick one of the chat rooms and glance at the
coincidences. We choose snopes.com because they choose to discuss the coincidences in a
manner befitting this discussion; they claim to have analyzed each coincidence thoroughly. Source:
http://www.snopes.com/history/american/linckenn.htm









Abraham Lincoln was elected to congress in 1846. John F. Kennedy was elected to congress in 1946.
Abraham Lincoln was elected president in 1860. John F. Kennedy was elected President in 1960
Lincoln's secretary was named Kennedy. Kennedy's secretary was named Lincoln.
Andrew Johnson, who succeeded Lincoln, was born in 1808. Lyndon Johnson, who succeeded Kennedy,
was born in 1908.
John Wilkes Booth, who assassinated Lincoln, was born in 1839. Lee Harvey Oswald, who assassinated
Kennedy was born in 1939.
Booth ran from the theater and was caught in a warehouse.
Oswald ran from a warehouse and was caught in a theater.
Booth and Oswald were assassinated before their trials.
The names Lincoln and Kennedy each contain seven letters.
- 44 -









Both names contain fifteen letters
Both wives lost their children while living in the White House.
Both Presidents were shot on a Friday.
Both were shot in the head.
Both were assassinated by southerners.
Both were succeeded by southerners.
Both successors were named Johnson.
Both assassins were known by three names.
Both were particularly concerned with civil rights.
Hain’t we got all the fools in town on our side? And hain’t that a big enough majority in any town?
- Mark Twain, Huckleberry Finn
Conspiracy Theories or Coincidences What do all these seemingly unrelated coincidences
mean? Anything? we say, “no.” Any two particular events have an extremely high number of
associated facts, events and meanings associated with them. The fact that some of these
occurrences appear to overlap or coincide should not surprise us. It is just a reflection of the
multiplication principle discussed in the previous chapter. The events that coincide are as the
term suggest, coincidences, not anything more. They do not conspire something dark and
dangerous, they are not conspiracies.
Take this example. Two people meet on a plane, they find they have a common acquaintance.
So what? I know 100 people, who know 100 people, who also know 100 people means
100x100x100 = 102 x 102 x 102  106  1,000,000 acquaintances. She knows 100
people who each know 100 people who also each know 100 people. And one person is an
acquaintance of two people sitting next to each other on the plane. Big deal. It’s just a
consequence of the multiplication principle. It does not mean there exists a special significance
between the two passangers.
Don’t believe it? Take any two seemingly unrelated events, ‘google’ the topics on the Internet,
and you will uncover just as many “coincidences” or “what’s the odds of that” as the
Kennedy/Lincoln coincidences. You try it. Start benignly. Pick any two unconnected
presidents, Millard Fillmore and Harry Truman or Grover Cleveland and James Buchanan.
Google. Then spin your own conspiracy theories about two presidents. Write a book, if you
want. We often thought a good book could be written if we employed the same notion with the
vice-presidents. Common thread or little known conspiracies about those really in charge.
From John Adams to Dick Cheney. Certainly, Dan Quayle and Levi Morton must have had
inconsequential commonalities that may be cleverly exploited into something more.
Enter these three seemingly unrelated items into google.com: December 5th,, boats, sinking. And
then search. And search. And search. The result: You could find that on December 5th, 1664, a
ship off north Wales with 81 passengers aboard, sank. There was one survivor. A man named
Hugh Williams. On the same date, December 5th, 1785, a ship sank with 60 passengers aboard.
There was one survivor. A man named Hugh Williams. On December 5th, 1860, a ship with 25
passengers abroad sank. There was one survivor. A man named Hugh Williams. Going into
this venture, we didn’t know the coincidence we would find, but we believed a coincidence
could be put together. By only telling you the three facts by themselves, we could easily extol
- 45 -
this coincidence as something meaningful. And if the point has yet to be made, coincidences are
not meaningful.
Numerology Closely associated with the notion of coincidence is the pseudo science
numerology. Through the dark annals of history, people have made veiled attempts at projecting
importance in the numerical coincidences that may be nothing more than the result of a simple
fact; seemingly rare events happen everyday. Collecting data for a single day from history and
assigning some hidden importance to random occurrences is more than likely meaningless.
Rather, what is more astounding is that someone actually took the time to tie together unrelated
facts.
Let’s look at September 11th, 2001. The tragic day will forever be seared into the very
conscience of the American people with images never to be forgotten. The death toll associated
with the terrorist attacks that day exceeded 3,000. Now, let’s examine this date through out
history:












September 11, 3 B.C. The Birth of Christ. In a book written in 1981 called "The Birth
of Christ Recalculated", the author, Dr. Ernest L. Martin, claims this as the date of the
birth of Jesus Christ's. (Yes, we know it is controversial.)
Sept. 11th , 1297. The Battle of Stirling Bridge that was portrayed in the film Braveheart
took place
September 11, 1609. The explorer Henry Hudson discovers Manhattan Island, the
location of the World Trade Center Towers.
Sept 11th, 1709. The first European War of the modern Era, The Spanish War of
Succession began. And on September 11, 1714, the war ends.
Sept 11th, 1772. The last battle of the American Revolution took place.
September 11, 1922. A British memorandum is approved for the state of Palestine,
securing a promise to create a Jewish national home. There was vast objection voiced
through out the Arab world.
September 11, 1941. Construction of the Pentagon officially starts, 60 years to the day
before the Sept 11th attacks would destroy one side.
September 11, 1944. The US Army crosses the border into German territory for the
first time during WWII.
September 11, 1972. The ending of the 1972 Munich Olympics, the Olympics which
introduced the world to terrorism. There were 121 participating countries (11x11=121),
and 11 Israeli athletes were killed.
September 11th, 1990. Then U.S. President George Bush addresses Congress on the
Persian Gulf crisis, vowing that "Sadam Hussein will fail" in his takeover of Kuwait.
September 11, 1992. Hurricane Iniki, the most powerful hurricane to strike the state of
Hawaii, causes six deaths.
September 11, 1999. The Jewish calendar is 6,000 years old on this date.
Source: napsterites web site http://www.p2p-zone.com/underground/showthread.php?t=20564 ,
September11thNews.com http://www.september11news.com/Sept11History.htm , and http://en.wikipedia.org/wiki/British_Mandate_of_Palestine
For some reason far too many of us are compelled to find significance when presented a string of
facts that appear to be related. Sadly, in reality, it is random coincidences that more often than
- 46 -
not govern our lives. Unfortunately, these numerological assumptions are more fascinating to us
and implications that much in life is random is probably a little scary to most.
Sports Every four years, the most important football game of the year is played by the
Washington Redskins amid the District of Columbia’s changing fall colors of late October or
early November. For the National Football League, there were times when this game was of
consequence, but usually for the NFL, due to the too often lean years of the Redskins, this game
was meaningless. But, for our nation, ‘the game’ dictated more than just the direction of two
football teams; it directed the direction of our nation for the next four years. Since 1936, in an
election year, if the Washington Redskins won their last home game prior to the presidential
election, the incumbent party won the election. If they lost, the challenger’s party won. The
streak began in 1933, when the Boston Braves were renamed the Redskins. Since then,
beginning with a Redskins home win prior to the FDR versus Alf Landon presidential race and
the consequent re-election of Franklin Roosevelt in 1936, the trend has held. In 2000, the
Redskins lost to the Tennessee Titans, and the incumbent’s party was ousted as George W. Bush
beat Al Gore. In 2004, the Redskins lost to the Packers 28-14. Correlation verses causation?
Political If you want to ingratiate someone to your way of thinking, then exploit the highly
known fact that most people are not comfortable with numbers. Politicians, spin doctors and the
like have been doing this for years. Just assign significance to a situation by extolling raw
numbers and leave out minor details such as the likelihood said situation would occur in the first
place.
Democratic US Representative from Massachusetts Edward Markey voiced a frightening
statement recently when he said, “This is a rapidly growing problem that will soar out of control
if the industry does not wake up to its responsibility … “ What was he referring to, what was the
problem that could spin out of control if something is not done by the highest echelon of the
industry? His statement was in response to these recently publicized raw numbers: in 2000,
there were 6,594 injuries at fixed site amusement parks, and there were 15 fatalities or serious
head injuries in the past ten years.
What we are about to say in no way minimizes the seriousness of this type of injury. In fact, just
the opposite. It serves to give us, the public, a better understanding of this issue. Articles
extolling this fear, with disturbing headlines like “Do breakneck speeds and high-g turns push
thrills to a new lethal level” are deceptively alarming. Yes, new roller coasters these days are
bigger, faster, wilder and have incredible ground-breaking latest up to the minute modernized
new twists and turns. But, to imply the industry is negligent in its attempt to secure the safety of
its passengers is outright misleading. The fact is when published alone, the number
6,594
stands out in a frightening way. But, let’s examine this number more closely. Roughly
317,000,000 or over 300 million people visit fixed site amusement parks in a given year. Now,
let’s compare this number to those injured each year in other recreational activities: 82,722
people are injured on trampolines, 62,812 are injured in swimming pools, 544,561 are injured on
bicycles, 20,000 are injured at music concerts, and 200,000 are injured in preschool and
elementary school playgrounds.
- 47 -
Why do people fear amusement parks so much? Why don’t they fear a trampoline, a bicycle or a
playground? Sure, maybe there are deep psychological reasons, like when one bounces on a
trampoline or rides a bike, they are in control and when you go on a ride in an amusement park,
you strap yourself into a ride, thus giving up your control over the motion. But, to be honest,
headlines extolling raw numbers, such as 6594 people injured each year with follow ups from
politicians are notoriously common instruments to instill public panic or at the least, public
apprehension.
We see alarming headlines all of the time associated with sensitive issues. Memorial Day
weekend, 2003. The headlines read, “Holiday weekend traffic death soar to nearly 500
nationwide.” Should you be alarmed? Reluctant to drive next weekend? 29 million people
traveled by car more than 50 miles from their homes that weekend.
Unfortunately, the public places too much emphasis on small numbers singled out by alarmists
and does not recognize the significance of large raw numbers that are quite significant. Consider
the state of Massachusetts, where US Representative Markey hails. In 2002, in Boston alone,
109,128 people live in poverty, up 7,036 from the 102,092 people in the city ten years earlier.
Massachusetts has 600,000 people living in poverty. 600,000 people is nearly one in ten people
from the good commonwealth. In Boston, 28,928 children live in poverty and statewide,
240,000 children are poor. These numbers pale compared to Hollywood’s dilemma. In Los
Angeles county alone, where entertainment routinely makes the headlines, 1,679,000 men,
woman and children are poor. 640,000 of these people are children. Where are the alarming
headlines here? Where is the public outcry? Is the issue of mass poverty, where some
35,800,000 United States citizens live and function below the poverty level too big to confront.
Health care, education, homeless, and unemployment are too complicated to tackle. Headlines?
We just don’t see them.
Take a lighter issue, where the numbers and the implications are largely ignored by our society.
In 2002, there were 288,368,698 people in the United States. Roughly one in eight were 65
years or older. By 2030, it is estimated that one in five will be ‘elderly’. Moreover, the fastest
growing segment of our population is the oldest of the old, the 85 and older crowd. In 2010,
there are projected to be 6 million, which is double the 1990 number of 85 and older. By 2050,
the number is projected to rise to 21 million.
- 48 -
Exercise Set
1. Answer this question, what impact does
the demographic statistics stated in the last
paragraph of this section concerning the
‘elderly’ have on politicians, policy makers
and planners. Start with the notion that the
people from this segment of the population
are more likely to be disabled than the
young elderly, from 65 to 85. Attack the
question by discussing the effects on
insurance, care, hospitals, voting alignments
and so on. To support your argument use
numbers. Research and gather data and then
develop a numerical argument to advocate
your position.
For questions 2 and 3 use the data below.
Taken from the US Census Bureau: http://www.census.gov/PressRelease/www/releases/archives/facts_for_features/001676.html
May 17, 1954 marked the U.S. Supreme
Court decision that ruled unanimously
segregation of public schools “solely on the
basis of race” denied African American
children equal educational opportunity, even
though “physical facilities and other
‘tangible’ factors may have been equal.” To
honor that landmark decision, the US
Census Bureau assembled the following data
below in the “*Special Edition* Brown v.
Board of Education: 50th Anniversary on
the educational attainment and school
enrollment of African Americans — then
and now.”
Enrollment: 1954 to 2002. Sixty nine
percent of African American children ages
5 and 6 were enrolled in school in 1954. By
2002, enrollment for African American
children of those ages had risen to 96
percent.
Twenty four percent of young, African
American adults ages 18 and 19 were
enrolled in school in 1954. In 2002,
enrollments of this age group increased to
58 percent.
High School Graduates: 1952 to 2002.
Fifteen percent of African Americans age
25 and over in 1952 were high school
graduates. By 2002, this percentage had
risen to 79 percent.
The raw numbers associated with these
percentages are 1.6 million African
American, 25 years of age and older, had
earned a high school diploma in 1957. This
number had risen to 16 million by in the
year 2002.
College Graduates: 1952 to 2002. Two
percent of African Americans age 25 and
over in 1952 were college graduates. By
2002, this percentage had risen to17 percent.
In 1957, 252,000 African Americans had
earned at least a bachelor’s degree. In 2002,
3.5 million African Americans had achieved
a bachelor’s degree.
Students: 1955 to 2002. In 1955, 4.5
million African Americans were enrolled in
schools (nursery through college). This
number had risen to 11.7 million by 2002.
There were 155,000 African American
college students in 1955 and by 2002, this
number had risen to 2.3 million.
There were 926,000 African American high
school students in 1955 and in 2002, this
number had rose to 2.6 million.
2. Causation. Certainly, dramatic social
changes occurred from 1955 to 2002 and are
reflected in the numbers above. No rational
person could construe these changes as
coincidences. The statistics reflect
causation. Construct an argument using the
above numbers provided that addresses a
“cause and effect relationship” for these
increases. In your argument, try to account
for as many of the statistics presented above.
Also, include any statistic or social change
you are presently aware of to support your
argument.
- 49 -
3. Causation. What impact did the changes
in the above demographic statistics have on
society over the past 50 years? Attack the
question by imagining you are currently a 75
year old who witnessed the changes first
hand. Write a narrative discussing the
perceived effects on education, finance,
business and entertainment/the arts.
Research and gather data and then develop a
numerical argument to advocate and support
your position.
4.
Taken from the US Census Bureau
http://www.census.gov/PressRelease/www/releases/archives/facts_for_features/001702.html
Is it difficult for you to imagine life without
a color television? March 25, 2004 marked
the 50th year anniversary of color TV
emergence in American society. On March
25, 1954, the Radio Corporation of America
began to manufacture color television sets at
its Bloomington, Ind., plant. Roughly 5,000
model CT-100 color receiver’s were built
11 whose parents in 2000 imposed at least
one rule for watching TV, such as types of
programs watched, how early or late the
children could watch and the number of
hours watched. The percentage dropped to
73 percent for children ages 12 to 17.
$255.18 - The projected spending per person
for cable and satellite TV in 2004.
$34.71 - The estimated average monthly rate
for cable TV in 2002. 154,000 - The
number of people employed in the
manufacture of television, radio and wireless
communications equipment in the United
States in 2001. 21,724 - The number of
stores that primarily sold televisions and
other electronic equipment in 2001.
$11.7 billion - The annual payroll for the
245,000 employees of 6,692 cable TV
networks and program distribution firms in
the United States in 2001. 1,937 - The
number of television broadcasting networks
and they retailed for $1,000 each. The US
Census Bureau assembled, as seen below, a
sampling of statistics from its publications
about television and the television industry
to celebrate this anniversary.
248 million - The number of television sets
in U.S. households in 2001.
98.2% - The percentage of households with
at least one TV in 2001. (Compared to
87.3%, the percentage of households with at
least one TV in 1960: From the 1960 census)
2.4 - The average number of televisions per
home in 2001.
1,669 - The projected number of hours that
adults (age 18 and older) will watch
television in 2004. This is the equivalent of
about 70 days.
94.3% - The percentage of people age 18
and over who said they watched television
in the spring of 2002. Older Americans (age
65 and over) were more likely to be glued to
the tube (97 percent) than any other age
group.
92% - The percentage of children ages 6 to
and stations in the United States in 2001.
31,235 - The number of people working
behind the lens as television, video and
motion picture camera operators and editors,
according to Census 2000. $10.7 billion The payments by television broadcasting
firms for broadcast rights and music license
fees in 2001. Such payments constitute the
biggest expense of TV broadcasters. The
next highest expense was the annual payroll,
$6.5 billion. $41.8 billion - Amount spent
on television advertising in 2002, up from
$38.9 billion in 2001. Research several of
the above statistics for the year 1990 (see
US Census Bureau http://www.census.gov/).
What differences do you observe between
the years 1990 and 2000? Which categories
changed the most? The least? Based on this
information, had you been an investor with a
crystal ball in 1990, which companies would
- 50 -
have been a good choice for investments for
the upcoming decade of the 90’s.
The following are projects that may be
assigned as “on-line projects, projects to
involve research on-line.”
5. Coincidences: Presidential Births and
Deaths Considering coincidences among
the presidents, we turn our attention to the
birth and death dates of American
presidents. Research the forty-one
presidential births, and thirty-six presidential
deaths. Were any presidents born and did
any presidents die on the same day of the
year? (Note: Later in this section, we will
examine if these common days were indeed
coincidences.)
8. Coincidence in Literature. Research
the following: In 1898, Morgan Robertson
wrote a book entitled Futility. It described
the maiden voyage of a transatlantic luxury
liner named the Titan. Although it was
flaunted as being unsinkable, it collides with
an ice burg and sinks. Of the 3,000
passengers, many perish. The boat had 24
lifeboats.
When did the Titanic sink? Was it touted as
a transatlantic luxury liner that was
unsinkable? Did it strike an ice berg? How
many passengers were on the Titanic? How
many lifeboats?
6. Presidential Coincidences Aside from
the aforementioned presidents Lincoln and
Kennedy, two of the other presidents who
were shot has coincidences associated with
their presidencies; Research James Garfield
and Ronald Reagan. Find as many
coincidences as you possibly can about these
presidents as well as their assassination
attempts.
One last note on this topic. When
interacting with numbers presented by the
media, it should be noted that an all too
common political ploy is to publicize either
percentages of the population or rates of
change instead of the raw data. Many times
the “fear” is that the numbers themselves are
so large that the public would become
distressed. For example, when the AIDS
epidemic swept the nation in the 1980’s and
1990’s, television news commentators
nightly quoted how fast the AIDS epidemic
was spreading instead of informing us as to
how many people were inflicted with the
virus. This was done to soften public
outcry, the actual numbers were thought to
frightening for the general public. Checkout
your own level of numeracy: When viewing
nightly news or reading the daily paper or
other media sources, be on the look out for
rates of change, percents, or numbers that
may be spun so that the presentation of the
facts are not as alarming as they first
appear.
7. Coincidence in Literature. Research
the following: In 1883, Edgar Allen Poe’s
book, The Arthur Gordon Pym of
Naugatuck, had a sequence where four men
are adrift in a boat, and they kill and eat the
cabin boy Richard Parker. When in the
future did this sequence of events actually
occur?
- 51 -
Probability
Poverty and Probability On Thursday, August 26, 2004, CNN.Com reported “Census: More
Americans living in poverty. WASHINGTON (AP) -- The number of Americans living in
poverty increased by 1.3 million last year, while the ranks of the uninsured swelled by 1.4
million, the Census Bureau reported Thursday. It was the third straight annual increase for both
categories. While not unexpected, it was a double dose of bad economic news during a tight reelection campaign for President Bush. Approximately 35.8 million people lived below the
poverty line in 2003, or about 12.5 percent of the population.”
Poverty, in case you are interested, was defined to be a household income under 18,400 dollars per year for
a family of 4, a household income under 15, 260 dollars per year for a family of 3, a household income under
12,120 dollars per year for a family of 2 and under 8,980 dollars per year for a person living alone. Do you
think this is a substantial number of people who live in poverty? What is the probability that
someone you meet lives in poverty? Well, if there are 293 million US citizens in August, 2004,
and 35.8 million live beneath the poverty level, then the ratio or percent of the population who
35,800,000
live in poverty is
 0.122 or roughly 1 in 8 Americans live in poverty. Because
293,000,000
people tend to socialize with others within their own economic status level; within their own
group is their perceived average slice of reality; this does not automatically imply one in eight
people you meet will live in poverty.
Probability is used to shine light on ‘the other side of the argument.’ For example, let’s look at
both sides of an argument using the same exact statistics. In 1990, there were 33.4 million
people who lived beneath the poverty level. It would be obvious to point out that between 1990
and 2003, 2 ½ million more US citizens lived beneath the poverty level. But, in 1990, the
population of the United States was roughly 248 million. So, the probability you will meet
33, 400, 000
someone who lives in poverty is
 0.135 . Between 1990 and 2003, the number of
248, 000, 000
people living on poverty increased by 2 ½ million, while the percent of US citizens living on
poverty dropped from 135 per 1000 to 122 per 1000.
The probability an event E is simply a measure of the likelihood event E will occur.
Intuitively, people have a “feel” for the concept of the likelihood that an event will occur. For
example, if we ask you what the probability is for you to obtain a heads up if you were to toss a
coin, you know it is just as likely to toss a head than to toss a tail. And you probably know the
probability associated with a heads up is ½. In mathematical terms, we call the tossing of the
coin the event and we call the result, be it heads or tails, the outcome.
The probability an event E will occur is
the number of desired outcomes divided
by the total possible number of outcomes.
* The outcomes need to be equally likely to occur.
- 52 -
P( E ) 
Number of desired outcomes
Total number of outcomes
For example, if you roll a die, the total number of outcomes is 6 because you may roll a 1, 2, 3,
4, 5 or 6. The set of all outcomes comprise what is called the sample space. If we ask what is
the probability of rolling a two, the event E is the rolling of the die, the number of desired
outcomes is 1, specifically, it is the desire to roll a two. We say the probability of rolling a two is
1/6, we write P(E) = P(roll a 2) = 1/6. Based on this thought process, we have P(2 or more) = 5/6
because there are the same 6 possible outcomes and 5 of these 6 outcomes, rolling a 2, 3, 4, 5 or
6 are desired.
Since we are using the language of probabilities, we interpret the meaning of these ratios as
follows. If you rolled a single die six times, on average you would see the two appear exactly
once. Is this exactly once each time you roll the dice six times? Possibly. Possibly not. If you
rolled a die a large number of times, say 1000, a good guess would be that about 160 to 170 rolls
would be recognized as a ‘roll of a two’. Try it. This is because the theoretical probability of
rolling a two is 1 chance in 6, which is 1 divided by 6, or 0.1666, but this is only what is
expected to happen. It is not guaranteed. So, when you really roll the die over and over again,
anything can happen. The result of any single roll is not predictable. But, the more you role, the
closer you get to the theoretical “1 out of 6 times” of rolling the two. Maybe after ten rolls you
might not get a two at all, but after two hundred rolls, our guess is that you will probably see the
two 0.166 x 200 or approximately 33 times. Roll the dice a thousand times and obtain a two
0.166 x 1000 166.7 , or around the interval of 160 to 170 times.
Example One
When rolling a single die, find the following probabilities:
a) P(4 or less) b) P(an even number) c) P(a number less than 11) d) P(a number greater than
11)
Solution
a) The probability of rolling a 4 or less is 4/6 or 2/3. There are 4 desired outcomes and they are
rolling a 1, 2, 3 or 4. The sample space consists of the same 6 possible outcomes, 1, 2, 3, 4, 5 or
6 as discussed above.
b) The probability of rolling an even number is 3/6 or 1/2. There are 3 successful outcomes and
they are rolling a 2, 4 or 6. Again, the sample space consists of the same 6 possible outcomes as
discussed above.
c) The probability of rolling a number less than 11 is 6/6 or 1. All 6 possible outcomes from the
sample space are numbers less than an 11 and thus would be a desired outcome. In other words,
every time you roll a single die, you will see a number less than 11. The probability of 1
implies the desired event is certain to occur.
d) The probability of rolling a number greater than 11 is 0/6 or 0. None of the 6 possible
outcomes from the sample space would be a success because none are greater than an 11. If you
- 53 -
roll a single die, you will never see a number greater than 11. A probability of zero means
your set of desired outcomes will never occur.
Numerically speaking, the range of numbers for the probability of any desired event to occur
spans from the smallest number, which is 0 for an impossible event to the largest number,
which is 1 for an event that will definitely occur. All probabilities are numbers between 0 and
1. The likelihood of an event occurring has a probability closer to 1 if it is more likely to happen
and closer to 0 if it is less likely to happen.
Once again, working with P(E) = P(roll a two) = 1/6 we can quickly determine the probability
__
__
for the compliment of E, written P( E ). P( E ) is the probability event E will not occur. We
have the event E is to “roll a two” and the compliment set is found by removing all the elements
of event E from the sample space, S. Using set notation, we have S\E = E . The probability
associated with E can be found by subtracting P(E) from P(S). We know P(S) = 1. Why?
__
P( E ) = P(roll a 1, 2, 3, 4, 5 or 6} = P(S) – P(roll a two) = 1 – 1/6 = 5/6. The probability of
rolling a number is not a 2 is 5/6. Let’s look at this from an alternative point of view. The basic
Number of desired outcomes
premise of determining a probability is to use the ratio of
.
Total number of outcomes
In our example, n( E ) = 5 and n(S) = 6. Thus, P( E ) =
n( E ) 5
 .
n( S ) 6
There are two basic strategies in determining a probability for an event E. We can apply the
ratio of the number of desired outcomes divided by total possible number of outcomes. Or if we
know the probability of an event E, we can find the probability of E compliment by subtracting
P(E) from 1. Often we employ the compliment strategy when the number of desired possible
outcomes is difficult and seemingly impossible to calculate. We will see this technique applied
often in our discussion of probabilities.
Example Two
Question - Is race is an important “filter” on who goes to prison? On New Year’s Eve of
2001, there were 1,955,705 people in U.S. prisons and jails. The population of the country that
year was approximately 275 million people.
a) To what degree does the US incarcerate it’s citizens?
Solution: The probability you were incarcerated was 1,955,705
275,000,000
or
711
 0.007 . This ratio implies nearly 7 out of every 1,000 Americans were
100,000
incarcerated in 2001.
In 2001, the United States population was comprised of the following proportions. There were
approximately 211 million Caucasians and 500,000 were incarcerated. There were
approximately 34 ½ million African Americans and 625,000 were incarcerated. There were
- 54 -
approximately 35 million Latin Americans and 213,000 were incarcerated, among the 10 million
Asians 9,900 were incarcerated and from the 2 ½ million Native Americans, 17,700 were
incarnated. Note: Approximately 600,000 of those incarcerated did not readily fit in any of the
above ethnic groups.
b) When you break down the statistics, is incarceration is an equal opportunity
punishment? For each ethnic group, find the proportion of these citizens who were
incarcerated. Write the result in terms of so many out of 100,000 US citizens.
In 2001, the probability you were incarcerated if you were Caucasian was
500, 000
 0.00237 or nearly 237 per 100,000 Caucasians. That same year, the
211, 000, 000
probability you were incarcerated if you were African American was
625, 000
 0.01812 or nearly 1812 per 100,000 African Americans. Again that same
34,500, 000
 0.00609 or
year, the probability you were incarcerated if you were Latino 213, 000
35, 000, 000
nearly 609 per 100,000 Latin Americans, the probability you were incarcerated if you were
 0.00099 or 99 per 100,000 Asians and the probability you were
Asian was 9,900
10, 000, 000
 0.00708 or nearly 708 per
incarcerated if you were Native American was 17, 700
2,500, 000
100,000 Native Americans.
The inferences you draw are based on your belief system, which has been evolving since you
were born. In the US in 2001, there were 211 million Caucasians and only 34 ½ million African
Americans. This means there were more than six times the number of Caucasians than African
Americans sprinkled through out the US population that year. Yet, there were more African
American inmates incarnated, 625,000, than Caucasian inmates, 500,000. Moreover, the rate per
100,000 within each respective ethnic group was over seven times greater in the African
American population compared to the Caucasian population. We’ll let you continue along this
train of thought as you consider ratios involving the other ethnic groups.
- 55 -
Determining the ‘reasonability of media’s statistics’
Remember numbers are not controversial, rather it is one’s interpretations that may be construed
as controversial. Let’s examine one case in point. We will use our probability knowledge mixed
with a little common sense to analyze a random media publication taken from the following
source: http://www.westernprisonproject.org/Publications/Factsheets/Prison_Index_Incarceration_Fact_Sheet.pdf.
The Prison Index
The Prison Index: Taking the Pulse of the Crime Control Industry is the first index of statistics about our nation’s
criminal justice system ever published. Containing 611 facts and 17 graphs and charts, this 48page volume presents, in black-and-white, the state of crime control in America. Below is just a sample of some of
the insightful and useful information, in easy to read index form.
Get the real facts about crime and punishment in the US today!
Incarceration is not an equal opportunity punishment.
On December 31, 2001, there were 1,955,705 people in U.S. prisons and jails.
As of December 31, 2001, the U.S. incarceration rate was 709 per 100,000 residents (7 out of every 1,000
Americans). But when you break down the statistics you see that incarceration is not an equal opportunity
punishment.
Select U.S. incarceration rates:
Whites: 235 per 100,000
Blacks: 1815 per 100,000
Latinos: 609 per 100,000
Asian: 99 per 100,000
Native: 709 per 100,000
Race is an important “filter” on who
goes to prison:
Males: 1,318 per 100,000
White males: 708 per 100,000
Black males: 4,848 per 100,000
Hispanic males: 1,668 per 100,000
Break it down by age and race, and you can see what is going on even clearer:
For Black males ages 25-29: 13,391 per 100,000. (That’s 13.4% of Black men in their late 20s)
Or you can make some international comparisons:
South Africa was internationally condemned for its racial policies under apartheid.
South Africa under apartheid (1993), incarceration rate for Black adult men: 851 per 100,000
U.S. under George Bush (2001), incarceration rate for Black adult men: 4,848 per 100,000
Select U.S. Incarceration rates from Bureau of Justice Statistics, Prison and Jail Inmates at Midyear 2000; Gender and Age &
Race statistics from BJS, Prison and Jail Inmates at Midyear 2001, Tables 1 and 15; Calculation for Black adult men uses data
from Tables 14 and 15 of the spreadsheet version of Prison and Jail Inmates at Midyear 2001 to count only Black men 18 and
older; Statistics as of December 31, 2001 from BJS, Prisoners in 2001. South Africa figures from Sentencing Project, Americans
Behind Bars: The International Use of Incarceration. The incarceration rates by race alone were calculated by Mother Jones for
the Prisons: Debt to Society, Racial Inequality page.
A Joint Project of The Western Prison Project and the Prison Policy Initiative
Western Prison Project POB 40085 Portland, OR 97240
Prison Policy Initiative POB 20038 Cincinnati, OH 45202
- 56 -
Is the above web page alarmist? Is the data presented on the page accurate? Are these different
questions? How were these numbers derived? What about the issue of coincidence versus
correlation (cause and effect)? Where do we start in our quest to place the information presented
above into some reasonable context within our mind?
We will confine our analysis to the International Comparisons made in the last paragraph, simply
because they appear to be the most “alarming” portion of the flyer. The flyer states, South
Africa was internationally condemned for its racial policies under apartheid. South Africa, under
apartheid (1993), had an incarceration rate for Black adult men of 851 per 100,000 and in the
US, under George W. Bush (2001) the incarceration rate for black adult men was 7,226 per
100,000.
We must ask ourselves before we do any serious number crunching, is the adult black man from
South Africa defined to be from the same age group as the African American male from the
United States. In other words, is the publication comparing equivalent demographics? If the
answer is yes, we continue. If not, the comparison is not justified. If we assume the answer is
in the affirmative, how or where do we start to analyze the argument?
We will need to make some assumptions in investigating the given data. As we examine the
statement “7,226 per 100,000” in terms of reasonableness, does this statistic even seem
reasonable? There are several ways to determine the reasonability of this statement. One
obvious route would be to track down the original source and verify the data. An alternative
would be to use our own numerical literacy and evaluate the reasonableness of the data our
selves. Back to our assumptions, is it reasonable to assume there are more male than female
African Americans incarcerated? Let’s say (for argument sake), 80 % of the African Americans
incarcerated are male. Since there are 625,000 African Americans incarcerated, 80 % of those is
0.80 x 625,000 = 500,000 or ½ a million African American males. Now, we test the statement
“7,226 per 100,000” or roughly 7.2 % of the African American males are incarcerated. Applying
some basic algebra, we have the equation, 0.072 x Number of adult African American males =
500,000. If we divide both sides of the equation by 0.072, we will identify the approximate
number of adult African American males in the U.S. After dividing, we have the Number of
adult African American males is approximately 7 million. This means there were nearly 7
million adult African American males in the United States in 2001. Does this seem reasonable?
We have seen earlier there are roughly 34.5 million African Americans in the US, so 7 million
out of 34.5 million is about 0.20 or 20 percent. It does seem reasonable that 20 percent of the
population of African Americans are adult males. If the number seems to be reasonable, why
does the article seem to be inflammatory? What “inferences” are the authors of the article
assuming the reader will make?
The phrase “under George W. Bush” needs to be considered. Is it logical to infer blame for this
alarming statistic and assign it to President George W. Bush? Were the adult African American
males incarcerated when the President took office? President Bush assumed the Office in 2000.
What proportion of the adult African Americans males were incarnated in either 2000 or 2001?
Could there be other explanations or other “directions to place blame” that would be more
rational or logical? If one investigated a possible correlation with welfare reform, possibly a
stronger and more rational argument could be constructed. Where in the summer of 1996,
- 57 -
Congress and the President at that time passed the "Personal Responsibility and Work
Opportunity Reconciliation Act of 1996." This act radically transformed the nation's welfare
system, a full four years before George W. Bush took the Oath of Office. Could this act have
been a major contributing factor for an escalating incarceration rate some five years later? Can
you think of other factors that would explain this statistic other than “under George W. Bush”?
Though these questions we present are rhetorical in nature, they are worth considering in
determining the reasonableness of the data presented in the flyer. Is the statement “under George
W. Bush” an example of nothing more than a coincidence or is he responsible for the
incarceration rate of adult African American males in the U.S. in 2001? A little numerical
literacy applied to data presented by media can go a long way in determining whether an
argument is a representation of a valid correlation or simply an observed coincidence.
Example Three
The Spinner
For the spinner below, if the wand is spun around, what is the probability…?
a)
b)
c)
d)
e)
f)
g)
it lands on a blue portion?
it lands on a 3?
it lands on a blue 3?
it lands on an even number?
it lands on a even number less than 6?
it lands on a number divisible by either 2 or 3?
What is the sample space?
Solution
a)
b)
c)
d)
e)
Of 8, there are 4 blue portions and 4/8 = ½ is the probability.
Of 8, there is 1 blue portion and 1/8 is the probability.
Of 8, there are no blue 3’s, 0/8 = 0 is the probability.
Of 8, there are four even numbered portions, and 4/8 = ½ is the probability.
Of 8, there are two even numbers, 2 and 4 , that are smaller than 6 and 2/8 = ¼ is the
probability.
f) Of 8, there are 5 numbers divisible by either a 2 or a 3. They are 2, 3, 4, 6, 8, thus 5/8 is the
probability.
g) {Blue 1, Blue 2, Blue 6, Blue 8, Red 3, Red 4, Red 5, Red 7}
Let’s go back to rolling a die. What happens if we roll two dice and observe the sum shown?
First, what is the sample space? The sample space is the collection of all possible outcomes.
Well, the smallest sum possible is a one on each die, and a one and a one yields a sum of a two.
The largest possibility is a six on both die, which gives a sum of twelve. All integers between
two and twelve are the possible sums when rolling two die. So, how many possible outcomes are
there? Below is the sample space:
- 58 -
1-1
1-2
1-3
1-4
1-5
1-6
2-1
2-2
2-3
2-4
2-5
2-6
3-1
3-2
3-3
3-4
3-5
3-6
4-1
4-2
4-3
4-4
4-5
4-6
5-1
5-2
5-3
5-4
5-5
5-6
6-1
6-2
6-3
6-4
6-5
6-6
The sample space consists of the 36 different possible combinations from rolling two die. Which
sum is least likely to occur?
Example Four
Lady Luck We roll two die and observe the sum shown. Use the sample space shown above to
find the probability of rolling the sum indicated and state whether the event is more likely to
occur or less likely to occur.
a) five
d) not a double or a six
b) not an eleven
e) a double and an eight
c) not a double
f) not a double and not a four
Solutions
a) There are 4 different ordered pairs from the sample space with a sum of 5 and you see them
along the diagonal of 4 –1, 3 –2. 2 –3, and 1 – 4. With 36 total outcomes possible, the
probability of rolling two die and obtaining a sum of 5 is denoted P(5) = 4/36 = 1/9.
b) This may be easier to answer using the compliment of the event. There are only two ways to
roll a sum of eleven, with a 5 – 6 or a 6 –5. So, P( 11 ) = 1 – 2/36 = 34/36 = 17/18.
c) Again, using the compliment, it is quicker since there are only six doubles, 1-1, 2-2, 3-3, 4-4,
5-5, 6-6. So, P(not a double) = 1 – 6/36 = 30/36 = 15/18.
d) We need to account for all of the non doubles as well as all of the sixes. The desired event
consists of the union of two sets of possibilities. The set of non-doubles and the set of ordered
pairs with a sum of six. The diagonal containing all the ordered pairs with a sum of six has 5
elements. There are 30 non-doubles and the cardinality of the union of the two sets is 30 + 5 – 4
= 31. thus, P(non-double or a six) = 31/36.
e) We need the desired outcome to be a double and an eight. There is only one outcome, 4 – 4,
that is both, a double and a sum of 8. P(double to equal a sum of 8) = 1/36.
f) We need the desired outcomes to be non doubles that are not fours. There are 30 non doubles,
and if we remove the non-doubles with a sum of a 4, we are left with 28 desired outcomes. So,
P(non double that is not a four) = 28/36 = 7/9.
- 59 -
Exercise Set
Before proceeding, recall that a standard
deck of playing cards consists of four each
of the numbers 2 though 10, 4 jacks, 4
queens, 4 kings and 4 aces. The face cards
or picture cards refer to the jacks, queens
and kings. There are 52 cards separated into
4 suits, clubs, diamonds, hearts, and spades
and within each suit are 13 distinct cards,
ace through 10, jack, queen and king. The
diamonds and hearts are traditionally red,
the clubs and spades are traditionally black.
15. What is the probability of tossing 1
head?
16. What is the probability of tossing at
least 1 head?
17. What is the probability of tossing no
heads?
For problems 18 to 22, we toss three coins.
18. What is the sample space?
For problems 1 to 12, assume 1 card is
drawn from a standard deck of 52 cards.
Find the probability of drawing a …
1. two 2. red two
19. What is the probability of tossing 2
heads?
20. What is the probability of tossing 1
head?
3. red two or a diamond
4. red two or not a diamond
5. red two and a diamond
21. What is the probability of tossing at
least 1 head?
22. What is the probability of tossing no
heads?
6. red two and not a diamond
7. a heart
For problems 23 to 29, we roll two die and
observe the sum shown.
8. face card or heart
23. What is the sample space?
9. a heart or a non-jack
24. What is the probability of rolling a sum
of a seven?
10. a heart and a non-jack
11. a red card or a black card
12. a red card and a black card
25. What is the probability of rolling a sum
of not a seven?
26. What is the probability of rolling a
double?
For problems 13 to 17, we toss two coins.
13. What is the sample space?
27. What is the probability of rolling a
double or a six?
14. What is the probability of tossing 2
heads?
28. What is the probability of rolling a
double and a six?
- 60 -
29. What is the probability of rolling a
double and not a six?
For question 30 to 35, we will discuss Gun
Control They say that Texans love their
guns. But, in Arizona, a century after the
wild-wild west days, about 67,000 people
legally carry guns. And in 2004 there are
approximately 5,456,453 people residing in
the state. Hence, the probability of
encountering someone who has a gun in
67, 000
Arizona is
 0.01228 . To
5, 456, 453
1
 0.0122 or 1 in
phrase it another way,
82
82 people in the state of Arizona tote a gun,
compared to Texas, where 1 out of every 92
people legally tote a gun. But, to quote
Gomer Pyle, surprise, surprise, Utah loves
their guns the most because 1 in every 40
people in Utah legally carry a gun.
30. 13,500 Arizonian women have gun
permits. What’s the probability of
encountering a gun toting women if you
meet someone from Arizona? Phrase your
chances of encountering a woman who
carries a gun in ordinary English.
31. 52 Arizonians are women over the age
of 80 who carry a gun. What’s the
probability of encountering a gun toting
women who is 80 years or older if you meet
some from Arizona? Phrase your chances of
encountering an elderly gun carrying woman
in ordinary English.
32. What is the probability that an
Arizonian who legally carries a gun is a
woman?
33. What is the probability that an
Arizonian who legally carries a gun is a
man?
34. In 2002, if 236,738 people in Texas and
57,907 people in Utah own guns, what is the
population of each state?
35. How many more times likely are you to
encounter a some one who has a legally
concealed gun if you run into a native from
Utah compared to a native from Arizona?
36. Earlier in the text we mentioned that in
2000, in this country, there were 6,594
injuries at fixed site amusement parks. We
also mentioned that roughly 317,000,000 or
over 300 million people visit fixed site
amusement parks in a given year. What is
the probability that someone who visited an
amusement park in 2000 would suffer an
injury? Then express the answer in 1 in
“how many” people were injured.
For questions 37 to 41, recall also that in
other recreational activities that same year,
82,722 people were injured on trampolines,
62,812 were injured in swimming pools,
544,561 were injured on bicycles and 20,000
were injured at music concerts We have no
way of knowing how many people bounced
on a trampoline, swam in a swimming pool,
peddled a bicycle or frequented a musical
concert that year. But it is safe to say the
chance of injury for any of those activities
was far higher than an amusement park visit.
To explore this another way, answer the
following question?
37. In 2000, how many people would have
to have bounced on a trampoline if the
probability (risk) of getting injured was
identical to the risk of visiting an
amusement park?
38. In 2000, how many people would have
to have swam in a swimming pool if the
probability of getting injured was identical
to the risk of visiting an amusement park?
- 61 -
39. In 2000, how many people would have
to have peddled a bicycle if the probability
of getting injured was identical to the risk of
visiting an amusement park?
40. In 2000, how many people would have
to have frequented a musical concert if the
probability of getting injured was identical
to the risk of visiting an amusement park?
41. Recalling how many people, 6.4 billion
people, live on this planet, explain the point
of exercises 37-40.
42. Project: Identify information presented
from any media source that has the
appearance of being “alarmist”. Use the
math skills we’ve developed to confirm the
argument or refute the argument.
Probability: Independent and Dependent Events
Now, let’s discuss the probability of two or more events occurring. We need to be careful
because we need to distinguish between whether or not the likelihood of one event influences the
likelihood of the other event.
The probability of a National Football League team winning the Super Bowl and a Democrat
winning the White House is a case where the outcome of event A, a football team winning the
Super Bowl, has no effect on the outcome of event B, which party wins a presidential election.
When considering two events, where outcome of one event has no affect on the outcome of the
other event, we call these independent events. Some other examples of independent events are
choosing a card from a standard deck of playing cards and getting a four, replacing it, shuffling
the deck and choosing a second card and getting an ace. Knowledge that you drew the four the
first time around does not affect the probability you will draw the ace. Whether or not you
pulled the four in the first event is irrelevant, because when you draw the second time, there are
still 4 aces in the 52 card deck. How does the probability of drawing an ace change if you leave
the first 4 drawn out of the deck? Tossing a coin and it landing on tails and rolling a pair of die
and getting a double are independent events, because whether or not your coin landed on heads
or tails, the chance of rolling a double is still 1/6. Another pair of independent events are rolling
a die and getting a 4, and then rolling a second die and getting a 5. In each of the previous
mentioned events, the knowledge of the first event occurring does not affect the likelihood of the
second event occurring.
When the likelihood of one event does affect the likelihood of the other event occurring, we say
the events are dependent on one another. Knowing a person weighs less than 150 pounds
decreases the likelihood the person is over 6 foot tall. Other examples include choosing a card
and getting a four, not replacing it and then choosing a card and getting an ace. Knowledge you
have drawn the four does affect the likelihood you will draw an ace, because this means there are
still 4 aces left in the remaining 51 cards (rather than the independent event with 52 cards).
Let’s see if we can visualize independent events. Toss two die. The result for the first die does
not affect the result for the second die. If first you roll a three, when you toss the second die, this
outcome does not depend on the value of the previous roll. So, if we asked what is the
probability of rolling a sum of a five when tossing two die, the answer is 4/36 or 1/9. This is
because there are only four possible ordered pairs with a sum of 5, a (1,4), (4,1), (2,3) or (3,2).
Independent events.
- 62 -
Let’s toss two coins in a row and ask what is the probability of tossing two heads in a row. Each
event’s outcome has no effect on the next event’s outcome. In other words, whether or not you
tossed a head the first time does not effect the likelihood you’ll toss a head the second time. The
probability of tossing a head is also ½ and the probability of tossing a head a second time is ½ so
the probability of tossing two heads in a row is ½ of a ½. In other words, the probability of
tossing the first head is ½, then to toss it twice in a row, it is ½ of the first ½. This is the
multiplication principle.
The multiplication principle says that to calculate
the probability two independent events will occur in a
row, just multiply the two respective probabilities.
Clearly, this can be extended to a sequence of events. The probability of tossing four heads in a
row is ½ x ½ x ½ x ½ = 1/16.
Example One
Rolling a single die repeatedly
A single die is rolled three times in a row. What is the probability …
a) all three rolls are either a 1 or a 2?
b) all three rolls are above 2?
c) all three rolls are not 2?
d) all are a 1 or all are a 2?
Solution
2 2 2
 
6 6 6
4 4 4
b)
 
6 6 6
5 5 5
c)
 
6 6 6
1 1 1
 
d)
6 6 6
a)
23
63
43
 3
6
53
 3
6
1 1 1 2
    3
6 6 6 6

- 63 -
Exercise Set
Probability of Independent Events
For problems 1 through 4, a card is chosen
at random from a deck of 52 cards. It is then
replaced, the deck shuffled and a second
card is chosen.
1. What is the probability of getting a jack
and an eight?
2. What is the probability of getting a jack
and then an eight?
3. A diamond and then a heart?
9. A nationwide survey found that 35 % of
adults in the United States like ice cream. If
two couples (4 adults) sit down for a meal,
what is the probability that all four like ice
cream?
10. Spin a spinner numbered 1 to 6, and toss
a die. What is the probability of getting a
number less than 5 on the spinner and a
number less than 5 on the die?
11. Spin a spinner numbered 1 to 6, and toss
a die. What is the probability of getting the
same number on the on the spinner that you
got on the die?
4. A non-diamond and then a red ace?
5. A dresser drawer contains one pair of
socks of each of the following colors: black,
brown, white, orange and stripped. Each
pair is folded together in matching sets. You
reach into the sock drawer and choose a pair
of socks without looking. The first pair you
pull out is stripped. Wanting something
different, you replace this pair and reach into
the drawer again. What is the probability
that you will get the stripped pair of socks
twice?
For problem 12 to 14, use a jar contains jelly
beans, and there are 100 pink, 200 purple,
50 white, 70 black, 30 orange and 300
yellow.
12. If two jelly beans are chosen from the
jar, with replacement, what is the probability
that both are green?
If two jelly beans are chosen from the jar,
with replacement, what is the probability
that
13. both are pink?
For problems 6 –8, use a coin that is tossed
and a single 6-sided die that is rolled.
6. Find the probability of getting a tail on
the coin and an even number on the die.
14. If two jelly beans are chosen from the
jar, with replacement, what is the probability
that both are not the same color?
7. Find the probability of getting a tail on
the coin or an even number on the die.
For number 15 to 18, use three cards that are
chosen from a standard deck of 52 playing
cards with replacement.
8. Find the probability of getting a tail on
the coin and an not a six on the die.
15. What is the probability of getting no
jacks or no diamonds?
16. What is the probability of getting 3
diamonds?
- 64 -
17. What is the probability of not getting 3
diamonds?
For problems 25 through 29, two cards are
chosen in succession from a deck of 52
cards.
18. What is the probability of getting aces?
For problems 19. and 20., use a 5-item
false quiz, where a student has
probability of correctly answering
question. Assume that answering
question are independent events.
true4/5
each
each
19. What is the probability that the student
will get all answers correctly?
20. What is the probability that the student
will get at least 4 answers correctly?
For problems 21 and 22, use a bowl of m
and m’s that contains 12 blue, 15 brown, 10
black 16 yellow, 10 red and 8 green m and
m’s. Reaching into the bowl, you grab a
single candy. It is brown. You toss it back.
Yuck. You reach in again. Brown. Again,
you toss it back.
25. What is the probability of getting a jack
and then an eight?
26. What is the probability of getting a jack
and an eight?
27. What is the probability of getting a
diamond and then a heart?
28. What is the probability of getting two
diamonds?
29. What is the probability of getting two
sixes?
For problems 30-35, use the following: a
bowl of m and m’s contains 12 blue, 15
brown, 10 black 16 yellow, 10 red and 8
green m and m’s. Reaching into the bowl,
you grab a handful of five pieces of candy.
21. What is the probability of getting all
brown m and m’s 3 times in a row?
30. What is the probability of getting all
brown m and m’s?
22. What is the probability of getting two
brown m and m’s and then an m and m that
is not brown?
31. What is the probability of getting 4
brown m and m’s?
For problem 23 and 24, use a school survey
that found that 8 out of 10 children like ice
cream. You are throwing a birthday party
for your kid. You are setting the table.
23. What is the probability that the first
three children you put out plates for like ice
cream?
24. What is the probability that the first
three children you put out plates for don’t
like ice cream?
32. What is the probability of getting 3
brown m and m’s?
33. What is the probability of getting 2
brown m and m’s?
34. What is the probability of getting 1
brown m and m’s?
35. What is the probability of getting no
brown m and m’s?
- 65 -
Probability Using Counting Theory
Many times to find either the number of desired outcomes or just to find the total number of
possible outcomes, the cardinality of the sample space requires counting techniques introduced
in Chapter One. Techniques like the Fundamental Counting Principle, Permutations or
Combinations become useful to us once again.
Example One
Suppose 12 men and 18 women apply for jobs. If four people are selected randomly to fill job
positions, a) find the probability only one man received the job b) find the probability everyone
who received the job was a woman.
Solution
Well, probability is defined as the number of desired outcomes divided by the number of
possible outcomes.
a) The number of outcomes where 1 man and 3 woman have been given jobs is characterized as
a combination problem because the order does not matter. Each person is given an equivalent
job. So, from 12 men, we choose 1, from 18 women, we choose 3. We want to identify all
possible combinations where 1 male and 3 females that can be formed into a selected group, so
we multiply these numbers together, 12 C1 18 C3 . Now, we have the number of possible 1 man
and 3 woman committees. The number of possible outcomes is found by finding the number of
ways four people can be chosen from a group of 30 people, choose 4 to get the job, or 30 C 4 . So,
C1 18 C3
0.357
C
30 4
b) The number of successful outcomes, 4 woman have be given the jobs is again a combination
problem. From 18 women, we choose 4 and from the 12 men, we choose 0. The number of
successful outcomes is again 30 people, choose 4 or 30 C 4 . So, the probability all the jobs go to
C  C
women is 12 0 18 4 0.112
30 C 4
the probability 1 man gets the job is
12
Example Two
Cards What is the probability that someone dealt a five card poker hand will get dealt a hand
with…
a) four aces
b) two aces and two sixes
c) not a straight flush
d) a pair
- 66 -
Solution
a) For the 4 aces, we choose all 4 of these aces, 4 choose 4, or 4 C4 and from the 48 other nonaces, we choose 1 of them 48 C1 . And we recall, there are 52 C5 possible 5 card poker hands. So,
C4 48 C1
 0.000018 .
52 C5
b) For the 2 aces, we choose 2 of the 4 or 4 C2 and for the 2 sixes, we again choose 2 of the 4
sixes or 4 C2 . Our fifth card could be any of the cards except for the 4 aces and the 4 sixes. So,
we have 44 C1 . And we recall, there are 52 C5 possible 5 card poker hands. So, we have
6(6)(44)
4 C 2  4 C 2  44 C1

 0.00061 .
52 C5
52 C5
c) For the non-straight flush, we have 2,598,960 possible 5 card hands – 4(9) possible straight
2,598,924
flushes. So, we
 0.997 .
2,598,960
d) For the pair, we need two of a kind and three other unique cards. For example, let’s consider
a pair of jacks. For two of a kind, we select two of the four jacks, 4 C2 . There are 13 different
cards that we could have as a pair, the aces, two’s, three’s and so on until we get to the kings.
So, now we have 13 4 C2 . For the other three cards, they must be different types, say a three,
five and six. For the three, we select 1 of the four three’s, 4 C1 , for the five, we select 1 of the
four five’s, 4 C1 , and for the six, e select 1 of the four six 4 C1 . Now, we must ensure the number
of ways to choose 3 different types of cards from the 12 remaining types that were not pairs.
13 4 C 2  12 C3  4 C1 4 C1 4 C1
This is 12 choose 3. So, we have
 0.423 .
52 C5
we have
4
Example Three
Birthday Problem You are having a party, some 20 guests are gathered, but there is tension in
the air. No one is mingling. No one is reaching out to get to know another. You need an ice
breaker. So, you say, I bet two of you have the same birthday? You go around the room, giving
each person the easiest topic of conversation of all, each person states their own birthday.
Maybe someone has a funny retort. Maybe someone has a funny story. Or maybe, just maybe,
two people have the same birthday. Something in common. And conversations begin.
Regardless of the year, what is the probability two people have the same birthday? How many
people would you have to gather to have more of a chance two people have the same birthday
than not? What is the probability that if 30 people attended the party, two would have the same
birthday? Or 40 party goers? How about 367 guests?
Solution
A general overview of the mathematical argument we are about to make is this: We will find the
probabilities associated with people who do not have the same birthdays. We will then find the
- 67 -
probability of the compliment set, the probability of at least two people having the same
birthday.
Our argument to find the solution to this question starts with assuming all birthdays are equally
likely to occur on any of the 365 days of the year and we will forgo the argument with respect to
leap year.
Starting with person 1, we record the birthday; one of 365 possible days in the year. Person 2
can have a birthday on any of the remaining 364 days that wasn’t the date of Person 1’s birthday.
Person 3 can have a birthday on any of the remaining 363 days that weren’t the birth dates of the
first two people. And so on.
The probability of person 2 having a birthday different than person 1 is found by realizing that
any of the other 364 days could be the birthday, just not the one day that belongs to Person 1. A
similar sentence could be echoed for Person 3, their birthday needs to be any of the remaining
363 days of the year that are different from the two already taken, so to speak.
Applying the multiplication principle, the probability of the first three people having different
365 364 363 365 P3
birthdays is



365 365 365 3653
Extending this pattern, the probability all 20 people have different birthdays is
365 364 363 346 365 P20


... 

365 365 365 365 36520
The complimentary probability for this statement is what we are seeking.
P20
365 364 363 346
1


... 
1 365 20
 0.41 . There is a 41 percent chance two people among a
365 365 365 365
365
group of twenty people will have the same birthday.
If you have 23 people at the party, 1 
P23
 0.507 or slightly over 50 percent chance two will
36523
365
have the same birthday.
P
 0.706 or 70 percent chance that two
365
individuals would have the same birthday. If you do not believe us, try it the next time such a
crowd is gathered. In case your curious, if you gathered 40 people, it would be highly likely that
two people had the same birthday. Hard to believe? Like we said, try it…
If 30 people had gathered at the party, there is 1 
365 30
30
- 68 -
Exercise Set
Probability Using Counting Theory
For problems 1 and 2, use a bowl of m and
m’s containing 12 blue, 15 brown, 10 orange
16 yellow, 10 red and 8 green m and m’s.
Reaching into the bowl, you grab five pieces
of candy.
1. What is the probability of getting 2
brown pieces and 3 red pieces?
2. What is the probability of getting only 1
brown piece?
For problems 3 to 12, find the probability of
obtaining each of the following 5 card poker
hands:
3. Royal Flush.
4. Straight Flush (5 cards in sequence in a
single suit, but not a royal flush)
5. Four Aces
6. Four of a Kind
7. Flush of Hearts
8. Flush
9. Four hearts
10. Three of a kind
11. Two Sixes and Three Aces
12. Two Aces and Two Kings
13. A pair
For questions 14 to 16, suppose license plate
"numbers" consist of all possible
arrangements of 3 letters followed by 2
digits.
14. What is the probability that a randomly
chosen license plate number has no repeated
symbol?
15. What is the probability that a randomly
chosen license plate number has the last
digit a one?
16. What is the probability that a randomly
chosen license plate number has no repeated
symbol and the last digit is a one?
For problems 17 and 18, in a shipment of
1,000 Play Stations, 46 are defective.
17. If a person buys two Play Stations, what
is the probability that both are defective?
18. If a person buys two Play Stations, what
is the probability that neither is defective?
For questions 19. through 21., suppose 4
men and 6 women are on a crowded
elevator. At the next floor, three people get
off the elevator.
19. Find the probability all are women.
20. Find the probability two are women
21. Find the probability none are women
22. What is the probability that in a group of
3 people at least two were born in the same
month?
23. What is the probability that in a group of
5 people at least two were born in the same
month?
24. What is the probability that in a group of
7 people at least two were born in the same
month?
25. What is the probability that in a group of
9 people at least two were born in the same
month?
26. What is the probability that in a group of
11 people at least two were born in the same
month?
27. What is the probability that in a group
of 20 people at least two were born in the
same month?
28. What is the probability that in a group
of 5 people at least two had the same
birthday?
29. What is the probability that in a group
of 10 people at least two had the same
birthday?
30. What is the probability that in a group
of 20 people at least two had the same
birthday?
- 69 -
31. What is the probability that in a group
of 30 people at least two had the same
birthday?
32. What is the probability that in a group
of 60 people at least two had the same
birthday?
33. What is the probability that in a group of
500 people at least two had the same
birthday?
For problems 34 through 38, two cards are
chosen in succession from a deck of 52
cards.
34. What is the probability of getting a jack
and then an eight?
35. What is the probability of getting a jack
and an eight?
36. What is the probability of getting a
diamond and then a heart?
37. What is the probability of getting two
diamonds?
38. What is the probability of getting two
sixes?
For problems 39-46, use the following: a
bowl of m and m’s contains 12 blue, 15
brown, 10 black 16 yellow, 10 red and 8
green m and m’s. Reaching into the bowl,
you grab a handful of five pieces of candy.
39. What is the probability of getting all
brown m and m’s?
40. What is the probability of getting 4
brown m and m’s?
41. What is the probability of getting 3
brown m and m’s?
42. What is the probability of getting 2
brown m and m’s?
43. What is the probability of getting 1
brown m and m’s?
44. What is the probability of getting no
brown m and m’s?
The Binomial Probability Profile
Often referred to as Bernoulli Trials, after John Bernoulli who first published in this area, the
probability studied in this section is the practical probability when one asks what’s the likelihood
you go to a party and meet five perspective mates, all of whom never married? Or meet five
perspective mates, all of whom are living alone? If you listen softly, this is the same question as
how many times can you roll snake eyes (a one on each die) five times in a row, or flip a coin
and have it land on heads five times in a row.
Let’s pick the first scenario, choosing a mate who has never been married and highlight this
example for you the Binomial Probability model. If you meet 5 perspective mates, this can be
considered the same experiment, meeting a mate, repeated over and over again, specifically here
five times. There are only two possible outcomes, the perspective mate has either never married,
or not. The trials are all independent because if you were to meet someone who was never
married, this does not affect the likelihood (probability) the next person you were to meet has
never been married. Finally, the outcomes for each experiment does not change, the probability
of someone in a certain age range who was never married is published by the US Census Bureau
for the year 2000.
- 42 -
These similarities, you may recognize, exist for our other scenarios we posed too. Those all
involve the same experiment, the quest for a perspective mate who lives alone, or the snake eyes
reoccurring as dice are tossed, the heads reappearing as a coin is flipped, repeated over and over
again, and again with five repetitions being specified for each event. In each question, there are
only two possible outcomes, like the perspective mate lives alone or they don’t live alone, the
snake eyes reoccur or doesn’t, the heads reappear or they don’t. The trails are all independent,
meeting a perspective mate who lives alone does not affect the probability the next person lives
alone. Rolling snake eyes does not enhance or deter the possibility snake eyes will show on the
next roll and clearly a flip of a coin showing a tail falls under the same guise. Finally, the
outcomes for each experiment does not change, the probabilities of each occurring remains the
same, each perspective mate has the same likelihood of living alone, according to the US Census
Bureau, that is, each roll of the dice has a probability of 1/36 that snake eyes will show, each flip
of the coin has a probability of ½.
Probabilistic events such as these are called Bernoulli Trials if the following are satisfied:
 the same experiment is repeated over and over again
 there are only two possible outcomes
 the trails are independent
 the probability for each outcome must remain the same.
Example One
According to the US Census Bureau, 2000, “the times they are a-changin’” as Bob Dylan used to
say. In 1970, the probability you encountered a 20 to 24 year old woman who never married was
0.36, but in the year 2000, the probability more than doubled to be 0.73. For a 20 to 24 year old
man, the probability also grew, from 1970, 0.55 to 2000, 0.84. For 34 year olds, the drastic
increase was even more acute. In 1970, the probability you encountered a 30 to 34 year old
woman who never married was 0.06, in the year 2000, the probability almost quadrupled to 0.22.
For 30 to 34 year old men, the results were nearly as striking because the probability you met a
man who never married in this age group more than tripled, from 0.09 in 1970 to 0.30 in the year
2000.
Assuming the probabilities have not changed much since the year 2000, let’s explore these
statistics. You’re at a social gathering and you encounter 5 women between 20 and 24 years old.
What is the probability …
a) three were never married
b) none of the 5 have ever been married
c) at least one has been married
d) at most 1 has been married
Let’s also ask the same question with regard to 20 to 24 year old men. What is the probability …
e) three were never married
f) none of the 5 have ever been married
g) at least one has been married
h) at most 1 has been married
- 79 -
Solution
a) Let’s clarify our problem. Slowly. We meet five women (n = 5) and if three of the women
were never married (k = 3), where the probability of encountering a 20-24 year old woman who
never married is 0.73 for each meeting (p = 0.73). We can now proceed. The long way first.
We want to explore the possible arrangements where three women were never married and two
were married . Assuming the first three women met were never married, the probability would
be (0.73)(0.73)(0.73). The probability a woman was married would then be 1 – 0.73 = 0.27. The
probability the remaining two women were married would be 0.27(0.27). By virtue of the
multiplication principle, we see the probability of meeting five women, the first three unmarried,
the last two married is calculated by (0.73)(0.73)(0.73)(0.27)(0.27) or more precisely
(0.733)(0.272). There are 10 different arrangements or orders in which someone encounters five
women, three unmarried, two married. To see this, use M for married and N for never married.
The ten various arrangements when meeting three women who never married and two who were
married. Or more concisely, NNNMM. The other nine ways would be NNMMN, NMMNN,
NMNMN, NNMMN, MNNMN, MNNNM, MMNNN, NNMNM, and NMMNN. What is a
shorter way of counting all of these possible arrangements? C(5,2), which we will write again
5
5(4)
3
more concisely as   
10. So, we multiply 10 with  0.73 (0.27) 2 to arrive at the
2
 2
answer, 0.28. This means that for every 5 women you meet in the age group 20 – 24, more than
one in four times you meet 5 women, 3 will have never been married.
Now, in terms of the Binomial Probability Profile, this pattern includes a repetitive train of
thought, that is, an experiment that is repeated over and over again. As politically incorrect as
this sounds, the trial is meeting women and a success is finding a woman who was never
married. We first consider the number of arrangements, which we can always find with a
combination. We multiply this combination with the probability of the success each time the
success occurs. This result is multiplied by the compliment of the probability’s success the
number of times it occurs. Always. Never to deviate. So, our formula is written:
n
k
b(n, k , p)     p  (1  p) n k
k
where n is the number of trials, k is the number of successes and p is the probability for the
success.
 5
3
a) b(5,3, 0.73)     0.73 (0.27)53  0.284
 3
b) If none of the women had ever married, then we want 5 successes out of 5 trials.
 5
5
b(5,5,0.73)   0.73 (0.27) 0  0.207
 5
c) If at least one women was once married, this means we have several cases to consider: one,
two, three, four or five married. If we speak in terms of a success, then we have several cases
here as well: If five were once married, then none were never married, if four were once
married, then one was never married, and continuing in this vain, two, three or four could also be
never married. So, we have 0, 1, 2, 3 or 4 successes to make the statement at least one woman
was once married true. Let’s use what we know about finding probabilities and use the
- 80 -
probability of the compliment of a set: 1 – the opposite probability, 1 – P(5 were never married.)
 5
5
1  b(5,5, 0.73)  1     0.73 (0.27)0  1  0.207  0.793 . Said in another way, if you were to
 5
meet 5 women 10 times, on the average, nearly eight of those times at least 1 of the five were
once married.
d) If at most 1 was once married, then either one or none were once married. Thus, 5 would
have been never married or 1 was once married, which implicitly states 4 were never married.
We have
 5
5
5
4
b(5,5, 0.73)  b(5, 4, 0.73)     0.73 (0.27)0     0.73 (0.27)1  0.207  0.383  0.59 The
 5
 4
next time you meet 5 women between the ages of 20-24, over half the time at most 1 was once
married.
 5
3
e) b(5,3, 0.84)     0.84  (0.16)53  0.152
 3
f) If none of the men were never married, we want 5 successes out of 5 trials.
 5
5
b(5,5, 0.84)     0.84  (0.16)0  0.418
 5
 5
5
g) 1  b(5,5, 0.84)  1     0.84  (0.16)0  0.582 . Over half the time you meet five men, at
 5
least 1 of them was once married.
 5
5
5
4
h) b(5,5, 0.84)  b(5, 4, 0.84)     0.84  (0.16)0     0.84  (0.16)1  0.816 . So, 8 out of
 5
 4
every 10 times you meet 5 men, at most 1 of them was once married.
Example Two
You roll a pair of dice, what is the probability you get snake eyes three times out of 5 rolls, and
oury next question is what is the probability of getting snake eyes five times in a row? What
does your intuition tell you? Pretty low likelihood, small probability; you are right.
Solution
a) The probability of getting snake eyes is 1/36, so we have 5 trials, 3 success, each with a
3
 5
   1
(35 ) 2  0.000203
probability of 1/36. So, we have b 5,3, 1
36
36
36
 3
b) The probability of getting snake eyes is 1/36, so we have 5 trials, 5 success, each with a
5
 5
   1
(35 )0  1.654 x108 . This tells us
probability of 1/36. So, we have b 5,5, 1
36
36
36
 5
0.00000001654 is the probability or the likelihood of tossing 5 pairs of snake eyes in a row is
less than 1 in 200 million.


 


 
Example Three
- 81 -
A fair coin is tossed 8 times. What is the probability of getting at most 7 heads?
Solution
Each reoccurring probability is simple enough, the probability of getting a head when you flip a
coin is ½. But, to calculate the answer directly, we would have to considered all of the options
possible that could yield at least 7 heads. So, we would add, loosely phrased,
P(0 H) + P(1 H) + P(2 H) + P(3 H) + P(4 H) + P(5 H) + P(6 H) + P(7 H). This calculation is
lengthy. Instead, we will use the strategy employing the probability of the compliment. This
would mean 1 - the opposite of (P(0 H) + P(1 H) + P(2 H) + P(3 H) + P(4 H) + P(5 H) + P(6 H)
8
 8
+ P(7 H)). This would be 1 – P(8 H) = 1b 8,8, 1  1    1 ( 1 )0  0.996
2
2
 8 2


 
Exercise Set
For all 29 questions from this section: Write
the probability in the form b(n,k,p) and then
find the probability to six decimal places.
8. no one lives alone.
9. there is at least one household with more
than one person.
For questions 1 to 9:The following data extracted from
For problems 10-13, The following data extracted from
the US Census Bureau Occupation 2000, Issues august, 2003 .
Bases on the most detailed level of
occupations available in Census 2000 from
509 occupations. Below is a table with the
Ten Occupations Employing the Most
Women for the United States, 2000. There
were 60,630,069 women 16 years and older
employed in the civilian capacity in 2000 in
the United States.
the US Census Bureau Population Profile of the United States:
2000, Chapter 5.
In 1970, the probability a
household consisted of someone living alone
was 0.17, while by the year 2000, the
probability had grown to 0.26. Suppose you
live a cul-de-sac where there are 8
households and suppose the probability of
some living alone has not changed very
much since the year 2000. Find the
probability using the data from the year
2000 where:
1. there are two house holds where someone
lives alone.
2. there are three house holds where
someone lives alone.
3. there are four house holds where
someone lives alone.
4. there are five house holds where
someone lives alone.
5. there is someone living alone on your
cul-de-sac.
6. there is one household where someone
lives alone.
7. there are no families on your cul-de-sac,
everyone lives alone.
Occupations
Secretaries, Administrative Assistants
Elementary, Middle School Teachers
Registered Nurses
Cashiers
Retail Salespersons
Bookkeeping, Accounting, Auditing
Nursing, Psychiatric, Home Health
Aides
Customer Service Representatives
Child Care Workers
Waiter, Waitress
Number
3,597,535
2,442,104
2,065,238
2,030,805
1,775,889
1,526,803
1,469,736
1,396,105
1,253,306
1,228,977
You meet six employed women (in the
civilian capacity) at a party.
- 82 -
10. What is the probability half of the
women are elementary or middle school
teachers?
11. What is the probability none of the six
are in any of the top ten occupations listed
above?
12. What is the probability at most two of
the six are secretaries or administrative
assistants?
13. What is the probability at least one of
them are either child care workers or
nursing, psychiatric or Home Health Aide
Specialists?
For problems 14-17, use a fair coin that has
been tossed six times.
14 What is the probability of tossing a tail
once?
15 What is the probability of tossing at
most 3 tails?
16 What is the probability of tossing at least
5 tails?
17. What is the probability of tossing at
most 5 tails?
For problem 18-20, we roll 2 dice seven
times and observe the sum.
18. What is the probability of rolling the
sum of a four exactly 5 times?
19. What is the probability of rolling the
sum of a five or seven exactly 5 times?
20. What is the probability of rolling the
sum of a four at least once?
Problems 21-26. The professional. A true
professional basketball player has nerves of
icy steel. If they miss their foul shot, they
are not affected the next time they shoot.
The likelihood of hitting or missing the foul
shot is not going to be affected by their prior
stint at the line. Independent probabilities.
If they miss three times in a row, they are
not affected the next time they shoot. Fives
misses in a row, still, they are not affected
the next time they shoot. The same is true
for a professional baseball player. If they do
not get a hit, they are not affected the next
time they are at bat, ten hitless at bats, no
problem. There is no such thing as a slump
in professional sports. In other words, the
probability of success remains the same, and
they are independent.
A basketball player has a shooting
percentage of 0.850 at the free throw line.
21. What is the probability he will hit 11
out of 12 free throws?
22. What is the probability he will hit 10 or
more out of 12 free throws?
23. What is the probability he will miss 12
free throws in a row?
Consider the baseball player who has a
batting average of 0.300.
24. What is the probability he will go hitless
in a game, say 0 for 4?
25. What is the probability he will get a hit
in all 4 at bats?
26. What is the probability he will go 2 for
4?
Problems 27-29. Woe is me. You order
blank Cd’s, 10 of them. You research and
learn that they probability of getting a
defective Cd is 0.06.
27. You think woe is me, just my luck and
you wonder what the probability is of
getting at least 1 defective Cd. Well, what is
it?
28. What is the probability of getting half
defective?
29. What is the probability of getting the
majority of Cd’s defective?
- 84 -
Conditional Probability
Let’s compare two questions:
 What is the probability that the sum of two dice tossed is smaller than a 4?
 If we know one of the die shows a one, what is the probability the sum of two
dice tossed is smaller than 4?
The answer to the first question is 3/36 or 1/12 because the possible successful outcomes
are (1,1), (1,2) and (2,1) from the 36 events that make up the sample space. But, if we
are privy to inside information like we already know that one die shows a 1, we call this a
conditional probability because we are given a condition for the probability of an event
to occur. The probability the sum on two dice tossed is smaller than a 4 is increased to
2/6 = 1/3 because we are given that a one occurred on the first roll and thus we are
reduced to six equally likely possibilities, (1,1), (1,2), (1,3), (1,4), (1,5) and (1,6). Two of
these six, (1,1) and (1,2) have a sum less than a four.
 We write “P(sum less than 4 a one was rolled first)”
 We read “the probability of a sum less than 4 given that a one was rolled first.”
In ordinary English, conditional probability asks what is the probability that one event
will occur given that another event has already occurred.
Which is more likely, the probability that some one who speaks Spanish is from Mexico or that
someone from Mexico speaks Spanish? In the first case, we are saying given the person speaks
Spanish, what’s the probability they are from Mexico. Written with probability notation, we
have P(a citizen of Mexico| person speaks Spanish). In other words, we are restricting our
sample space to be those who speak Spanish, and we are asking of these people, what is the
probability they are from Mexico. In the second case, we have a sample population of citizens
from Mexico, and we want to know, of those people, what is the probability a random person
speaks Spanish and we denote it P(person speaks Spanish| a citizen of Mexico).
Rephrased, we are asking you to decide if it is more likely a person from Mexico speaks Spanish
or is it more likely if a person who speaks Spanish, then they are from Mexico. It is more likely
that a person who is from Mexico speaks Spanish. Employing conditional probability notation,
we would write
 P(a citizen of Mexico| person speaks Spanish) < P(person speaks Spanish| a citizen of
Mexico).
Remember, probabilities are just numbers with feelings and emotions like any number.
If you are from Mexico, then you speak Spanish or if you speak Spanish, then you are from
Mexico are logically speaking, neither always true or always false. Conditional probabilities
help us determine which is more likely.
- 70 -
Example One
Test Results A teacher gave her class two tests. 60 % of the class passed both tests, and 80 % of
the class passed the first test. What percent of those that passed the first test also passed the
second test?
Solution:
P(second | first ) 
P(first and second ) 0.60

 0.75
P(first )
0.80
Example Two
US Army Demographics
The following information was taken from the ARMY DEMOGRAPHICS, FY03, Army Profile, US Army, Headquarters, Dr. Betty D. Maxfield,
Office of Army Demogrpahics.: [email protected]
Race by Gender - Enlisted
Male
Female
White
208,698 23,667
Black
81,024 27,317
Hispanic 37,882 7,176
Asian
11,925 2391
Other
11,225 2392
Total
350,754 62,943
Total
232,365
108,341
45,058
14,316
13,617
413,697
Find the following probabilities and explain the significance in a complete sentence.
a) P(male)
b) P(black female)
c) P(female| black)
d) P(black| female)
e) P(Asian| male)
f) P(male| white or black)
g) P(white or black| male)
Solution
350,754
 0.848 or nearly 85 out of 100 of the enlisted in the Army are men.
413,697
27,317
b) P(black female) =
 0.066 or nearly 6 2/3 out of 100 of the enlisted in the Army are
413,697
black females.
a) P(male) =
c) Note: Because it is a conditional probability, we must restrict our sample space to only those
27,317
enlisted personal who are black. P(female| black) =
 0.252 or nearly 25 out of 100 of
108,341
the Army’s enlisted personnel that are black are women.
- 71 -
27,317
 0.434 or nearly 43 out of 100 of the Army’s enlisted personnel
62,943
that are women are black.
d) P(black| female) =
11,925
 0.034 or nearly 3 ½ out of 100 of the Army’s enlisted personnel
350,754
that are men are Asian.
e) P(Asian| male) =
208,698  81,024
 0.851 or men comprise nearly 85 out of 100 of
232,365  108,341
the Army’s enlisted personnel that are either white or black.
f) P(male| white or black) =
208,698  81024
 0.826 or nearly 83 out of 100 of the Army’s
350,754
enlisted men are either white or black.
g) P(white or black| male) =
Exercise Set
Conditional Probability
For problems 1 through 3, an opinion poll
was administered to the students at the
University of America. Use the table below
because it reflects the response to the
following question, "Are you for or against
the death penalty?"
For
Against Not Sure Total
Men
20,000 10,000
5,000
35,000
Women
10,000 5,000
5,000
20,000
Total
30,000 15,000 10,000
55,000
A student is selected at random, find the
probability the student was ...
1. male
2. male, given that the student was against
the death penalty
3. for the death penalty given that
the student was a woman
For problems 4 to 10, use the
following data. A fox News poll in
2003 indicated 69 % of Americans
favor the death penalty. This is 7
percentage points down from 6 years
earlier. Moreover, according to a
new Gallup Poll, there is an even
closer split over the issue of what
should be done with a convicted
murderer. When asked if you were
in favor of the death penalty, a
whopping 74 % of answering the
poll responded ‘yes’, but the support
dropped to 53 % when people were
offered the alternative “LWOP –
Life Without the Possibility of
Parole.” Below is a mock table
reflecting what the demographics
would look like for such a poll,
separated by education. HS stands
for High School Education or Less,
C stands for College Graduate – 2 or
4 year degrees, and P stands for postbachelor’s, that is, Master’s Degree
or above. Similarly, DP stands for
Death Penalty, LWOP stands for
Life Without the Possibility of
- 72 -
Parole and No stands for No
Opinion.
HS
C
P
DP
1251
799
176
LWP
758
780
310
No
91
21
14
A person was selected at random.
Find the probability the person …
4. was for the death penalty.
5. was against the death penalty.
6. had a 2 or 4 year education, given
that the person responded LWP.
7. responded LWP given the person
had a Master’s degree or higher.
8. was for the death penalty, given
that the person had a high school
education or less.
9. was for the death penalty, given
the person has a 2 or 4 year degree.
10. was for the death penalty, given
the person had a Master’s degree or
higher.
For question 11 to 15, use the following
data: In early 2004, many polls, such as
those published by the PEW Research
Center on March 16th of that year, focused
their questions on nation of Iraq. Below is
representative mock table reflecting the
relative demographics of some of these
polls. The question asked was “After
Saddam Hussein, the Iraqi People will be” a)
Worse Off b) Better Off c) Don’t know.
The results are the number of respondents to
the question, by nationality.
Worse Better
Know Total
Off
Off
US
9
84
3
96
Britain
11
82
2
95
Germany
27
65
8
100
Jordan
70
25
3
98
Pakistan
61
8
10
79
Total
178
264
26
468
A person was selected at random.
Find the probability the person …
11. answered “Worse Off”
12. answered “Worse Off”, given
that they were from Jordan
13. was from Jordan, given that they
answered “Worse Off”
14. answered “Worse Off”, given
they were from the US
15. was from the US, given that they
answered “Worse Off”
16. A jar contains pink and white
jelly beans. Two jelly beans are
chosen without replacement. If the
probability of selecting a pink jelly
bean and a white marble is 0.50, and
the probability of selecting a pink
jelly on the first draw is 0.60, find
the probability of selecting a white
marble on the second draw, given
that the first marble selected was
pink.
17. At an Italian restaurant, 60 % of the
customers order a wine. If 50 % of the
customers order a pasta and wine, what is
the probability that a customer who orders a
wine will also order pasta?
18. The probability that a student is absent
and that it is Friday is 0.05. Since there are 5
school days in a week, the probability that it
is Friday is 0.2. What is the probability that
a student is absent given that today is
Friday?
For problems 19 to 23, which is greater?
19. P(he or she speaks English| he or she is
American) or P(he or she is American| he or
she speaks English)?
20. P(he or she is wealthy| he or she is a
orthodontist) or P(he or she is a orthodontist|
he or she is wealthy)?
21. P(it is raining| humidity is high) or
P(humidity is high| it is raining)?
22. P(person is in the NBA | person is
considered tall) or P(person is considered
tall | person is in the NBA)?
- 73 -
23. P(vegetable is green | vegetable is
broccoli) or P(vegetable is a broccoli |
vegetable is green)?
Miscellaneous Problems Using
Probability
1. If a number is chosen at random from the
following set,{2,3,4,5,6,7,8,9,10} what is
the probability that it is not prime?
2. A number from 1 to 5 is chosen at
random. What is the probability that the
number chosen is not odd?
For questions 3 through 6, the experiment is
rolling a single die and tossing 2 coins.
3. Describe the sample space.
4. Find the probability of rolling a three on
the die.
5. Find the probability of tossing at least
one head.
6. Find the probability of rolling a three on
a die or tossing at least one head.
For questions 7 to 8, according to a recent
elementary school survey, a child waiting in
line in a school cafeteria has a probability of
1/7 of not selecting a drink during lunch.
Assuming independence by making believe
there is no such thing as peer pressure, find
the probability ...
7. three children in a row won't select a
drink.
8. the first child will select a drink, the next
two children to follow won't.
For problems 9 and 10, two cards are drawn
from a deck. The first card is replaced back
into the deck after it is drawn. Find the
probability ...
9. of selecting a two on the second draw if a
two is drawn on the first draw.
10. of selecting a two on the first draw and
a two on the second draw.
For problems 11 to 15, a pair of dice are
rolled, find the probability …
11. of rolling a sum of a 2
12. of rolling a sum of a 5
13. of rolling a sum of not a double
14. of rolling a sum of a 6 given that a 2
appeared on the first die.
15. of rolling sum less than 10 if we know a
5 or 6 was not rolled on the first die.
Bayes Theorem
Certainly, we all have some concern about the accuracy of those polls we see every day
presented by the media. We routinely ask ourselves, are the polls biased or are they merely just
trying to locate the collective pulse of their audience? We often assume there is some hidden
agenda, be it political or social, behind the wording of the questions. We know the words chosen
when creating the polls can be, and often are chosen to solicit specific answers. But, maybe
instead, we should ask ourselves, what information is provided for us and what information is
deleted from us? And we do have the capacity to find the deleted information. The following
two stories are taken straight from viewing television one evening.
“According to a recent Gallop Poll, when asked which side of the political debate on abortion do
you sympathize with more: the right-to-life movement that believes abortion is the taking of
human life and should be outlawed, OR, the pro-choice movement that believes a woman has the
right to choose what happens to her body, including the right to decide to have an abortion, 43 %
of those polled answered Right-to-life.” Interestingly, half of the republicans polled sided with
the Right-Life argument. Does this mean we cannot draw along party lines on this issue and that
- 74 -
the abortion debate transcends party lines, polarizing us in directions we may not want to admit,
as the media would like us to believe?
“Here in Winstonboro County, 15% of the population developed lung cancer. And while 40% of
the county’s residents are smokers, 80% of those developing lung cancer are smokers.” While
this is informative, doesn’t it seem natural that you would want to know what the probability of
the exact opposite? Wouldn’t you want to know what the probability was for someone who
smokes to develop lung cancer? If you smoke, you may want to know the risks. Said another
way, you do not know right off if you have lung cancer or not, but you certainly know whether
or not you smoke. What are your chances, as a smoker, of developing lung? Isn’t this a more
useful question. It seems the report is dancing around this core point. Why not tell us directly?
Next, we will develop the skill to answer the question directly.
Let’s explore these two questions. In the first scenario, the Gallop Poll, 43% answered Right-tolife, and half of the republicans answered Right-to-life. When trying to connect party lines to the
abortion issue, the more relevant question is “if someone is opposed to
abortions, what is the likelihood they are republican?” Or more precisely, in mathematical
terms, of those respondents polled who aligned themselves with the Right to Life movement,
what is the probability they are republican? More abstractly, does the issue of abortion divide us
along party lines or does it transcend party lines as the media implies?
Let us begin by obsessing over the poll itself by defining the type of poll we have here. One
aspect of the poll is to place someone in a definitive position, either your are “for” or you are
“against” the Right to Life movement. How was this done? The poll was worded so that the
respondents were forced to declare themselves either Pro-choice or Right-to-life. Not both. Poll
participants were then asked to check off the party that most closely reflects their views, thus
they had to declare themselves to be either democrat, republican or independent. No one was left
sitting on the proverbial fence.
Now, let’s slow these sentences down. Inherent in the poll itself are three characteristics. Those
answering our poll comprise the sample space. Each individual from the sample space declared
themselves pro Right-to-life or pro-choice, and then further declared themselves democrat,
republican or independent, there was no overlap between these sets. No one was both Pro-life
and Pro-choice. No one was a member of two political parties. Two sets are said to be mutually
exclusive where there is no overlap between the two sets. Next, the poll consisted of everyone
who participated. You’re probably reading this and saying ‘no kidding’. But this means
something to us, that the union of the sets comprise the whole sample space. In other words, all
of the Pro-choice and all of the Right-to-life respondents comprise everyone. Similarly, all of
the respondents from the three political parties comprise all of the respondents too. Finally,
every set has at least one member. In other words, if no one was an Independent, we would not
include it in our sample space. When sample spaces are separated into non-empty, mutually
exclusive sets, we say the space has been partitioned.
A partition is where the sample space it divided into sets where
1. the sets are mutually exclusive
2. the union of all sets is the whole sample space
- 74 -
3. there is no empty set
 Bayes Rule can be applied to partitioned sample spaces.
Now, let’s examine all of the data so that we may answer our real question, which is ‘does
abortion divide us along party lines?’ According to the poll, 43 % of the respondents declared
themselves to be Right-to-life. So, everyone else declared themselves to be Pro-choice. That is
57 %, right?
Now, we will dig deeper into the data taken from the poll. According to the poll, of those who
declared themselves to be Right-to-life, 40 % percent were democrats, 55 % were republicans
and 5 % were independent. Of the respondents who declared themselves to be Pro-choice, half
were democrats, 45 % were republicans and again, 5 % were independent. Lot’s of data crying
out for some order. Let’s use a tree diagram to re-examine the data.
Now, let’s selectively read the tree diagram. It seems that 40% of the Right-to-life respondents
were democrat. Thus, 40 % of 45 % of the population were democrats who labeled themselves
Right-to-life. That is 0.40(0.45) = 0.18 or 18 percent of the population. Continuing in this
manner, 55 % of the 45 % of the respondents were Republicans who declared themselves Rightto-Life, of 0.55(0.45) = 0.2475 or 24.75 %. Let’s re-examine our tree diagram now.
- 75 -
Adding up these percentages, we have the sum 0.18 + 0.2475 + 0.0225 + 0.275 + 0.2475 +
0.0275 or 1, thus 100 percent of the whole population. The republicans comprise 0.2475 +
0.2475 or 49.5 % of the respondents. Notice, half of the Republicans declared themselves Rightto-life, the other half were Pro-choice.
Now, let’s ask some of our own questions.
1. Given a democrat, what was the probability he or she was Pro-Choice? Right-to Life?
Solution
Written in terms of conditional probability, we have
P( PC D)
0.275
P( PC | D) 

 0.604 and
P ( D)
0.18  0.275
0.18
P( RTL | D) 
 0.396 or 1  0.694  0.396
0.18  0.275
2. Given a republican, what was the probability he or she was Pro-Choice? Right-to-Life?
Given a democrat, what was the probability he or she was Pro-Choice? Right-to Life?
Solution
Written in terms of conditional probability, we have P( PC | R) 
P( RTL | R) 
0.2475
 0.5 and
.2475  0.2475
0.2475
 0.5
.2475  0.2475
3. Given some one who was Right-to-Life, what was the probability he or she was a democrat?
A republican?
Given a democrat, what was the probability he or she was Pro-Choice? Right-to Life?
Solution
Written in terms of conditional probability, we have
P( D RTL) (0.45)(0.40)
P( D | RTL)

 0.40 and P( R | RTL )  0.55
P( RTL)
0.45
4. Given someone was Pro-Choice, what was the probability he or she was a democrat? A
republican?
Given a democrat, what was the probability he or she was Pro-Choice? Right-to Life?
Solution
Written in terms of conditional probability, we have P( D | PC )  0.50 and P( R | PC )  0.45
- 76 -
It seems now you are armed with enough information to ask your self if the abortion issue falls
along party lines or is bi-partisan (transcends party lines).
Using the mathematics developed in the previous example we can quickly assess the smoking
example. Recall, 15% of the population developed lung cancer and 40 % of the county’s
residents were smokers, we knew that 80% of those who developed lung cancer were smokers.
But, we wanted to know given the probability for someone who smoked to develop lung cancer?
You want to know the precise risk involved in smoking.
Solution
We want to know, of those who smoke (given someone smokes), what is the probability from
this group that someone will develop ling cancer.
P( A  B)
We will use P( B | A) 
. We want
P( A)
P( S  LC )
P( Lung Cancer | Smo ker)  P( LC | S ) 
P( S )
What do we know? Well, 15% of the population developed lung cancer can be written P(LC) =
0.15 and 40 % of the county’s residents were smokers can be written P(S) = 0.40. Also, 80% of
those who developed lung cancer were smokers can be written
P(S | LC) = 0.80. So,80% of 15% of the populations developed lung cancer.
P( S  LC ) (0.15)(0.80)
P( LC | S ) 

 0.30 or 30 % risk.
P( S )
0.40
- 77 -
Exercise Set
For problems 1-7, use the following a
popular opinion poll about the upcoming
presidential election, where 30 % responded
they were Democrat (D), 45 % responded
they were Republican (R), and surprisingly
25% said they were Independent (I). After
the election, it turned out 65 % of the
Democrats voted (v) against the ban of
liquor on Sundays, 80 % of the Republicans
voted against the ban and only 20 % of the
Independents voted against the ban.
1. What is the probability that someone
voted against the ban?
2. What is the probability that someone who
does not vote against the ban is a Democrat?
3. What is the probability that who was
Independent did not vote against the ban?
4. What is P( D | v) ?
5. What is P ( D | v ) ?
6. What is P (v | D ) ?
7. What is P (v | R ) ?
For problems 8-14, use the following a
popular opinion poll given to students at
Southwestern college regarding the death
penalty, where 20 % of the respondents were
Freshman, (F), 30 % were Sophomores (So),
35 % were Juniors (J) and 15 % were
Seniors (Sr). Of the Seniors, 70 % said they
opposed the death penalty (d ) , of the
Juniors only 30 % opposed the death
penalty, and for the Sophomores, 40 % were
in favor of the death penalty, while half of
the freshmen were in favor and half were
against.
8. What is the probability that a freshman
was against the death penalty?
9. What is the probability that someone who
was against the death penalty was not a
sophomore?
10. What is the probability some who was
in favor of the death penalty was a senior?
11. What is P( So | d ) ?
12. What is P (d | Sr ) ?
13. What is P(d | F or So ) ?
14. What is P( J | d ) ?
15. “In a recent poll, 35 % said that Bobby
Bonds (b) was the greatest baseball player
ever. Of those who thought Bonds was the
greatest ever, 80 % felt that the wide spread
use of steroids has not cheapened (c ) the
very fabric of baseball and that the records,
once hallowed, are still ‘legit’. It should be
noted, though, that 60 % of all polled did
think that the records were cheapened by the
wide spread us of steroids among today’s
athletes.” Is this just an ‘old school, new
skool’ issue, case of the younger generation
feeling one way, the older another. Or is
this reflective of a larger problem is our
society, an ever widening divergence
between generations in what we adhere to as
morality. It is clear that a large percent of
those who thought that someone other than
Barry Bonds was the greatest ever also
thought steroid use among today’s athletes
has cheapened baseball. What was that
percent?
- 78 -
Expected Values
Suppose you are considering purchasing a small insurance company. Is it a profitable venture?
How would you answer such a question? For ease, let us suppose each policy costs $500 per
year. You check the yearly records from the past 20 years and it appears the numbers don’t
really change all that much. Yearly, 9995 out of 10,000 policy holders never result in a claim, 2
out of 10,000 result in a claim of an average of $2,000, 1 out of 10,000 result in a claim of an
average of $5,000 and 2 out of 10,000 result in huge payoffs, claims that average half a million
dollars. Should this scare us off from purchasing the company?
In ordinary English, we are searching for the expected value, and we may find this with respect
to the policy holders or for the company. Let’s find the expected value for the company.

if it is positive, then we should buy the company, because in the long run, for each policy
holder, we will take in more money than we pay out.

If it is negative, then we should not buy the company, because in the long run, for each
policy holder, we will pay out more money than we take in.
Expected values, weighted averages or expected winnings are mathematically all one and the
same. The expected value is found multiplying the event (often stated as an amount) by the
probability that event (or the amount stated) will occur and then adding each subsequent similar
product. The expected value is the result of the sum of all these products.
n
a p a p a
i 1
i
i
1 1
2
p2  a3 p3  ...  an pn , 1  i  n,
n is the number of different events (or outcomes )
where ai is eachindividualevent , and
pi is the probability that " i th " event will occur
So, to find the expected value, we take each claim amount and multiply it by the probability that
claim amount could occur. We then sum together each result.
9995
2
1
2
0
2, 000 
5, 000 
500, 000 100.90
10, 000 10, 000
10, 000
10, 000
We see, annually, for each policy holder, the average to pay off is $100.90 and the policy holder
paid $500. We expect to make $500 - $100.90 = $399.10 per policy holder each year. The next
question would be, is that enough to run the business?
If we had only three clients, the expected value tells 3 x $ 399.10 = $1,197.30 is the predicted
amount our company could expect to net. Who knows if the company will be profitable because
maybe one of the unlikely half a million dollar claims would come through and all we would
- 85 -
have is $1,500 (3 x $500) to draw from. But, if we had 10,000 clients, we could reasonably
expect to take in $399.10 x 10,000 = $3,991,000 or nearly 4 million dollars per year. If we had
25,000 clients, our company could reasonably expect to take in $399.10 x 25,000 = $9,977,500
or nearly 10 million dollars annually. Even though we pay out an average of 2 claims equaling a
half a million dollar each for every 10,000 policy holders, we make money because most of our
policy holders result in us making $500 per year. The more clients we have, the more money we
can expect to flow into the company. But, we need to be aware that to find the expected value, it
does not matter how many policy holders we have, the expected value refers to what is the
expected amount of money the company will either take in or pay off per client.
In general, the theoretical probabilities of experiments or events are experienced after many,
many trials have been conducted. So, if you are playing a con game, play as many games as
possible, if you are taking in clients, get as many as possible, if you are answering questions on
an exam, answer them all with the same strategy. The rule to remember is that the expected
value is more likely to be approached in the long haul.
Example One
Con game On a street corner, you come across a folding table, two chairs and a bag. A street
vendor is smiling. In the bag, there are 10 one dollar bills, 15 five dollar bills, 5 twenty dollar
bills, 2 one hundred dollar bills and 15 blank pieces of paper. You are charged $10 to play. The
player, blind folded, reaches into the bag and withdraws a single item. You keep what draw.
What is the expected value for the player? Would you play?
Solution
10
15
5
2
15
1 5 
20  100 
0  8.19 . You expected value is $8.19 - $10 = - $1.81.
47
47
47
47
47
This number means you may expect to lose $1.81 every time you play the game, even though
there may be occasions where you will win $20 or $100.
CHECK: Note, a good check is too add across the numerator, 10+15+5+2+15 = 47, which
matches the number of pieces in the bag, thus matching the denominator, thus insuring you have
accounted for all probabilities associated with each dollar amount.
Example Two
Standardized Tests We have all taken these huge standardized exam, SAT, ACT, MCAT,
LSAT, ABCTE, PCAT, TDEFL, GRE, GMAT and so on. Often, their scores on these
standardized exams are based on weighted averages, as well as a comparison of your score to
others performance. We will concentrate on the weighted average portion of the exam. The
question, when should one guess if they can eliminate one or more choices is crucial to your
success. This test taking strategy can separate two equal prepared students on a grade scale.
On a two hundred question multiple choice exam, with each question having choices a through e,
this standardized test is graded as follows, for each correct answer you get a point, for each
incorrect answer you lose ½ a point. For each question you leave blank, you gain or lose no
- 86 -
points. What is the expected value and would you guess if there is only one correct answer per
question and if
a) you couldn’t eliminate any answers
b) you could eliminate one answer
c) you could eliminate two answers
d) you could eliminate three answers
Solution
First, the number of questions, 200, is irrelevant because we need the expected value per
question.
4 1
1

 (1)   0.2
a)
2 5
5
You could expect to lose – 0.2 points for each question if you simply guess. Or stated more
clearly, you could expect to lose 1 point for every 5 questions you guess when you can not
eliminate a choice. This would be a poor choice on your part, it is not to your advantage. Don’t
employ this strategy.
3 1
1

 (1)   0.125
b)
2
4
4
You could expect to lose – 0.125 points for each question if you when you can eliminate a
choice. Or stated more clearly, you could expect to lose 1 point for every 8 questions you guess
when you can eliminate a single choice. Don’t employ this strategy, it is not to your advantage.
2 1
1

 (1)  0
c)
2
3
3
You could expect to lose no points for each question. Or stated more clearly, you neither gain
nor lose if you can eliminate two choices. Should you then make an educated guess if you can
safely eliminate two choices, thus knowing the answer is one of three choices? No. Do not
employ a strategy unless it is to your advantage. This is a rule for life, not just math.
1 1
1

 (1)  0.25
d)
2 2
2
You could expect to gain 0.25 points per question. Or stated more clearly, you could expect to
gain 1 point for every 4 questions you guess when you can eliminate three of the five choices.
Should you do this? Absolutely.








But, if you employ this test taking strategy, this is where a 200 question exam does matter.
Employ the strategy consistently. Every time you can eliminate 3 or more choices, then guess.
Remember, expected values are more likely to reach their expected value the more the game is
played. When the game is a standardized test, there is no “fun loving game” connotation to it. It
is very important to you. This translates into the words this is a serious strategy, take it
seriously.
Every time you can eliminate at least three choices, then guess.
It is advantageous to do so.
- 87 -
Example Three
Con Game 2 The game revolves around a standard deck of cards. The game is as follows, the
player draws one card. If the player draws a two through nine, the player gets $5. If the player
draws a ten, the player pays $100. If the player draws a face card (J, Q or K) the player pays $ 2.
If the player draws a red ace, the players get $50, if the player draws the ace of clubs, the player
pays $5 and if the player draws the ace of spades, the player pays $25. What is the expected
value of the game for the player?
Solution
32
4
12
2
1
1
5  (100)  (2)  50  (5)  (25)   3.73 or the player could expect to
52
52
52
52
52
52
lose $3.73 every time the player plays.
Example Four
Your final grade You exam grades are 78, 90, 75 and 92. Each exam is 15 percent of your
grade. Your homework average is a 50, it is 10 percent of your grade. Your project earned you
a 100, which is 5 percent of your grade. You final exam is 25 percent of your grade. You earned
a 75. What is your average? What would your homework average had to have been to earn the
next letter grade higher than the one you earned?
Solution
Your grade is the weighted average or expected value:
78(.15)  90(.15)  75(.15)  92(.15)  50(.10)  100(.05)  75(.25)  79 , a C.
To earn a B, replace the homework grade of a 50 with an x and we replace the overall final
grade with an 80. 78(.15)  90(.15)  75(.15)  92(.15)  x(.10)  100(.05)  75(.25)  80
Solving for x,
0.10 x  80  (78(.15)  90(.15)  75(.15)  92(.15)  100(.05)  75(.25))  6
6
 60
0.10
To earn a B, the homework average would have to be 60 percent.
x
- 88 -
Exercise Set
1. You have the following grades on your
three exams, 75, 84, 91. Your homework
average is a 40. You want to make a C in
the course. What do you need to make on
the final exam if the Test Average is half
your grade, 25 % of the grade is your
homework, and 25 % of your grade is your
final exam.
For problems 2 and 3, use the game where a
bag contains 3 red and 4 blue balls. You pay
7 dollars to play the game:
2. you draw one ball. If it is red I pay you
$5 and if it is blue I pay you $8. Find your
expected gain or loss for this game.
3. you may play further the following way.
Draw another ball without replacing the
first. If both are red I pay you $11, if both
balls are blue I pay you $4 and if the balls
are different colors then I pay you 6 dollars.
Find your expected gain or loss for this
second type of game.
4. On a street corner, you come across a
folding table, two chairs and a bag. A street
vendor is smiling. In the bag, there are 20
one dollar bills, 10 five dollar bills, 2 ten
dollar bills, 6 twenty dollar bills, 2 one
hundred dollar bills and 25 blank pieces of
paper. You charge $ 12 to play. The player,
blind folded, reaches into the bag and
withdraws a single item. You keep what
draw. What is the expected value for the
player? Would you play?
5. The game revolves around a standard
deck of cards. The game is as follows, the
player draws one card. If the player draws a
two or three, the player gets $ 2, four
through nine, the player gets $100. If the
player draws a ten, the player pays $50. If
the player draws a face card (J, Q or K) the
player pays $ 25. If the player draws a red
ace, the players get $200, if the player draws
the ace of clubs, the player pays $5,000 and
if the player draws the ace of spades, the
player pays $ 1. What is the expected value
of the game for the player?
For problems 6 and 7, use the following
data: In the past five years, your life as a
realtor in the summer months (June, July
and August) has been quite predictable.
Below is a table of the expected
commissions and the probability each
commission will occur.
Expected Commissions for the summer months and
the probability they will occur.
$1,000 $2,000 $3,000 $4,000 $5,000
40 %
20%
15 %
15 %
10%
6. What is your expected commission per
sale for those hot summer months?
7. If you made 1 sale all summer, what your
income be for the summer? How about if
you made 4 sales? 10 sales? 20 sales?
For problems 8 and 9, use the following
data: In the past five years, your life as a
realtor in the winter months (December,
January and February) has been just as
predictable. Below is a table of the expected
commissions and the probability each
commission will occur.
Expected Commissions for the winter months and the
probability they will occur.
$1,000 $2,000 $3,000 $4,000 $5,000
30 %
10%
5%
25 %
30%
8. What is your expected commission per
sale for those cold winter months?
9. If could vacation in the summer or the
winter, when would you vacation?
For problems 10 – 12, five cards are dealt
from a standard 52 card deck. Find the
expected value and state whether or not you
would play the game.
1
10. You get $ 1,000,000 for a flush. You
pay $ 10 for any other hand dealt to you.
- 89 -
11. You get $ 1,000,000 for a royal flush.
You pay $ 1 for any other hand dealt to you.
12. You get $ 20 for a straight flush. You
get $ 100 for a pair. You pay 0.25 for any
other hand dealt to you.
13. On a 155 question multiple choice
exam, with each question having choices a
through f, this standardized test is graded as
follows, for each correct answer you get 3
points, for each incorrect answer you lose 2
points. For each question you leave blank,
you gain or lose no points. What is the
expected value and would you guess if there
is only one correct answer per question if
a) you couldn’t eliminate any answers
b) you could eliminate one answer
c) you could eliminate two answers
d) you could eliminate three answers
For problems 14 – 17. Your grade is
calculated on a standard scale: 0 – 59 is an
F, 60 – 69 is a D, 70 – 79 is a C, 80 – 89 is a
B and 90 – 100 is an A. You final grade is
calculated where your exam average is
worth 50 % of your grade, your homework
is worth 25 %of your grade, and the final
exam is 25 % of your grade. You had 3
exams, your scores were 89, 90 and 85.
Your final exam grade was a 92. Your
homework grades were the average of the
nine homework assignments, where you
earned a 50, 60, 30, 40, 70 and did not
complete the other 4 assignments.
14. What is your final course average?
15. What would your final course average
have been if your homework grades were
80, 70, 80, 70, 80, 70, 90, 90, 70?
16. What would your final course average
be if your homework grades were 80, 90, 80,
90, 80, 100, 100, 100, 100?
17. Is there a point to questions 14 – 17?
- 90 -