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PART
Randomness
and Probability
CHAPTER 13
Let Me Count the Ways
CHAPTER 14
What Are the Chances?
CHAPTER 15
Probability Rules!
CHAPTER 16
Probability Models
IV
CHAPTER
13
Let Me Count
the Ways
E
arly humans saw a world filled
with random events. To help
them make sense of the chaos
around them, they sought out seers, consulted oracles, and read tea leaves. As
science developed, we learned to recognize some events as predictable. We can
now forecast the change of seasons, tell
when eclipses will occur precisely, and
even make a reasonably good guess at
how warm it will be tomorrow. But many
other events are still essentially random.
Will the stock market go up or down
today? When will the next car pass this
corner? And we now know from quantum
mechanics that the universe is in some sense random at the most fundamental levels of
subatomic particles.
But we have also learned to understand randomness. The surprising fact is that in the
long run, even truly random phenomena settle down in a way that’s consistent and predictable.
It’s this property of random phenomena that makes the next steps we’re about to take in
Statistics possible.
284
DEALING WITH RANDOM PHENOMENA
285
Dealing with Random Phenomena
Light
% Green so far
1
2
3
4
5
6
o
Green
Red
Green
Green
Red
Red
o
100
50
66.7
75
60
50
o
100.0
75.0
Percent Green
Day
Every day you drive through the intersection at College and Main. Even
though it may seem that the light is never green when you get there, you
know this can’t really be true. In fact, if you try really hard, you can recall
just sailing through the green light once in a while.
What’s random here? The light itself is governed by a timer. Its pattern isn’t
haphazard. In fact, the light may even be red at precisely the same times each
day. It’s the pattern of your driving that is random. No, we’re certainly not
insinuating that you can’t keep the car on the road. At the precision level of the
30 seconds or so that the light spends being red or green, what time you arrive
at the light is random. Even if you try to leave your house at exactly the same
time every day, whether the light is red or green as you reach the intersection
is a random phenomenon.1
Is the color of the light completely unpredictable? When you stop to
think about it, it’s clear that you do expect some kind of regularity in your
long-run experience. Some fraction of the time, the light will be green as you
get to the intersection. How can you figure out what that fraction is?
Suppose you record what happens at the intersection each day. The
first day, it was green. Then on the next day it was red, the day after that green
again, then green, red, and red. If you plot the percentage of green lights
against days, the graph would start at 100% (because the first time, the light
was green, so 1 out of 1, for 100%). Then the next day it was red, so the
accumulated percentage dropped to 50% (1 out of 2). The third day it was
green again (2 out of 3, or 67% green), then green (3 out of 4, or 75%), then red
twice in a row (3 out of 5, for 60% green, and then 3 out of 6, for 50%), and so
on. A graph of the accumulated percentage of green lights looks like this:
50.0
25.0
0
20
40
60
Day Number
80
100
FIGURE 13.1 Green lights over time. The overall percentage
of times the light is green settles down as you see more outcomes.
As you collect a new data value for each day, each new outcome becomes
a smaller and smaller fraction of the accumulated experience, so, in the long
run, the graph settles down. As it settles down, you can see that, in fact, the
light is green about 35% of the time.
1
If you somehow managed to leave your house at precisely the same time every day and
there was no variation in the time it took you to get to the light, then there wouldn’t
be any randomness, but that’s not very realistic.
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CHAPTER 13
Let Me Count the Ways
The Law of Large Numbers
PROBABILITY
For any event A,
P(A) =
# times A occurs
total # of trials
in the long run.
What’s the probability of a green light at College and Main? Based on the graph,
it looks like the relative frequency of green lights settles down to about 35%, so
saying that the probability is about 0.35 seems like a reasonable answer.
This illustrates a principle called the Law of Large Numbers (LLN) for
events that are independent. (Informally, independent means that the outcome
of one trial doesn’t affect the outcomes of the others.) The LLN says that as
the number of independent trials increases, the long-run relative frequency of
repeated events gets closer and closer to some value.
The LLN guarantees that relative frequencies settle down in the long run,
and we call the value that they approach the probability of the event. If the
relative frequency of green lights at that intersection settles down to 35% in
the long run, we say that the probability of encountering a green light is 0.35,
and we write P(green) = 0.35.
The Nonexistent Law of Averages
Don’t let yourself think
that there’s a Law of
Averages that promises
short-term compensation
for recent deviations
from expected behavior.
A belief in such a “Law”
can lead to money lost in
gambling and to poor
business decisions.
“Slump? I ain’t in no slump.
I just ain’t hittin’.”
—Yogi Berra
Even though the LLN seems natural, it is often misunderstood because the
idea of the long run is hard to grasp. Many people believe, for example, that
an outcome of a random event that hasn’t occurred in many trials is “due” to
occur. Many gamblers bet on numbers that haven’t been seen for a while,
mistakenly believing that they’re likely to come up sooner. A common term
for this is the “Law of Averages.” After all, we know that in the long run, the
relative frequency will settle down to the probability of that outcome, so now
we have some “catching up” to do, right?
Wrong. The Law of Large Numbers says nothing about short-run behavior. Relative frequencies even out only in the long run. And, according to the
LLN, the long run is really long (infinitely long, in fact).
The so-called Law of Averages doesn’t exist at all. But you’ll hear people
talk about it as if it does. Is a good hitter in baseball who has struck out the
last six times due for a hit his next time up? If you’ve been doing particularly
well in weekly quizzes in Statistics class, are you due for a bad grade? No.
This isn’t the way random phenomena work. There is no Law of Averages for
short runs.
The lesson of the LLN is that sequences of random events don’t compensate in the short run and don’t need to do so to get back to the right long-run
probability. If the probability of an outcome doesn’t change and the events
are independent, the probability of any outcome in another trial is always
what it was, no matter what has happened in other trials.
Coins, keno, and the law of averages You’ve just flipped a fair coin
and seen six heads in a row. Does the coin “owe” you some tails? Suppose you spend
that coin and your friend gets it in change. When she starts flipping the coin, should
she expect a run of tails? Of course not. Each flip is a new event. The coin can’t
“remember” what it did in the past, so it can’t “owe” any particular outcomes in the
future.
MODELING PROBABILITY
287
Just to see how this works in practice, we ran a computer simulation of 100,000
flips of a fair coin. We collected 100,000 random numbers, letting the numbers 0 to
4 represent heads and the numbers 5 to 9 represent tails. In our 100,000 “flips,”
there were 2981 streaks of at least 5 heads. The “Law of Averages” suggests that
the next flip after a run of 5 heads should be tails more often to even things out.
Actually, the next flip was heads more often than tails: 1550 times to 1431 times.
That’s 51.9% heads, hardly what the “Law of Averages” suggests (and very close
to the theoretical 50%).
Keno is a simple casino game in which numbers from 1 to
80 are chosen. The numbers, as in most lottery games, are supposed to be equally likely. Payoffs depend on how many of
those numbers you match on your card. A group of graduate
students from a Statistics department decided to take a field
trip to Reno. They (very discreetly) wrote down the outcomes
of the games for a couple of days, then drove back to test
whether the numbers were, in fact, equally likely. It turned out
that some numbers were more likely to come up than others. Rather than bet on
the Law of Averages and put their money on the numbers that were “due,” the
students put their faith in the LLN—and all their (and their friends’) money on the
numbers that had come up more often before. After they pocketed more than
$50,000, they were escorted off the premises and invited never to show their faces
in that casino again.
THE LAW OF AVERAGES
IN EVERYDAY LIFE
“Dear Abby: My husband
and I just had our eighth
child. Another girl, and I
am really one disappointed
woman. I suppose I should
thank God she was healthy,
but, Abby, this one was
supposed to have been a
boy. Even the doctor told
me that the law of averages
was in our favor 100 to
one.” (Abigail Van Buren,
1974. Quoted in Karl
Smith, The Nature of Mathematics. 6th ed. Pacific
Grove, CA: Brooks/Cole,
1991, p. 589)
The Law of Large Numbers.
Watch the relative frequency
of a random event approach
the true probability in the
long run.
Just Checking
1. One common proposal for beating the lottery is to note which numbers
have come up lately, eliminate those from consideration, and bet on numbers that have not come up for a long time. Proponents of this method argue that in the long run, every number should be selected equally often, so
those that haven’t come up are due. Explain why this is faulty reasoning.
(Check your answers on page 304.)
Modeling Probability
Maybe you don’t feel like waiting around for an infinite number of trials to
find out a probability? Thankfully, there are other ways. Probability was first
studied extensively by a group of French mathematicians who were interested
in games of chance.2 Rather than experiment with the games (and risk losing
their money), they developed mathematical models of theoretical probability.
To make things simple (as we usually do when we build models), they started
by looking at games in which the different outcomes were equally likely. Fortunately, many games of chance are like that. Any of 52 cards is equally likely to
be the next one dealt from a well-shuffled deck. Each face of a die is equally
likely to land up (or at least it should be).
2
Ok, gambling.
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CHAPTER 13
Let Me Count the Ways
A phenomenon consists
of trials. Each trial has
an outcome. A sample
space is a list of all possible outcomes. Outcomes
combine to make events.
Let’s define some terms. In general, each time we observe a random
phenomenon is called a trial. We call that result the trial’s outcome. (If this
language reminds you of simulations, that’s not unintentional.) Rolling a die
is a trial. Seeing it land a 5 is an outcome.
We sometimes talk about the collection of all possible outcomes and call
that event the sample space.3 We’ll denote the sample space S. For the die,
S = 51, 2, 3, 4, 5, 66.
Often we’re more interested in a combination of outcomes rather than in
the individual ones. Suppose you roll a die, needing a 5 or 6 to win a game.
When we combine outcomes like that, the resulting combination is an event.4
It’s easy to find probabilities for events that are made up of several equally
likely outcomes. We just count all the outcomes that the event contains. The
probability of the event is the number of outcomes in the event divided by
the total number of possible outcomes. We can write
P1A2 =
# outcomes in A
.
# of possible equally likely outcomes
For example, the probability of drawing a face card (JQK) from a deck is
NOTATION ALERT
We often use capital
letters—and usually from
the beginning of the
alphabet—to denote events.
We always use P to denote
probability. So,
P(A) = 0.35
means “the probability of
the event A is 0.35.”
P(face card) =
# face cards
12
3
=
=
.
# cards
52
13
This may seem like a pretty straightforward way to find a probability, but
there’s a critical issue lurking here: the outcomes you count must be equally
likely. Consider: either there’s an earthquake today where you live or there
isn’t. One out of the two possible outcomes is “Earthquake”. That doesn’t
mean the probability you experience an earthquake today is 50%! Before you
count outcomes to find a probability, be sure to check the
Equally Likely Condition: The outcomes being counted are all
equally likely to occur.
For Example
Sample Spaces and Probability
In lots of games you roll two dice, then do something based on the
total. It’s easy to determine the probabilities of various outcomes.
Question: What’s the probability of rolling a total of 7?
First I’ll make a sample space showing all the possible
outcomes when rolling two dice. I could list the totals—
{2,3,4,5,6,7,8,9,10,11,12}—but they’re not all equally likely.
After all, there’s only one way to get a total of 2, but lots
of ways to get a total of 7.
3
Mathematicians like to use the term “space” as a fancy name for a set. Sort of like
referring to that closet colleges call a dorm room as “living space.” But remember
that it’s really just the set of all outcomes.
4
Each individual outcome is also an event.
H O W T O C O U N T ( C L E V E R LY ! )
✔ Equally Likely Condition: Assuming the dice are fair, all
36 possible outcomes are equally likely.
There are 6 ways to roll a total of 7, so P(7) =
6
1
= .
36
6
Dice
Total
1
2
Second Die
3 4 5
6
1
2
3
4
5
6
2
3
4
5
6
7
3
4
5
6
7
8
4 5 6
5 6 7
6 7 8
7 8 9
8 9 10
9 10 11
7
8
9
10
11
12
First Die
It’ll be more helpful to list all the pairs of numbers the two dice
could show—as in {(1,1), (1,2), (1,3), and so on, through (6,6)}—
but that’s a pain. This table is a much easier way to see the
sample space.
289
Just Checking
2. Suppose a family has two children. In order of birth, they may be a boy
and a boy, a boy and then a girl, a girl and then a boy, or a girl and a
girl. Let’s represent those outcomes as the sample space
S = 5BB, BG, GB, GG6.
a) What are we assuming in thinking that these four outcomes are
“equally likely”?
b) What’s the probability a 2-child family has two girls?
c) What’s the probability there’s at least one girl?
d) What’s the probability both children are the same sex?
3. Now let’s think about families with three children.
a) Create a sample space of equally likely outcomes.
b) What’s the probability a 3-child family has at least one girl?
c) What’s the probability that there are both boys and girls in the
family?
(Check your answers on page 304.)
How to Count (Cleverly!)
Finding the probability of any event when the outcomes are equally likely is
straightforward, but not necessarily easy. Sure, if you need to roll a 5 or 6 on
a die to win a game, P1win2 = 26 because the die has 6 equally likely faces
and 2 of them are winners. Simple enough. It gets hard though when the
number of outcomes in the event (and in the sample space) gets big. Think
about flipping two coins. The sample space is S = 5HH, HT, TH, TT6 and
each outcome is equally likely. So, what’s the probability of getting exactly
one head and one tail? Let’s call that event A. Well, there are two outcomes
in the event A = 5HT, TH6 out of the 4 possible equally likely ones in S, so
P(A) = 24, or 12 .
OK, now flip 100 coins. What’s the probability of exactly 67 heads? Well,
first, how many outcomes are in the sample space? S = {HHHHHHHHHHH . . .
H, HH . . .T, . . .} Hmm. A lot. In fact, there are 1,267,650, 600,228,229,
401,496, 703,205,376 different outcomes possible when flipping 100 coins.
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Let Me Count the Ways
To answer the question, we’d still have to figure out how many ways there are
to get 67 heads!
Yikes! If you’re thinking (hoping?) there has to be an easier way, you’re
right. Let’s not go to the trouble of listing the outcomes in the sample space
and then counting in the usual 1, 2, 3, . . . way. There are other ways of
counting that are quite clever, and far more efficient.
Suppose you stop at a local café and find that the Lunch Special menu
lists 4 salads and 5 sandwiches. For $5 you can have any salad or sandwich
for lunch. We don’t need to list all the options one by one to know that you
have 4 + 5 = 9 choices. The key word here was “or”; that’s what told us we
could add the number of salads and the number of sandwiches to see how
many outcomes were possible. The basic rule is “or means add,” but math
folks like to make these things look way more complicated:
Fundamental Counting Principle (Part 1: OR)
If event A has m outcomes and event B has n different outcomes,
then the number of outcomes in event A or B is m + n.
For bigger eaters this café offers the Hungry Special: any salad and any
sandwich for $8. How many different lunches are possible now? Imagine
starting to make the list: spinach salad & roast beef sandwich, spinach salad
& BLT, and so on. There are 5 different lunches that start with the spinach
salad, and then another 5 with the fruit salad paired with each sandwich,
and so on. Because we’ll need to include every salad/sandwich combo, we’ll
end up listing each of the 4 salads with each of the 5 sandwiches; 4 groups of
5 means we’d have 4 # 5 = 20 different possibilities.
The key word this time was “and”; that’s what led us to multiply the number of salads times the number of sandwiches to see how many outcomes
were possible. Here the basic rule is “and means multiply.” We know you just
can’t wait to see the formal statement, so here you are:
Fundamental Counting Principle (Part 2: AND)
If event A has m outcomes and independent event B has n outcomes,
then the number of outcomes in event A and B is mn.
For Example
Counting License Plates
Colorado license plates like the one shown consist of three digits and then three letters.
Question: Using this pattern of letters and numbers, how many
different license plates are possible?
PUBLIC
SCHOOLS
SUPPORT E D U C AT I O N
The first space could be any of 10 different digits (0, 1, 2,
3, . . . , 9) and the second space any of 10, and the third,
too. The fourth space and the fifth space and the sixth could each be any of 26 letters. Since “and”
means multiply, the total number of possibilities is
10 # 10 # 10 # 26 # 26 # 26 = 17,576,000 different license plates.
ON WINNING THE LOTTERY
291
Just Checking
4. Chances are you use a password of some kind
every day, whether you’re getting cash from an
ATM, logging in to get your email, or checking up
on your friends on Facebook. One factor that
helps make these codes secure is the large number
of possibilities. In each of the situations below,
how many different passwords are there?
c) The email login code can alternate letters and
digits with either a letter or a digit first (examples: a7r3w or 8m8k3).
d) That email login code can be any mixture of letters and digits in any order, and is case-sensitive
(example: mp27g isn’t the same as mP27g).
a) The access code for an ATM is 4 digits
(example: 3551).
b) The login code for an email site is 5 characters,
alternating letter-digit-letter-digit-letter (example: a7r3w).
(Check your answers on page 304.)
On Winning the Lottery
NOTATION ALERT
An exclamation point
always means “factorial.”
For example,
3! = 3 # 2 # 1 = 6
States run lotteries to make money. They get people to pay (or “play,” as the
ads like to say) by offering bazillion dollar payouts. “All” you have to do5 is
pick a few numbers. Hit the right ones, and you’re a millionaire. It’s easy,
right—just “a dollar and a dream”? But what are your chances, really? To
find out, we’ll need to do some even cleverer counting.
Step 1: Permutations: Permu—what? Don’t panic, “permutations” is just
a big word for “orders” or “arrangements.” Let’s say there are 8 runners in a
race to determine who advances to the State Championships. We wonder, in
how many different orders can they cross the finish line? Our Fundamental
Counting Principle can answer that question. Any one of the 8 runners
could be first, and then any one of the other 7 second, and any one of the
remaining 6 third, and so on. Remember, and means multiply, so there
are 8 # 7 # 6 # 5 # 4 # 3 # 2 # 1 = 5040 different orders in which the runners can
finish.
A product like this, starting with some number n and multiplying by each
smaller number down to 1, is called n factorial. Factorials are used so often
for counting that there’s a special notation, too: n!. Yes, that’s an exclamation
point—as in 8! = 5040. For some practice with factorials, visit the Math Box
on the next page. (Then come back!)
At a track meet we’re usually not interested in the entire order of finish
for a race, just who comes in first, second, and third. To count those possibilities, we start out just like a factorial, but stop after just 3 factors:
8 # 7 # 6 = 336. That’s called “permutations of 8 things taken 3 at a time,” and
yes, there’s special math notation for it: 8P3. Our factorial notation works so
well that we can’t resist using it here, too. If we start with 8! we need to chop
off the 5 # 4 # 3 # 2 # 1 part. Dividing 8! by 5! cancels out the factors we don’t
want, leaving just what we were after:
8P3
5
=
8!
8#7#6#5#4#3#2#1
= 8 # 7 # 6 = 336.
=
5!
5#4#3#2#1
They make it sound so easy.
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Let Me Count the Ways
Do The Math
You’ll often rely on counting methods to find probabilities, and that means
you’ll be using factorials a lot. For practice, try evaluating these:
1.
2.
3.
4.
5.
6.
7.
8.
9.
4!
6!
5!
3!
10!
(Hint: Don’t multiply; lots of factors cancel out.)
9!
20!
18!2!
10!
4!6!
100!
(Express the answer as a factorial.)
100
Look again at problem 7. What happens when you divide the factorial of a
51! 6!
? ?
number by the number itself? As a factorial, what’s
51 6
1!
Based on your answer to problem 8, what’s as a factorial, and as a number?
1
(Check your answers on page 304.)
NOTATION ALERT
nPr is pronounced “permutations of n objects r at a
time” and represents the
number of different orders
in which we could arrange
any r items out of a group
of n different items.
We call the number of ways to arrange any r items from a group of n different
n!
items permutations of n objects r at a time, and calculate nPr =
.
(n - r)!
Evaluating permutations is usually easier than it looks, because with factorials in both numerator and denominator, lots of numbers cancel out.
There’s just one thing you may find a little strange at first. In our first look at
the race we cared about the order of finish for all 8 runners. Think of that as
permutations of 8 people all 8 at a time, and apply our new formula:
8P8
=
8!
8!
= .
(8 - 8)!
0!
Hmmm . . . what’s 0! mean? Look at it this way: there’s only one way to
8!
8!
arrange no objects: don’t! Therefore 0! = 1, and 8P8 =
=
= 8!, as we
0!
1
found earlier.
Just Checking
5. You just downloaded 5 new albums to your iPod,
and are eager to listen to them. How many different orders can you play them in?
6. You take your iPod to the beach, and select a
playlist of your 8 favorite albums to be played in
random order. You’ll be there long enough to hear
5 of them. How many different orders of albums
could you get to listen to?
(Check your answers on page 304.)
ON WINNING THE LOTTERY
293
Step 2: Combinations. Sometimes we don’t care what order things are
in; all that matters is which ones are selected. Take that qualifying race with
8 runners, for example. If the fastest 3 competitors will move on to the State
Championships, it really doesn’t matter whether someone finishes first,
second, or third—a person just wants to be in the top 3.
We figured out that there could be 8P3 = 336 possible orders of the first
three finishers. Let’s say the winners were Anne, Bonita, and Celeste.6 Egos
aside, the orders ABC, ACB, BAC, BCA, CAB, and CBA are all the same: these
3 girls advance to the next meet. In fact, every possible group of three qualifiers
is counted 3! = 3 # 2 # 1 = 6 times. By dividing the total number of permutations
336
8P3
by this number of repetitions we learn that there are
=
= 56 different
3!
6
possible selections.
NOTATION ALERT
nCr is pronounced “combinations of n objects choosing r at a time” and
represents the number of
different ways to select any
r items out of a group of
n different items, in any
order.
In general, we call the number of ways to select any r items from a group of
n different items combinations of n objects chosen r at a time, and calculate
n!
.
nCr =
r!(n - r)!
Again, lots of numbers in these calculations cancel out because of the factorials. For our track example, it works like this:
8C3
=
8!
8#7#6#5#4#3#2#1
8#7#6
= # # # # # # # = # # = 56
3!5!
3 2 1 5 4 3 2 1
3 2 1
And now we’re finally ready to answer the question we started with: What are
your chances of winning the lottery? (Here’s a hint: don’t go spending that
jackpot yet!)
Wait! Should I use permutations or combinations? To count correctly, you need to be able to make that decision. Here are the key questions to ask
yourself:
• Do I care what order things happen in?
If the arrangement matters, use permutations.
• Or do I just care which ones are chosen?
If it’s only the selection that matters, use combinations.
For Example
Lotteries (and lightning)
The betting slips for New Jersey’s Pick-6 Lotto game offer a field
of numbers from 1 to 49. Each bettor chooses any 6 of them. You
win the grand prize if your six numbers match those randomly
chosen on TV by the NJ Lottery.
Question: How many different selections are possible?
To win, it only matters that I select the right numbers, not
what order they’re in, so I’ll use combinations. The number of
ways to choose any 6 numbers from the 49 offered is
49C6
=
49 # 48 # 47 # 46 # 45 # 44
49!
=
= 13,983,816
6!43!
6#5#4#3#2#1
That means I have about 1 chance in 14 million of winning the Pick-6 Lotto!
(continued)
6
Isn’t it strange how people in math books always have names starting with A, B, and C?
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Let Me Count the Ways
(For comparison, the National Weather Service reports that in any year about 400 of the 300 million
Americans—about 1 in every 700,000—are struck by lightning. That makes you about 20 times as likely
to be zapped as to win the lottery! The ads say you can’t win if you don’t buy a ticket. That’s true, of
course, but the ads don’t say that buying a ticket improves your chance of winning by only a ridiculously
tiny amount.)
Just Checking
7. Don’t do any arithmetic—just tell whether you’d
count these outcomes using permutations or
combinations.
a) How many 13-card bridge hands could you be
dealt from a deck of cards?
b) How many 3-turn combinations could a
padlock have if there are 60 numbers on the
dial and combinations can’t have repeated
numbers?
c) How many batting orders could a baseball manager create using the 9 players in his starting
lineup?
d) How many 6-player volleyball squads could a
coach put on the floor if there are 9 women on
the team?
8. A county legislature consists of 13 elected
representatives, 8 Democrats and 5 Republicans.
They’re setting up a 4-person committee to study
the proposal to build a new library. How many different committees could be formed if the group
will consist of:
a) 4 Republicans?
b) 4 Democrats?
c) 2 Democrats and 2 Republicans? (Remember,
“and” means multiply.)
d) 3 Democrats and 1 Republican?
e) 2 or 3 Democrats (and the rest Republicans).
(Check your answers on page 304.)
Combinations and Probability
In our quest to create models for probability we’ve developed two important
insights:
• When a sample space consists of equally likely outcomes, the probability
of an event is the number of ways the event can happen divided by the
total number of possible outcomes.
• We don’t have to actually list all the outcomes to find out how many there
are. It may be easier to just count them using permutations, combinations,
or the Fundamental Counting Principle.
Together these provide us with some clever ways to determine probabilities.
In the last chapter we asked whether there was something fishy about a
college dorm lottery. A total of 57 students, including 20 people on the same
varsity team, applied for just 3 very desirable rooms. When all 3 winners
turned out to be team members, the other students cried foul. Is it reasonable
to believe this could have happened by chance? Or would it be really unusual
if the process were fair?7 Earlier we used a simulation to help us think about
this issue. Now let’s find the probability that all the rooms would go to the
teammates if everything were done fairly.
7
Read that “Is it statistically significant?”
C O M B I N AT I O N S A N D P R O B A B I L I T Y
295
The dorm lottery sample space would consist of all the ways to choose
57!
any 3 students from the 57 applicants. There are 57C3 =
= 29,260
3!54!
8
possible outcomes, all equally likely if the selection were truly random.
Of those outcomes, how many consist of team members only? That event
includes all the ways to pick any 3 athletes from the 20 varsity applicants:
20!
= 1,140. And therefore:
20C3 =
3!17!
P(3 team winners) =
# of ways to choose 3 team members
=
total # of possible choices
20C3
57C3
1140
=
L 0.039
29260
So, was the lottery fair? Well, there’s only about a 4% chance that all 3 rooms
would have gone to team members by chance. This makes the outcome look
very suspicious, but the probability is high enough that we can’t prove for
sure that someone cheated.
STEP-BY-STEP EXAMPLE
Using Combinations to Find a Probability
A video rental store is running a “DVD Grab Bag” special. For $20
you can buy a bag containing 6 used DVDs randomly selected from
15 recently popular movies. The possible titles listed in the store’s ads
included 4 comedies, 8 dramas, and 3 animated features.
Question: If you buy one of the random Grab Bags, what’s the probability you get two comedies, three dramas, and one animated flick?
Plan Organize the information given and
what is asked for.
I am being asked to find the probability that my
Grab Bag contains
THINK
2 of the 4 comedies
3 of the 8 dramas, and
1 of the 3 animations, for a total of
6 of the 15 DVDs.
Decide which counting method you’ll
use. Remember:
permutations = arrangements,
combinations = choices.
I don’t care what order these DVDs come in,
just which ones were selected for my Grab Bag.
I’ll use combinations to count the number of
possibilities.
Check the conditions.
✔
Equally Likely Condition. The movies are
being chosen at random, so all the
outcomes are equally likely.
(continued)
8
Aren’t you glad we don’t have to list them all?
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CHAPTER 13
Let Me Count the Ways
Mechanics Show your work.
The probability of an event A is:
SHOW
P(A) =
# of outcomes in A
.
# of possible outcomes
I am asked to find a probability so I’ll write my
answer as a fraction. The denominator is the
total number of ways to choose any 6 of the 15
movies. That’s 15C16.
The numerator is the number of selections
that include any 2 of 4 comedies, and 4 of 8
dramas, and 1 of the 3 animated movies:
Remember: “and” means multiply.
#
The probability is:
If the messy arithmetic annoys you,
relax. You’ll love the TI Tips, coming
next.
#
4C2 8C3 3C1
P(A) =
=
#
#
4C2 8C3 3C1
15C6
6 # 56 # 3
5005
L 0.201
Conclusion Interpret your results in the
proper context.
TELL
TI Tips
There’s about a 20% chance that my Grab Bag
will have two comedies, three dramas, and one
animated movie.
Counting
Tired of writing down 4, 3, 2, 1 over and over, then canceling and multiplying?
You’ll be happy to know your TI can do the work for you. Let’s try it out.
Factorials.
The qualifying race at the track meet had 8 runners who could finish in 8!
orders.
• Type the 8.
• Hit the MATH button and go all the way to the right to the PRB menu. (You
can “wrap” there using the left arrow.)
• Choose 4:!, then hit ENTER.
Note that factorials can get quite large. With only 15 runners the number
of arrangements is so large the answer requires scientific notation! Try it.
Permutations.
The number of different orders for the top 3 runners could finish in is 8P3. To
evaluate that:
• Type the 8.
• Go to the MATH PRB menu again. This time choose 2:nPr.
• Type the 3, then hit ENTER.
W H AT C A N G O W R O N G ?
297
Combinations.
The number of possible groups of 3 qualifying runners (order of finish doesn’t
matter) is 8C3. Evaluate that the same way you did 8P3, only this time choose
3:nCr from the MATH PRB menu.
Probability.
You can even put a bunch of these calculations in the same expression. Try calculating the probability that those 3 dorm rooms all went to varsity teammates
20C3
by entering
.
57C3
Pretty nice, eh? Take these buttons out for a test drive with the following
Just Checking questions.
4 comedies, 8 dramas, and 3 animated features. Find
the probability you get:
Just Checking
9. The DVD Grab Bag described in the Step-ByStep Example is to contain 6 movies chosen at
random from 15 recently popular titles, including
a) nothing but dramas.
b) 3 of the comedies and 3 of the dramas.
c) all 3 of the animated flicks and any 3 other movies.
(Check your answers on page 304.)
WHAT CAN GO WRONG?
■
■
Don’t mistake the Law of Large Numbers for the so-called “Law of
Averages.” The LLN tells us that in the long run the frequency with which
things happen will approach the true probability. But the “long run”
means forever, not today! People who reason by the “Law of Averages”
think that things have to even out right away. Seeing 6 heads in a row
doesn’t mean a coin is more likely to land tails on the next toss. If you
were to flip the coin 94 more times and get around 47 heads and 47 tails,
you’d end up with a total of 53 heads and 47 tails in 100 tosses. There’s
nothing strange about that. The coin doesn’t compensate for the 6 extra
heads. Rather, whatever happens in the short run gets drowned out in the
long run. There’s no such thing as the “Law of Averages.”
Don’t think that random events are always equally likely. The chance of
winning a lottery—especially lotteries with very large payoffs—is small.
Regardless, people continue to buy tickets. In an attempt to understand
why, an interviewer asked someone who had just purchased a lottery
ticket, “What do you think your chances are of winning the lottery?” The
reply was, “Oh, about 50–50.” The shocked interviewer asked, “How do
you get that?” to which the response was, “Well, the way I figure it,
either I win or I don’t!” The moral of this story is that you must be sure
the events in your sample space are equally likely before thinking about
probability.
(continued)
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Let Me Count the Ways
■
■
Don’t mix up permutations and combinations. These two counting methods seem pretty similar, but the difference is critical. If you need to know
how many orders something can happen in, or how many different ways
you can arrange things, those are permutations. When you don’t care
what order things are in, combinations count the number of possible
choices or selections.
Don’t think a “personal probability” is mathematically valid. What’s the
probability that your grade in this Statistics course will be an A? You may
be able to offer a number that seems reasonable. How did you come up
with this probability? Do you plan on taking the course over and over
(and over . . .), calculating the proportion of times you get an A? Or do
you assume the outcomes are equally likely?
People use the language of probability in everyday speech to express a
degree of uncertainty about an outcome. We call this kind of probability a
subjective or personal probability. Although personal probabilities may
be based on experience, they’re not based either on long-run relative
frequencies or on equally likely events. So they don’t display the kind of
consistency that we’ll need probabilities to have. For that reason, we’ll stick
to formally defined probabilities. You should be alert to the difference.
COUNTING IN YOUR WORLD
Massachusetts Needs New
License Plates
There is only one state in the union with older license
plates on the street than Massachusetts . . . If you
don’t retire the old [green] plates, and restart a new
series of plates, you exhaust many millions of available letter-number combinations.
Consider the math. The first series of license
plates had three letters and three numbers, and the first
digit was never a 0. The letters I, O, and Q were never
used. For math geeks, these rules create 12,167 different letter combinations. Each three digit number can
9
generate 900 plates (numbers 100-999), so the 12,167
letter combinations generate 10,950,300 possible
plates. When Massachusetts exhausted the three-letter
three-number combinations, they simply reversed the
order, generating another 10,950,300 possibilities.
In 1987 the Registry started manufacturing
the red “Spirit of Massachusetts” plates . . . Soon
after the red plates began to appear, the NNNLLL9
combinations were exhausted, and the state went to a
series of two letter-four number plates. Given the
same rules, four number-two letter plates generate
4,761,000 combinations. Reversing the order generates another 4,761,000 plates.
number-number-number-letter-letter-letter
W H AT H A V E W E L E A R N E D ?
Those combinations were exhausted, and the
state then placed the two letters in the middle of the
number combination. That generates another
4,761,000 plates.
This numbering scheme is now exhausted, and
we appear to be going to a scheme of three digits,
two letters, followed by another number. So, here
we are in a state of 6.4 million people, and we have
exhausted more than 37 million possible plate
combinations.
299
http://sacredcod.blogspot.com/2008/09/massachusetts-needs-newlicense-plates.html
WHAT HAVE WE LEARNED?
We’ve learned that probability is based on long-run relative frequencies. We’ve
thought about the Law of Large Numbers and noted that it speaks only of longrun behavior. Because the long run is a very long time, we need to be careful
not to misinterpret the Law of Large Numbers. Even when we’ve observed a
string of heads, we shouldn’t expect extra tails in subsequent coin flips.
Also, we’ve learned some basic tools for modeling probabilities of
outcomes and more complex events. These include
씰 sample spaces,
씰 the Fundamental Counting Principle,
씰 permutations, and
씰 combinations.
Terms
Random phenomenon
A phenomenon is random if we know what outcomes could happen, but not which
particular values will happen.
Law of Large Numbers
The Law of Large Numbers states that the long-run relative frequency of repeated
independent events gets closer and closer to the true relative frequency as the
number of trials increases.
Probability
The probability of an event is a number between 0 and 1 that identifies the likelihood that the event happens. We write P(A) for the probability of the event A.
Independence (informally)
Two events are independent if learning that one event occurs does not change the
probability that the other event occurs.
Trial A single attempt or realization of a random phenomenon.
Outcome
Event
Sample Space
Fundamental Counting Principle
The outcome of a trial is the observed result.
A collection of outcomes. Usually, we identify events so that we can attach probabilities to them. We denote events with bold capital letters such as A, B, or C.
The collection of all possible outcomes.
If event A has m outcomes and event B has n outcomes, then
씰 the number of outcomes in event A or B is m + n;
씰 the number of outcomes in event A and B is mn.
Permutations
We call the number of ways to arrange any r items from a group of n different items
n!
.
permutations of n objects r at a time, calculated nPr =
1n - r2!
(continued)
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Let Me Count the Ways
Combinations
We call the number of ways to select any r items from a group of n different items
combinations of n objects chosen r at a time, calculated
nCr
Equally Likely Condition
THINK
n!
.
r!1n - r2!
The outcomes in a sample space (whether listed or just counted) are all equally
likely to occur. If this condition is satisfied, the probability of an event A is
P1A2 =
Skills
=
number of outcomes in A
number of possible outcomes
씰 Understand that random phenomena are unpredictable in the short term but
show long-run regularity.
씰 Be able to recognize random outcomes in a real-world situation.
씰 Know that the relative frequency of a random event settles down to a value
called the probability. Know that this is guaranteed for independent events by
the Law of Large Numbers.
씰 Recognize when outcomes in a sample space are equally likely.
씰 Recognize the difference between permutations (arrangements) and combina-
tions (selections).
씰 Be able to create a sample space for a random phenomenon.
SHOW
씰 Be able to count how many outcomes an event may have using the Fundamen-
tal Counting Principle, permutations, or combinations.
씰 Be able to use lists or counts of equally likely outcomes to find the probability
of an event.
씰 Know and be able to use the terms “sample space,” “outcomes,” and “events”
TELL
correctly.
씰 Be able to use statements about probability in describing a random phenomenon.
씰 Be able to use probability as evidence that an observed outcome may be un-
usual (statistically significant).
EXERCISES
A
1. Sample spaces. For each of the following, list the
sample space and tell whether you think the events are
equally likely:
a) Draw a card from a well-shuffled deck; record the
suit.
b) Roll two dice; record the larger number.
c) Toss 2 coins; record the order of heads and tails.
d) Flip a coin until you get a head or 3 consecutive
tails; record each flip.
2. Sample spaces. For each of the following, list the
sample space and tell whether you think the events are
equally likely:
a) Shake a coin out of a piggy bank containing a
penny, a nickel, a dime, and a quarter; record the
value.
b) Shake two coins out of the piggy bank in part a;
record the total value.
c) A family has 3 children; record each child’s sex in
order of birth.
d) Toss a coin 10 times; record the length of the
longest run of heads.
3. Roulette. A casino claims that its roulette wheel is
truly random. What should that claim mean?
4. Ping pong. Many state lotteries determine winners by
using a machine that rapidly mixes numbered pingpong balls, eventually allowing some to escape from
the container. What assumptions make this method of
selecting the winning numbers truly random?
5. Survival. A doctor tells a patient just diagnosed with a
serious disease that there’s a 60% chance she’ll live at
least 5 years. Where do you think that probability
comes from?
EXERCISES
6. Rain. The weather reporter on TV makes predictions
such as a 25% chance of rain. What do you think is
the meaning of such a phrase?
7. Winter. Comment on the following quotation:
“What I think is our best determination is it will be a
colder than normal winter,” said Pamela Naber Knox, a
Wisconsin state climatologist. “I’m basing that on a
couple of different things. First, in looking at the past
few winters, there has been a lack of really cold weather.
Even though we are not supposed to use the law of averages, we are due.” (Associated Press, fall 1992, quoted
by Schaeffer et al.)
8. Snow. After an unusually dry autumn, a radio
announcer is heard to say, “Watch out! We’ll pay for
these sunny days later on this winter.” Explain what
he’s trying to say, and comment on the validity of his
reasoning.
9. Cold streak. A batter who had failed to get a hit in
seven consecutive times at bat then hits a gamewinning home run. When talking to reporters
afterward, he says he was very confident that last time
at bat because he knew he was “due for a hit.”
Comment on his reasoning.
10. Fouls. Near the end of a basketball game the team
that’s behind often commits a foul in order to get the
ball back quickly. Of course, they hope the other
team’s player won’t make his free throws. Whom
should they foul? Here are three possible strategies:
• Foul the player who has missed the highest
percentage of his foul shots tonight, because he’s a
cold shooter.
• Foul the player who has made the highest percentage
of his foul shots tonight, because he’s due to miss.
• Foul the player with the lowest season-long foulshooting percentage, because he’s the one most
likely to miss.
Based on your understanding of probability, which is
the best strategy, and why?
11. Crash. Commercial airplanes have an excellent safety
record. Nevertheless, there are crashes occasionally,
with the loss of many lives. In the weeks following a
crash, airlines often report a drop in the number of
passengers, probably because people are afraid to risk
flying. A travel agent suggests that since the law of averages makes it highly unlikely to have two plane crashes
within a few weeks of each other, flying soon after a
crash is the safest time. What do you think? Explain.
12. Crash, part II. If the airline industry proudly
announces that it has set a new record for the longest
period of safe flights, would you be reluctant to fly?
Are the airlines due to have a crash? Explain.
13. Literature.Your American Lit class will read 4 novels
this year, chosen by class vote from a list of 12 possible books offered by the teacher.
a) How many different ways could the course unfold,
given that it probably matters what order you read
the books in.
b) How many different choices of books could the
class make?
301
14. Relay team. A swim coach is considering 7 swimmers
as possible members of a 4-person relay team.
a) How many different choices of 4 swimmers does
the coach have?
b) Given that the relay team’s performance may
depend on what order the athletes swim in, how
many different teams could the coach create?
15. Experiment. A researcher needs to split 12 plants into
two treatment groups of 6 plants each. How many different groupings can be formed?
16. Customer satisfaction. A furniture store routinely calls
some customers to inquire about their buying experience in the showroom, the delivery and setup, and
their overall satisfaction with the purchase. Today the
calls will go to 4 homes randomly selected from last
week’s 21 customers. How many different groups of
customers could the store call?
17. Dice product. You are playing a game that involves
rolling two dice, but instead of using the sum, you’ll
get paid in tokens equal to the product of the numbers
you roll. (For example, if you roll a 3 and a 4, you win
12 tokens.)
a) Create a sample space for the number of tokens
you may win. Be careful to do this in a way that
makes all the outcomes you list equally likely.
b) What’s the probability you win an odd number of
tokens?
18. Telemarketing. Some companies who make those
annoying phone calls to try to sell you something generate the phone numbers randomly.
a) Given that seven-digit phone numbers NNN-NNNN
can’t have a 0 or a 1 as the first digit, how many
different phone numbers are possible (within one
area code)?
b) One company is working your area code tonight,
10 people each making an average of 20 calls per
hour. If it takes your family a half hour to eat
supper, what’s the probability you get a call during
mealtime?
19. Poker. You’re playing poker with some friends, and
are about to be dealt 5 cards from a well-shuffled
standard deck.
a) How many different hands are possible?
b) How many hands contain only hearts?
c) What’s the probability you’ll be dealt a flush
(5 cards all the same suit)?
20. Hearts. You’re playing hearts with some friends, and
are about to be dealt 7 cards from a well-shuffled deck.
a) How many different hands are possible?
b) How many of those hands don’t contain any hearts
at all?
c) What’s the probability that there will be some
hearts in your hand?
B
21. Fire insurance. Insurance companies collect annual
payments from homeowners in exchange for paying to
rebuild houses that burn down.
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CHAPTER 13
Let Me Count the Ways
a) Why should you be reluctant to accept a $300 payment from your neighbor to replace his house
should it burn down during the coming year?
b) Why can the insurance company make that offer?
22. Jackpot. On January 20, 2000, the International Gaming Technology company issued a press release:
(LAS VEGAS, Nev.)—Cynthia Jay was smiling ear to ear
as she walked into the news conference at The Desert
Inn Resort in Las Vegas today, and well she should. Last
night, the 37-year-old cocktail waitress won the world’s
largest slot jackpot—$34,959,458—on a Megabucks machine. She said she had played $27 in the machine when
the jackpot hit. Nevada Megabucks has produced 49
major winners in its 14-year history. The top jackpot
builds from a base amount of $7 million and can be
won with a 3-coin ($3) bet.
a) How can the Desert Inn afford to give away
millions of dollars on a $3 bet?
b) Why did the company issue a press release? Wouldn’t
most businesses want to keep such a huge loss quiet?
23. Blood. There are four basic human blood types, in
order of frequency: O, A, B, and AB.
a) Make a sample space showing the possible pairings
of blood types for a mother and father.
b) Why can’t you use this sample space to find the
probability that a baby has parents whose blood
types match?
24. Cereal. Recall the cereal company that was putting
photos of athletes in boxes of cereal? It puts pictures
of Serena Williams in half the boxes and pictures of
either LeBron James or David Beckham in the others.
Hoping to get a complete set, you buy three boxes of
cereal.
a) Make a sample space showing the possible outcomes
as you open the boxes one at a time.
b) Why can’t you use this sample space to find the
probability that you get one picture of each athlete?
25. Family. A family has 3 children.
a) Create a sample space for the number of boys the
family could have.
b) Explain why counting outcomes in that sample
space can’t tell you the probability that there are
2 boys.
c) Use a different sample space to find the probability
that 2 of the 3 children are boys.
d) What assumptions did you make in using this
sample space to find the probability?
26. Coins. You’re going to toss 4 coins.
a) Create a sample space for the number of tails you
could get.
b) Explain why counting outcomes in that sample
space can’t tell you the probability that you’ll get
exactly 2 tails.
c) Use a different sample space to find the probability
that you’ll toss 2 heads and 2 tails.
27. Bigger die. In the game you’re playing, you’ll roll 2
dice. The number of spaces you get to move will be
the larger of the two numbers you roll.
a) Create a sample space showing the number of
spaces you may get to move.
b) You need to move 5 spaces in order to win the
game. Why can’t you count outcomes in this sample space to find the probability of winning?
c) Use a different sample space to find the probability
that you win the game.
28. Three dice. You are playing a game that involves
rolling a die 3 times in a row.
a) The sample space of all possible outcomes is a long
list, so don’t bother creating it—but how many outcomes are there?
b) You’ll win the game if each roll is a bigger
number than the roll before. (For example, 2-3-6
is a winner, but 1-5-3 and 3-4-4 are not.) List the
outcomes that are winners.
c) What’s the probability that you win?
29. Lottery. A scratch-off lottery ticket has 12 concealed
spaces among which are 4 symbols saying “Win!”. The
person who bought the ticket scratches the black yucky
stuff off 3 spaces, winning an instant $10 if all three are
winners. What’s the probability of winning this game?
30. Second chance lottery. On the scratch-off lottery
ticket described in Exercise 29, 5 of the spaces conceal
symbols saying “Free Play.” Buyers who reveal 3 of
those get another ticket for free. What’s the probability
this happens?
31. Audit. As a check against possible accounting errors,
each day a building supplies company audits 5
recently active accounts. Yesterday they made sales to
18 retail customers and 8 contractors.
a) If they select the 5 accounts to be audited as a simple
random sample from all accounts, what’s the probability that they don’t check any contractor accounts?
b) If they want to audit 3 retail accounts and 2
contractor accounts, what’s a better sampling
design?
c) How many ways are there to select the 3 retail
accounts and 2 contractor accounts?
d) What’s the probability that an SRS would produce
this more representative sample by chance?
32. Border inspections. Three buses, 9 cars, and 5 trucks
are currently waiting in line to enter the United States
at a border checkpoint. Customs agents will select 5
vehicles at random for a more thorough inspection.
a) What’s the probability that a simple random sample
of 5 vehicles doesn’t include any of the buses?
b) If agents want to be sure to inspect one bus, 2 cars,
and 2 trucks, what’s a better sampling design?
c) How many different ways are there to create a
sample of one bus, 2 cars, and 2 trucks?
d) What’s the probability that an SRS would produce
this more representative sample by chance?
33. Poker again. What’s the probability that a 5-card
poker hand contains exactly 3 or 4 queens?
34. Another poker hand. What’s the probability that a
5-card poker hand contains two pairs, one of them a
pair of aces?
EXERCISES
35. Legislature. In a state legislature the elected representatives include 17 Democrats, 13 Republicans, and 4
Independents. What’s the probability that a random
selection of 6 legislators would include 2 of each?
36. Team. On a high school varsity baseball team, 5 players are sophomores, 11 are juniors, and 10 are seniors.
The team selects 4 honorary co-captains by drawing
names from a hat. What’s the probability they pick 2
sophomores and 2 seniors?
37. Legislature revisited. What’s the probability that the
randomly selected group of legislators described in
Exercise 35 would end up being 6 members of the
same party?
38. Second team. What’s the probability that as a result of
the process described in Exercise 36 all 4 co-captains
come from the same class?
39. Motorcycles. A state’s motorcycle license plates are a
random combination of three letters followed by two
numbers (such as BLE-12). Because of possible confusion with numbers 0 and 1, they never use the letters
I, O, or Q.
a) How many different motorcycle plates are possible?
b) What’s the probability that a plate chosen at
random contains all vowels? (Include Y as a vowel.)
40. Trailers. License plates for trailers in one state are a
random combination of two letters followed by four
numbers (such as PN-5423). Because of possible confusion with numbers 0 and 1, they never use the
letters I, O, or Q.
a) How many different license plates are possible?
b) What is the probability that a plate chosen at
random contains all even digits?
C
41. Lotto winners. To win the New Jersey Pick-6 Lotto
jackpot (described in the chapter’s For Example) you
have to successfully match 6 numbers chosen from
the numbers 1 through 49. The lottery also pays
smaller prizes if you match only 3, 4, or 5 of the
numbers.
a) What’s the probability that your 6 choices match
3 of the 6 winning numbers?
b) What’s the probability your ticket wins some prize?
42. Quality control. Each hour a shipping inspector at an
Internet retailer spot-checks 5 outgoing packages
to be sure they are completely sealed. While some
occasional problems are to be expected, he will shut
down the process to look for serious trouble if more
than one of the packages he checks is improperly
sealed. He’s unaware of the fact that the machine stapling the packages shut has begun to misfire sporadically, leaving 7 of the 50 packages that were prepared
this hour unsatisfactorily sealed.
a) What’s the probability that 2 of the defective packages show up in his sample?
b) What’s the probability he detects the problem?
303
43. Job opening. A city recently held interviews for 4 jobs
available. Even though there were 12 Republicans
among the 22 applicants, none of them was hired.
Noting that the mayor is a Democrat, the local Republican Party chairperson has challenged the fairness of
what was supposed to be a non-political process.
What do you think? Assuming that the qualifications
of the applicants don’t depend on their political affiliation, is it reasonable to believe that this outcome
could simply have happened by chance? Support your
conclusion with a statistical argument.
44. Kittens. Are people equally likely to abandon male
kittens as female kittens? If so, you’d think that among
50 kittens recently turned in at a local animal shelter
there should be half males and half females. However, a
friend of yours who works there just told you that the
staff fed the kittens today, all 7 of the ones she picked up
(essentially at random) were female. Does this make you
suspect that female kittens are more likely to be turned
in, or could this have happened purely by chance?
Support your conclusion with a statistical argument.
45. More about permutations. In the chapter you learned
that there are 6! ways to arrange any 6 different
objects—the letters in the word MEDIAN, say. But what
if some of the letters are the same? How many ways
can we arrange the letters in MAXIMUM?
Well, there are 7 letters, so that would be 7! if
they were all different. But that counts MAXIMUM
as a new arrangement (we switched the M’s around
when you weren’t looking). To avoid recounting these
“arrangements,” note that every ordering appears 3! = 6
times because of the matching Ms. We can eliminate all
those duplicates from consideration by simply dividing
7!
by 3!, revealing that there are actually
= 840
3!
distinct permutations of the letters in MAXIMUM.
7!
Similarly, there would be
= 1260 different ways
2!2!
to arrange the letters in DOTPLOT: 7! if they were all
different, reduced by a factor of 2! for the matching
Os and 2! for the matching Ts.
How many ways could the letters in these words
be arranged?
a) SKEWED
b) EXPERIMENT
c) PROBABILITY
d) MINIMUM
e) STATISTICS
f) ASSOCIATION
46. States. Use the counting method described in Exercise
45 to find out how many ways you could arrange the
letters in the names of each of these states.
a) OHIO
b) ALASKA
c) HAWAII
d) ARKANSAS
e) INDIANA
f) MISSISSIPPI
304
CHAPTER 13
Let Me Count the Ways
Answers
Do The Math
1. 24
2. 720
4. a)
b)
c)
d)
3. 20
10 # 10 # 10 # 10 = 10,000
26 # 10 # 26 # 10 # 26 = 1,757,600
1,757,600 + 10 # 26 # 10 # 26 # 10 = 2,433,600
26 capitals ⫹ 26 lowercase letters ⫹ 10 digits ⫽
62 possible characters in each of the 5 positions;
625 = 916,132,832
4. 10
5. 190
6. 210
7. 99!
8. You get the next lower number, factorial:
51!
n!
6!
= 50!;
= 5!
= (n - 1)!.
n
51
6
9.
1!
1!
= 0!, and
= 1, so 0! = 1.
1
1
Just Checking
1. The LLN works only in the short run. The random
methods for selecting lottery numbers have no
memory of previous picks, so there is no change in
the probability that a certain number will come up.
2. a) We assume that babies are 50-50 boy-girl, and
that sexes of babies born in the same family are
independent.
1
3
1
b)
c)
d)
4
4
2
3. a) S = 5BBB, BBG, BGB, GBB, BGG, GBG,
GGB, GGG6
7
3
b)
c)
8
4
5.
5P5
=
5!
= 5 # 4 # 3 # 3 # 1 = 120
0!
6.
8P5
=
8!
= 8 # 7 # 6 # 5 # 4 = 6,720
3!
7. a)
b)
c)
d)
combinations
permutations
permutations
combinations
5!
5#4#3#2
= # # # = 5
4!1!
4 3 2 1
8!
8#7#6#5
= # # # = 70
8C4 =
4!4!
4 3 2 1
8! # 5!
8#7 5#4
#
= # # # = 280
8C2 5C2 =
2!6! 2!3!
2 1 2 1
8! # 5!
#
= 280
8C3 5C1 =
3!5! 1!4!
#
#
8C2 5C2 + 8C3 5C1 = 280 + 280 = 560
8. a) 5C4 =
b)
c)
d)
e)
9. a)
b)
c)
8C6
15C6
#
L 0.0056
4C3 8C3
15C6
#
3C3 12C3
15C6
L 0.045
L 0.044