BOCK C17 0321570448 pp3 - Math 50mathbriones.weebly.com/uploads/8/3/4/0/8340232/chapter_17.pdf · Tiger Woods cards, then when you find one, you’ve reduced the number of remaining

388

CHAPTER

17Probability Models

Suppose a cereal manufacturer puts pictures of famous athletes on cards in boxes ofcereal, in the hope of increasing sales. The manufacturer announces that 20% ofthe boxes contain a picture of Tiger Woods, 30% a picture of David Beckham, andthe rest a picture of Serena Williams.

Sound familiar? In Chapter 11 we simulated to find the number of boxes we’dneed to open to get one of each card. That’s a fairly complex question and onewell suited for simulation. But many important questions can be answered moredirectly by using simple probability models.

Searching for TigerYou’re a huge Tiger Woods fan. You don’t care about completing the whole sportscard collection, but you’ve just got to have the Tiger Woods picture. How manyboxes do you expect you’ll have to open before you find him? This isn’t the samequestion that we asked before, but this situation is simple enough for a probabil-ity model.

We’ll keep the assumption that pictures are distributed at random and we’lltrust the manufacturer’s claim that 20% of the cards are Tiger. So, when you openthe box, the probability that you succeed in finding Tiger is 0.20. Now we’ll callthe act of opening each box a trial, and note that:

u There are only two possible outcomes (called success and failure) on each trial.Either you get Tiger’s picture (success), or you don’t (failure).

u In advance, the probability of success, denoted p, is the same on every trial.Here for each box.

u As we proceed, the trials are independent. Finding Tiger in the first box doesnot change what might happen when you reach for the next box.

Situations like this occur often, and are called Bernoulli trials. Common exam-ples of Bernoulli trials include tossing a coin, looking for defective products rollingoff an assembly line, or even shooting free throws in a basketball game. Just as wefound equally likely random digits to be the building blocks for our simulation, wecan use Bernoulli trials to build a wide variety of useful probability models.

p = 0.20

BOCK_C17_0321570448 pp3.qxd 12/1/08 3:32 PM Page 388

The Geometric Model 389

Back to Tiger. We want to know how many boxes we’ll need to open to findhis card. Let’s call this random variable boxes, and build a probabilitymodel for it. What’s the probability you find his picture in the first box of cereal?It’s 20%, of course. We could write .

How about the probability that you don’t find Tiger until the second box?Well, that means you fail on the first trial and then succeed on the second. With theprobability of success 20%, the probability of failure, denoted q, is .Since the trials are independent, the probability of getting your first success on thesecond trial is .

Of course, you could have a run of bad luck. Maybe you won’t find Tiger un-til the fifth box of cereal. What are the chances of that? You’d have to fail 4 straighttimes and then succeed, so .

How many boxes might you expect to have to open? We could reason thatsince Tiger’s picture is in 20% of the boxes, or 1 in 5, we expect to find his picture,on average, in the fifth box; that is, boxes. That’s correct, but noteasy to prove.

The Geometric ModelWe want to model how long it will take to achieve the first success in a series ofBernoulli trials. The model that tells us this probability is called the Geometricprobability model. Geometric models are completely specified by one parameter,p, the probability of success, and are denoted Geom(p). Since achieving the firstsuccess on trial number x requires first experiencing failures, the probabili-ties are easily expressed by a formula.

x - 1

E1Y2 =1

0.2 = 5

P1Y = 52 = 10.82410.22 = 0.08192

P1Y = 22 = 10.8210.22 = 0.16

1 - 0.2 = 80%

P1Y = 12 = 0.20

Y = #

Daniel Bernoulli (1700–1782) was thenephew of Jacob, whom you saw inChapter 14. He was the first to workout the mathematics for what we nowcall Bernoulli trials.

Calvin and Hobbes © 1993 Watterson. Reprinted with permission of UNIVERSAL PRESS SYNDICATE. All rights reserved.

Activity: Bernoulli Trials.Guess what! We’ve beengenerating Bernoulli trials allalong. Look at the RandomSimulation Tool in a new way.

GEOMETRIC PROBABILITY MODEL FOR BERNOULLI TRIALS: Geom(p)p probability of success (and of failure)X number of trials until the first success occurs

Expected value:

Standard deviation: s = Aq

p2

E1X2 = m =

1p

P1X = x2 = qx-1 p

=

q = 1 - p = probability=

NOTATION ALERT:

Now we have two morereserved letters. Whenever wedeal with Bernoulli trials, prepresents the probability ofsuccess, and q the probability offailure. (Of course, .)q = 1 - p

Geometric probabilities. See whathappens to a geometric model as you change the probability ofsuccess.


390 CHAPTER 17 Probability Models

Spam and the Geometric modelFOR EXAMPLE

Postini is a global company specializing in communications security. The company monitors over 1 billion Internet messages per day and recently reported that 91% of e-mails are spam!

Let’s assume that your e-mail is typical—91% spam. We’ll also assume you aren’t using a spam filter, so every message gets dumped in your inbox. And, since spam comes from many different sources, we’ll consider your messages to be independent.

Questions: Overnight your inbox collects e-mail. When you first check your e-mail in the morning, about how many spam e-mails should you expect to have to wade through and discard before you find a real message? What’s the probability that the 4th message in your inbox is the first one that isn’t spam?

There are two outcomes: a real message (success) and spam (failure). Since 91% of e-mails are spam, the probabilityof success .Let X = the number of e-mails I’ll check until I find a real message. I assume that the messages arrive independentlyand in a random order. I can use the model Geom(0.09).

On average, I expect to have to check just over 11 e-mails before I find a real message. There’s slightly less than a 7%chance that my first real message will be the 4th one I check.

Note that the probability calculation isn’t new. It’s simply Chapter 14’s Multiplication Rule used to find P1spam ¨ spam ¨ spam ¨ real2.

P(X = 4) = (0.91)3(0.09) = 0.0678

E(X) =

1p

=

10.09

= 11.1

p = 1 - 0.91 = 0.09

MATH BOX

We want to find the mean (expected value) of random variable X, using a geometric model withprobability of success p.

First, write the probabilities:

The expected value is:Let Simplify:That’s an infinite geometric series, with first term 1 and common ratio q:

So, finally . . . E1X2 =

1p

.

=

11 - q

= 1 + q + q2+ q3

+Á

= 1 - q + 2q - 2q2+ 3q2

- 3q3+ 4q3

- 4q4+

Á

= 11 - q2 + 2q11 - q2 + 3q211 - q2 + 4q311 - q2 +Áp = 1 - q:

E1X2 = 1p + 2qp + 3q2p + 4q3p +Á

x 1 2 3 4 Á

P1X = x2 p qp q2p q3pÁ

IndependenceOne of the important requirements for Bernoulli trials is that the trials be inde-pendent. Sometimes that’s a reasonable assumption—when tossing a coin orrolling a die, for example. But that becomes a problem when (often!) we’re look-ing at situations involving samples chosen without replacement. We said thatwhether we find a Tiger Woods card in one box has no effect on the probabilities


Independence 391

in other boxes. This is almost true. Technically, if exactly 20% of the boxes haveTiger Woods cards, then when you find one, you’ve reduced the number of remaining Tiger Woods cards. If you knew there were 2 Tiger Woods cards hidingin the 10 boxes of cereal on the market shelf, then finding one in the first box youtry would clearly change your chances of finding Tiger in the next box. With a fewmillion boxes of cereal, though, the difference is hardly worth mentioning.

If we had an infinite number of boxes, there wouldn’t be a problem. It’s se-lecting from a finite population that causes the probabilities to change, makingthe trials not independent. Obviously, taking 2 out of 10 boxes changes theprobability. Taking even a few hundred out of millions, though, makes very little difference. Fortunately, we have a rule of thumb for the in-between cases.It turns out that if we look at less than 10% of the population, we can pretendthat the trials are independent and still calculate probabilities that are quite accurate.

The 10% Condition: Bernoulli trials must be independent. If that assumption isviolated, it is still okay to proceed as long as the sample is smaller than 10% ofthe population.

People with O-negative blood are called “universal donors” because O-negative blood can begiven to anyone else, regardless of the recipient’s blood type. Only about 6% of people have O-negative blood.

Questions:1. If donors line up at random for a blood drive, how many do you expect to examine before

you find someone who has O-negative blood?2. What’s the probability that the first O-negative donor found is one of the first four people

in line?

Working with a Geometric ModelSTEP-BY-STEP EXAMPLE

I want to estimate how many people I’ll need tocheck to find an O-negative donor, and the prob-ability that 1 of the first 4 people is O-negative.

Plan State the questions.

I can model X with Geom(0.06).Model Specify the model.

Let X number of donors until one is O-negative.

=Variable Define the random variable.

Ç There are two outcomes:success O-negativefailure other blood types

Ç The probability of success for each personis , because they lined up randomly.

Ç 10% Condition: Trials aren’t independentbecause the population is finite, but thedonors lined up are fewer than 10% of allpossible donors.

p = 0.06

=

=

Check to see that these are Bernoulli trials.



Blood drives such as this one expect to exam-ine an average of 16.7 people to find a universaldonor. About 22% of the time there will be onewithin the first 4 people in line.

Conclusion Interpret your results in context.

L 0.2193(0.94)2(0.06) + (0.94)3(0.06)

= (0.06) + (0.94)(0.06) +

P(X = 3) + P(X = 4)P(X … 4) = P(X = 1) + P(X = 2) +

E(X ) =

10.06

L 16.7Mechanics Find the mean.

Calculate the probability of success onone of the first four trials. That’s the prob-ability that or 4.X = 1, 2, 3,

TI TIPS Finding geometric probabilities

Your TI knows the geometric model. Just as you saw back in Chapter 6 with theNormal model, commands to calculate probability distributions are found inthe 2nd DISTRmenu. Have a look. After many others (Don’t drop the courseyet!) you’ll see two Geometric probability functions at the bottom of the list.

• geometpdf(.

The “pdf” stands for “probability density function.” This command allowsyou to find the probability of any individual outcome. You need only specifyp, which defines the Geometric model, and x, which indicates the number oftrials until you get a success. The format is geometpdf(p,x).For example, suppose we want to know the probability that we find our firstTiger Woods picture in the fifth box of cereal. Since Tiger is in 20% of the boxes,we use and , entering the command geometpdf(.2,5). Thecalculator says there’s about an 8% chance.

• geometcdf(.

This is the “cumulative density function,” meaning that it finds the sum ofthe probabilities of several possible outcomes. In general, the commandgeometcdf(p,x) calculates the probability of finding the first success onor before the xth trial.Let’s find the probability of getting a Tiger Woods picture by the time we openthe fourth box of cereal—in other words, the probability our first successcomes on the first box, or the second, or the third, or the fourth. Again we spec-ify p 0.2, and now use x 4. The command geometcdf(.2,4) calcu-lates all the probabilities and adds them. There’s about a 59% chance that ourquest for a Tiger Woods photo will succeed by the time we open the fourth box.

==

x = 5p = 0.2

The Binomial ModelWe can use the Bernoulli trials to answer other questions. Suppose you buy 5boxes of cereal. What’s the probability you get exactly 2 pictures of Tiger Woods?Before, we asked how long it would take until our first success. Now we want tofind the probability of getting 2 successes among the 5 trials. We are still talkingabout Bernoulli trials, but we’re asking a different question.


The Binomial Model 393

This time we’re interested in the number of successes in the 5 trials, so we’ll callit of successes. We want to find This is an example of aBinomial probability. It takes two parameters to define this Binomial model: thenumber of trials, n, and the probability of success, p. We denote this modelBinom(n, p). Here, trials, and the probability of finding a TigerWoods card in any trial.

Exactly 2 successes in 5 trials means 2 successes and 3 failures. It seems logi-cal that the probability should be . Too bad! It’s not that easy. That cal-culation would give you the probability of finding Tiger in the first 2 boxes andnot in the next 3—in that order. But you could find Tiger in the third and fifthboxes and still have 2 successes. The probability of those outcomes in that par-ticular order is (0.8)(0.8)(0.2)(0.8)(0.2). That’s also . In fact, the proba-bility will always be the same, no matter what order the successes and failuresoccur in. Anytime we get 2 successes in 5 trials, no matter what the order, theprobability will be . We just need to take account of all the possibleorders in which the outcomes can occur.

Fortunately, these possible orders are disjoint. (For example, if your two suc-cesses came on the first two trials, they couldn’t come on the last two.) So we canuse the Addition Rule and add up the probabilities for all the possible orderings.Since the probabilities are all the same, we only need to know how many ordersare possible. For small numbers, we can just make a tree diagram and count thebranches. For larger numbers this isn’t practical, so we let the computer or calcu-lator do the work.

Each different order in which we can have k successes in n trials is called a

“combination.” The total number of ways that can happen is written or and pronounced “n choose k.”

For 2 successes in 5 trials,

So there are 10 ways to get 2 Tiger pictures in 5 boxes, and the probability of eachis . Now we can find what we wanted:

In general, the probability of exactly k successes in n trials is .

Using this formula, we could find the expected value by adding up for all values, but it would be a long, hard way to get an answer that you alreadyknow intuitively. What’s the expected value? If we have 5 boxes, and Tiger’s pictureis in 20% of them, then we would expect to have success. If we had 100trials with probability of success 0.2, how many successes would you expect? Canyou think of any reason not to say 20? It seems so simple that most people wouldn’teven stop to think about it. You just multiply the probability of success by n. In otherwords, . Not fully convinced? We prove it in the next Math Box.

The standard deviation is less obvious; you can’t just rely on your intuition.Fortunately, the formula for the standard deviation also boils down to somethingsimple: . (If you’re curious about where that comes from, it’s inthe Math Box too!) In 100 boxes of cereal, we expect to find 20 Tiger Woods cards,with a standard deviation of pictures.

Time to summarize. A Binomial probability model describes the number ofsuccesses in a specified number of trials. It takes two parameters to specify thismodel: the number of trials n and the probability of success p.

1100 * 0.8 * 0.2 = 4

SD1X2 = 1npq

E1X2 = np

510.22 = 1

xP1X = x2

pkqn-k¢n

k≤

P1#success = 22 = 1010.22210.823 = 0.2048

10.22210.823

¢52≤ =

5!2!15 - 22!

=

5 * 4 * 3 * 2 * 12 * 1 * 3 * 2 * 1

=

5 * 42 * 1

= 10.

¢n

k≤ =

n!k!1n - k2!

where n! 1pronounced “n factorial”2 = n * 1n - 12 *Á

* 1

nCk¢n

k≤

10.22210.823

10.22210.823

10.22210.823

p = 0.2,n = 5

P1X = 22.X = numberActivity: The BinomialDistribution. It’s more interestingto combine Bernoulli trials.Simulate this with the RandomTool to get a sense of howBinomial models behave.

NOTATION ALERT:

Now punctuation! Throughoutmathematics n!, pronounced “n factorial,” is the product ofall the integers from 1 to n. Forexample, 4! � 4 . 3 . 2 . 1 � 24.



BINOMIAL PROBABILITY MODEL FOR BERNOULLI TRIALS: Binom(n, p)number of trialsprobability of success (and probability of failure)number of successes in n trials

Mean: Standard Deviation: s = 1npq

m = np

P1X = x2 = nCx pxqn-x, where nCx =

n!x!1n - x2!

X =

q = 1 - p =p =

n =

MATH BOX

To derive the formulas for the mean and standard deviation of a Binomial model we start withthe most basic situation.

Consider a single Bernoulli trial with probability of success p. Let’s find the mean and vari-ance of the number of successes.

Here’s the probability model for the number of successes:

Find the expected value:

And now the variance:

What happens when there is more than one trial, though? A Binomial model simply counts thenumber of successes in a series of n independent Bernoulli trials. That makes it easy to find themean and standard deviation of a binomial random variable,Y.

So, as we thought, the mean is And since the trials are independent, the variances add:

Voilà! The standard deviation is SD1Y2 = 1npq.

Var1Y2 = npq

= pq + pq + pq +Á

+ pq 1Again, n terms.2 = Var1X12 + Var1X22 + Var1X32 +

Á+ Var1Xn2

Var1Y2 = Var1X1 + X2 + X3 +Á

+ Xn2

E1Y2 = np.

= p + p + p +Á

+ p 1There are n terms.2 = E1X12 + E1X22 + E1X32 +

Á+ E1Xn2

E1Y2 = E1X1 + X2 + X3 +Á

+ Xn2

Let Y = X1 + X2 + X3 +Á

+ Xn

Var1X2 = pq

= pq112 = pq1p + q2

= p2q + q2p

Var1X2 = 10 - p22q + 11 - p22 p

E1X2 = p

E1X2 = 0q + 1p

x 0 1P1X = x2 q p

Binomial probabilities.Do-it-yourseif binomial models!Watch the probabilities change asyou control n and p.


The Binomial Model 395

Spam and the Binomial modelFOR EXAMPLE

Recap: The communications monitoring company Postini has reported that 91% of e-mail messages are spam. Suppose your inbox contains 25 messages.

Questions: What are the mean and standard deviation of the number of real messages you should expect to find in your inbox? What’s the probabilitythat you’ll find only 1 or 2 real messages?

I assume that messages arrive independently and at random, with the probability of success (a real message). Let X = the number of real messages among 25. I can use the model Binom(25, 0.09).

Among 25 e-mail messages, I expect to find an average of 2.25 that aren’t spam, with a standard deviation of 1.43 messages. There’s just over a 50% chance that 1 or 2 of my 25 e-mails will be real messages.

= 0.5117 = 0.2340 + 0.2777

= a251b(0.09)1(0.91)24

+ a252b(0.09)2(0.91)23

P(X = 1 or 2) = P(X = 1) + P(X = 2) SD(X ) = 1npq = 125(0.09)(0.91) = 1.43

E(X) = np = 25(0.09) = 2.25

p = 1 - 0.91 = 0.09

Suppose 20 donors come to a blood drive. Recall that 6% of people are “universal donors.”

Questions:1. What are the mean and standard deviation of the number of universal donors among them?2. What is the probability that there are 2 or 3 universal donors?

Working with a Binomial ModelSTEP-BY-STEP EXAMPLE

I want to know the mean and standard devia-tion of the number of universal donors among20 people, and the probability that there are 2or 3 of them.

Plan State the question.

I can model X with Binom(20, 0.06).Model Specify the model.

Let X number of O-negative donors among people.n = 20

=Variable Define the random variable.

Ç There are two outcomes:

success O-negativefailure other blood types

Ç , because people have lined up atrandom.

Ç 10% Condition: Trials are not independent,because the population is finite, but fewerthan 10% of all possible donors are lined up.

p = 0.06

=

=

Check to see that these are Bernoulli trials.



= 0.3106L 0.2246 + 0.0860

+ a203b(0.06)3(0.94)17

= a202b10.0622(0.94)18

P(X = 2 or 3) = P(X = 2) + P(X = 3)

SD(X ) = 1npq = 120(0.06)(0.94) L 1.06E(X ) = np = 20(0.06) = 1.2Mechanics Find the expected value and

standard deviation.

In groups of 20 randomly selected blooddonors, I expect to find an average of 1.2 univer-sal donors, with a standard deviation of 1.06.About 31% of the time, I’d find 2 or 3 universaldonors among the 20 people.

Conclusion Interpret your results incontext.

TI Tips Finding binomial probabilities

Remember how the calculator handles Geometric probabilities? Well, the com-mands for finding Binomial probabilities are essentially the same. Again you’llfind them in the 2nd DISTR menu.

• binompdf(

This probability density function allows you to find the probability of anindividual outcome. You need to define the Binomial model by specifyingn and p, and then indicate the desired number of successes, x. The format isbinompdf(n,p,X).

For example, recall that Tiger Woods’ picture is in 20% of the cereal boxes.Suppose that we want to know the probability of finding Tiger exactly twiceamong 5 boxes of cereal. We use , and , entering the com-x = 2n = 5, p = 0.2mand binompdf(5,.2,2). There’s about a 20% chance of getting twopictures of Tiger Woods in five boxes of cereal.

• binomcdf(

Need to add several Binomial probabilities? To find the total probability ofgetting x or fewer successes among the n trials use the cumulative Binomialdensity function binomcdf(n,p,X).For example, suppose we have ten boxes of cereal and wonder about theprobability of finding up to 4 pictures of Tiger. That’s the probability of 0, 1,2, 3 or 4 successes, so we specify the command binomcdf(10,.2,4).Pretty likely!Of course “up to 4” allows for the possibility that we end up with none.What’s the probability we get at least 4 pictures of Tiger in 10 boxes? Well,“at least 4” means “not 3 or fewer.” That’s the complement of 0, 1, 2, or 3 suc-cesses. Have your TI evaluate 1-binomcdf(10,.2,3). There’s about a12% chance we’ll find at least 4 pictures of Tiger in 10 boxes of cereal.

M17_BOCK0444_03_SE_C17.QXD 12/25/09 3:46 AM Page 396

The Normal Model to the Rescue! 397

The Normal Model to the Rescue!Suppose the Tennessee Red Cross anticipates the need for at least 1850 units of O-negative blood this year. It estimates that it will collect blood from 32,000 donors.How great is the risk that the Tennessee Red Cross will fall short of meeting its need?We’ve just learned how to calculate such probabilities. We can use the Binomialmodel with and . The probability of getting exactly 1850

units of O-negative blood from 32,000 donors is . No

calculator on earth can calculate that first term (it has more than 100,000 digits).1

And that’s just the beginning. The problem said at least 1850, so we have to do itagain for 1851, for 1852, and all the way up to 32,000. No thanks.

When we’re dealing with a large number of trials like this, making direct cal-culations of the probabilities becomes tedious (or outright impossible). Here anold friend—the Normal model—comes to the rescue.

The Binomial model has mean and standard deviation . We could try approximating its distribution with a Normal model, using

the same mean and standard deviation. Remarkably enough, that turns out to bea very good approximation. (We’ll see why in the next chapter.) With that approx-imation, we can find the probability:

There seems to be about a 5% chance that this Red Cross chapter will run short ofO-negative blood.

Can we always use a Normal model to make estimates of Binomial probabili-ties? No. Consider the Tiger Woods situation—pictures in 20% of the cereal boxes.If we buy five boxes, the actual Binomial probabilities that we get 0, 1, 2, 3, 4, or 5pictures of Tiger are 33%, 41%, 20%, 5%, 1%, and 0.03%, respectively. The first his-togram shows that this probability model is skewed. That makes it clear that weshould not try to estimate these probabilities by using a Normal model.

Now suppose we open 50 boxes of this cereal and count the number of TigerWoods pictures we find. The second histogram shows this probability model. It iscentered at pictures, as expected, and it appears to be fairlysymmetric around that center. Let’s have a closer look.

The third histogram again shows Binom(50, 0.2), this time magnified some-what and centered at the expected value of 10 pictures of Tiger. It looks close toNormal, for sure. With this larger sample size, it appears that a Normal modelmight be a useful approximation.

A Normal model, then, is a close enough approximation only for a largeenough number of trials. And what we mean by “large enough” depends on theprobability of success. We’d need a larger sample if the probability of successwere very low (or very high). It turns out that a Normal model works pretty wellif we expect to see at least 10 successes and 10 failures. That is, we check theSuccess/ Failure Condition.

The Success/ Failure Condition: A Binomial model is approximately Normal ifwe expect at least 10 successes and 10 failures:

np Ú 10 and nq Ú 10.

np = 5010.22 = 10

P1X 6 18502 = Paz 6

1850 - 192042.48

b L P1z 6 -1.652 L 0.05

42.481npq Lnp = 1920

a320001850

b * 0.061850* 0.9430150

p = 0.06n = 32,000

1 If your calculator can find Binom(32000,0.06), then it’s smart enough to use an approxi-mation. Read on to see how you can, too.

Activity: NormalApproximation. Binomialprobabilities can be hard tocalculate. With the SimulationTool you’ll see how well theNormal model can approximatethe Binomial—a much easiermethod.

0 1 2 3 4 5

0 30 40 502010

0 5 10 15 20

How close to Normal? How welldoes a Normal curve fit a binomialmodel? Check out the Success/Failure Condition for yourself.

M17_BOCK0444_03_SE_C17.QXD 6/28/10 2:10 PM Page 397


MATH BOX

It’s easy to see where the magic number 10 comes from. You just need to remember how Nor-mal models work. The problem is that a Normal model extends infinitely in both directions. Buta Binomial model must have between 0 and n successes, so if we use a Normal to approximate aBinomial, we have to cut off its tails. That’s not very important if the center of the Normal modelis so far from 0 and n that the lost tails have only a negligible area. More than three standard de-viations should do it, because a Normal model has little probability past that.

So the mean needs to be at least 3 standard deviations from 0 and at least 3 standard devia-tions from n. Let’s look at the 0 end.

We require: Or in other words:For a Binomial, that’s:Squaring yields:Now simplify: Since we can require:

For simplicity, we usually require that np (and nq for the other tail) be at least 10 to use the Nor-mal approximation, the Success/Failure Condition.2

np 7 9q … 1,np 7 9q

n2p27 9npq

np 7 31npq

m 7 3sm - 3s 7 0

Spam and the Normal approximation to the BinomialFOR EXAMPLE

Recap: The communications monitoring company Postini has reported that 91% of e-mail messages are spam. Recently, you installed a spam filter.You observe that over the past week it okayed only 151 of 1422 e-mails you received, classifying the rest as junk. Should you worry that the filtering istoo aggressive?

Question: What’s the probability that no more than 151 of 1422 e-mails is a real message?

I assume that messages arrive randomly and independently, with a probability of success (a real message) .The model Binom(1422, 0.09) applies, but will be hard to work with. Checking conditions for the Normal approximation,I see that:

Ç These messages represent less than 10% of all e-mail traffic.Ç I expect real messages and spam messages, both

far greater than 10.

It’s okay to approximate this binomial probability by using a Normal model.

Among my 1422 e-mails, there’s over a 98% chance that no more than 151 of them were real messages, so the filtermay be working properly.

= 0.9834 = P(z … 2.13)

P(x … 151) = P az …

151 - 127.9810.79

b

s = 1npq = 11422(0.09)(0.91) L 10.79m = np = 1422(0.09) = 127.98

nq = (1422)(0.91) = 1294.02np = (1422)(0.09) = 127.98

p = 0.09

151127.98

2.13

2 Looking at the final step, we see that we need in the worst case, when q (or p) isnear 1, making the Binomial model quite skewed. When q and p are near 0.5—say between0.4 and 0.6—the Binomial model is nearly symmetric and ought to be safe enough.Although we’ll always check for 10 expected successes and failures, keep in mind that forvalues of p near 0.5, we can be somewhat more forgiving.

np 7 5

np 7 9


What Can Go Wrong? 399

Continuous Random VariablesThere’s a problem with approximating a Binomial model with a Normal model.The Binomial is discrete, giving probabilities for specific counts, but the Normalmodels a continuous random variable that can take on any value. For continuousrandom variables, we can no longer list all the possible outcomes and their proba-bilities, as we could for discrete random variables.3

As we saw in the previous chapter, models for continuous random variablesgive probabilities for intervals of values. So, when we use the Normal model, weno longer calculate the probability that the random variable equals a particularvalue, but only that it lies between two values. We won’t calculate the probabilityof getting exactly 1850 units of blood, but we have no problem approximatingthe probability of getting 1850 or more, which was, after all, what we reallywanted.4

3 In fact, some people use an adjustment called the “continuity correction” to help with thisproblem. It’s related to the suggestion we make in the next footnote and is discussed inmore advanced textbooks.4 If we really had been interested in a single value, we might have approximated it by find-ing the probability of getting between 1849.5 and 1850.5 units of blood.

JUST CHECKINGAs we noted a few chapters ago, the Pew Research Center (www.pewresearch.org) reports that

they are actually able to contact only 76% of the randomly selected households drawn for a telephonesurvey.

1. Explain why these phone calls can be considered Bernoulli trials.

2. Which of the models of this chapter (Geometric, Binomial, Normal) would you use to model thenumber of successful contacts from a list of 1000 sampled households? Explain.

3. Pew further reports that even after they contacted a household, only 38% agree to be interviewed,so the probability of getting a completed interview for a randomly selected household is only 0.29.Which of the models of this chapter would you use to model the number of households Pew has tocall before they get the first completed interview?

WHAT CAN GO WRONG?u Be sure you have Bernoulli trials. Be sure to check the requirements first: two possible

outcomes per trial (“success” and “failure”), a constant probability of success, andindependence. Remember to check the 10% Condition when sampling without re-placement.

u Don’t confuse Geometric and Binomial models. Both involve Bernoulli trials, but the issuesare different. If you are repeating trials until your first success, that’s a Geometricprobability. You don’t know in advance how many trials you’ll need—theoretically,it could take forever. If you are counting the number of successes in a specified num-ber of trials, that’s a Binomial probability.

u Don’t use the Normal approximation with small n . To use a Normal approximation in placeof a Binomial model, there must be at least 10 expected successes and 10 expectedfailures.



CONNECTIONSThis chapter builds on what we know about random variables. We now have two more probabilitymodels to join the Normal model.

There are a number of “forward” connections from this chapter. We’ll see the 10% Conditionand the Success/Failure Condition often. And the facts about the Binomial distribution can helpexplain how proportions behave, as we’ll see in the next chapter.

We’ve learned that Bernoulli trials show up in lots of places. Depending on the random variable ofinterest, we can use one of three models to estimate probabilities for Bernoulli trials:

u a Geometric model when we’re interested in the number of Bernoulli trials until the next success;u a Binomial model when we’re interested in the number of successes in a certain number of

Bernoulli trials;u a Normal model to approximate a Binomial model when we expect at least 10 successes and

10 failures.

TermsBernoulli trials, if . . . 388. 1. there are two possible outcomes.

2. the probability of success is constant.3. the trials are independent.

Geometric probability model 389. A Geometric model is appropriate for a random variable that counts the number of Bernoullitrials until the first success.

Binomial probability model 393. A Binomial model is appropriate for a random variable that counts the number of successes ina fixed number of Bernoulli trials.

10% Condition 391. When sampling without replacement, trials are not independent. It’s still okay to proceed aslong as the sample is smaller than 10% of the population.

Success/Failure Condition 397. For a Normal model to be a good approximation of a Binomial model, we must expect at least10 successes and 10 failures. That is, .

Skillsu Know how to tell if a situation involves Bernoulli trials.

u Be able to choose whether to use a Geometric or a Binomial model for a random variable involvingBernoulli trials.

u Know the appropriate conditions for using a Geometric, Binomial, or Normal model.

u Know how to find the expected value of a Geometric model.

u Be able to calculate Geometric probabilities.

u Know how to find the mean and standard deviation of a Binomial model.

u Be able to calculate Binomial probabilities, perhaps approximating with a Normal model.

u Be able to interpret means, standard deviations, and probabilities in the Bernoulli trial context.

np Ú 10 and nq Ú 10

WHAT HAVE WE LEARNED?


Exercises 401

THE BINOMIAL AND THE GEOMETRIC ON THE COMPUTER

Most statistics packages offer functions that compute Binomial probabilities, and many offer functions forGeometric probabilities as well. Some technology solutions automatically use the Normal approximation for theBinomial when the exact calculations become unmanageable.

EXERCISES

1. Bernoulli. Do these situations involve Bernoulli trials?Explain.a) We roll 50 dice to find the distribution of the number

of spots on the faces.b) How likely is it that in a group of 120 the majority

may have Type A blood, given that Type A is found in43% of the population?

c) We deal 7 cards from a deck and get all hearts. Howlikely is that?

d) We wish to predict the outcome of a vote on the schoolbudget, and poll 500 of the 3000 likely voters to seehow many favor the proposed budget.

e) A company realizes that about 10% of its packages arenot being sealed properly. In a case of 24, is it likelythat more than 3 are unsealed?

2. Bernoulli 2. Do these situations involve Bernoulli trials? Explain.a) You are rolling 5 dice and need to get at least two 6’s

to win the game.b) We record the distribution of eye colors found in a

group of 500 people.c) A manufacturer recalls a doll because about 3% have

buttons that are not properly attached. Customers return 37 of these dolls to the local toy store. Is themanufacturer likely to find any dangerous buttons?

d) A city council of 11 Republicans and 8 Democratspicks a committee of 4 at random. What’s the proba-bility they choose all Democrats?

e) A 2002 Rutgers University study found that 74% ofhigh school students have cheated on a test at leastonce. Your local high school principal conducts a survey in homerooms and gets responses that admitto cheating from 322 of the 481 students.

3. Simulating the model. Think about the Tiger Woodspicture search again. You are opening boxes of cereal oneat a time looking for his picture, which is in 20% of theboxes. You want to know how many boxes you mighthave to open in order to find Tiger.a) Describe how you would simulate the search for Tiger

using random numbers.b) Run at least 30 trials.c) Based on your simulation, estimate the probabilities

that you might find your first picture of Tiger in thefirst box, the second, etc.

d) Calculate the actual probability model.e) Compare the distribution of outcomes in your simula-

tion to the probability model.

4. Simulation II. You are one space short of winning achild’s board game and must roll a 1 on a die to claim victory. You want to know how many rolls it might take.a) Describe how you would simulate rolling the die until

you get a 1.b) Run at least 30 trials.c) Based on your simulation, estimate the probabilities

that you might win on the first roll, the second, thethird, etc.

d) Calculate the actual probability model.e) Compare the distribution of outcomes in your simula-


5. Tiger again. Let’s take one last look at the Tiger Woodspicture search. You know his picture is in 20% of the ce-real boxes. You buy five boxes to see how many picturesof Tiger you might get.a) Describe how you would simulate the number of pic-

tures of Tiger you might find in five boxes of cereal.b) Run at least 30 trials.c) Based on your simulation, estimate the probabilities that

you get no pictures of Tiger, 1 picture, 2 pictures, etc.d) Find the actual probability model.e) Compare the distribution of outcomes in your simula-


6. Seatbelts. Suppose 75% of all drivers always weartheir seatbelts. Let’s investigate how many of the driversmight be belted among five cars waiting at a traffic light.a) Describe how you would simulate the number of

seatbelt-wearing drivers among the five cars.b) Run at least 30 trials.c) Based on your simulation, estimate the probabilities

there are no belted drivers, exactly one, two, etc.d) Find the actual probability model.e) Compare the distribution of outcomes in your simula-


7. On time. A Department of Transportation report aboutair travel found that, nationwide, 76% of all flights are ontime. Suppose you are at the airport and your flight is oneof 50 scheduled to take off in the next two hours. Can youconsider these departures to be Bernoulli trials? Explain.


8. Lost luggage. A Department of Transportation reportabout air travel found that airlines misplace about 5 bagsper 1000 passengers. Suppose you are traveling with agroup of people who have checked 22 pieces of luggageon your flight. Can you consider the fate of these bags tobe Bernoulli trials? Explain.

9. Hoops. A basketball player has made 80% of his foulshots during the season. Assuming the shots are inde-pendent, find the probability that in tonight’s game hea) misses for the first time on his fifth attempt.b) makes his first basket on his fourth shot.c) makes his first basket on one of his first 3 shots.

10. Chips. Suppose a computer chip manufacturer rejects2% of the chips produced because they fail presale testing.a) What’s the probability that the fifth chip you test is

the first bad one you find?b) What’s the probability you find a bad one within the

first 10 you examine?

11. More hoops. For the basketball player in Exercise 9,what’s the expected number of shots until he misses?

12. Chips ahoy. For the computer chips described in Exercise 10, how many do you expect to test before find-ing a bad one?

13. Customer center operator. Raaj works at the cus-tomer service call center of a major credit card bank.Cardholders call for a variety of reasons, but regardlessof their reason for calling, if they hold a platinum card,Raaj is instructed to offer them a double-miles promo-tion. About 10% of all cardholders hold platinum cards,and about 50% of those will take the double-miles pro-motion. On average, how many calls will Raaj have totake before finding the first cardholder to take the double-miles promotion?

14. Cold calls. Justine works for an organization commit-ted to raising money for Alzheimer’s research. From pastexperience, the organization knows that about 20% of allpotential donors will agree to give something if contactedby phone. They also know that of all people donating,about 5% will give $100 or more. On average, how manypotential donors will she have to contact until she getsher first $100 donor?

15. Blood. Only 4% of people have Type AB blood.a) On average, how many donors must be checked to

find someone with Type AB blood?b) What’s the probability that there is a Type AB donor

among the first 5 people checked?c) What’s the probability that the first Type AB donor

will be found among the first 6 people?d) What’s the probability that we won’t find a Type AB

donor before the 10th person?

16. Colorblindness. About 8% of males are colorblind. Aresearcher needs some colorblind subjects for an experi-ment and begins checking potential subjects.a) On average, how many men should the researcher

expect to check to find one who is colorblind?b) What’s the probability that she won’t find anyone

colorblind among the first 4 men she checks?

c) What’s the probability that the first colorblind manfound will be the sixth person checked?

d) What’s the probability that she finds someone who iscolorblind before checking the 10th man?

17. Lefties. Assume that 13% of people are left-handed. Ifwe select 5 people at random, find the probability of eachoutcome described below.a) The first lefty is the fifth person chosen.b) There are some lefties among the 5 people.c) The first lefty is the second or third person.d) There are exactly 3 lefties in the group.e) There are at least 3 lefties in the group.f) There are no more than 3 lefties in the group.

18. Arrows. An Olympic archer is able to hit the bull’s-eye80% of the time. Assume each shot is independent of theothers. If she shoots 6 arrows, what’s the probability ofeach of the following results?a) Her first bull’s-eye comes on the third arrow.b) She misses the bull’s-eye at least once.c) Her first bull’s-eye comes on the fourth or fifth arrow.d) She gets exactly 4 bull’s-eyes.e) She gets at least 4 bull’s-eyes.f ) She gets at most 4 bull’s-eyes.

19. Lefties redux. Consider our group of 5 people fromExercise 17.a) How many lefties do you expect?b) With what standard deviation?c) If we keep picking people until we find a lefty, how

long do you expect it will take?

20. More arrows. Consider our archer from Exercise 18.a) How many bull’s-eyes do you expect her to get?b) With what standard deviation?c) If she keeps shooting arrows until she hits the bull’s-

eye, how long do you expect it will take?

21. Still more lefties. Suppose we choose 12 people in-stead of the 5 chosen in Exercise 17.a) Find the mean and standard deviation of the number

of right-handers in the group.b) What’s the probability that

i) they’re not all right-handed?ii) there are no more than 10 righties?

iii) there are exactly 6 of each?iv) the majority is right-handed?

22. Still more arrows. Suppose our archer from Exercise 18shoots 10 arrows.a) Find the mean and standard deviation of the number

of bull’s-eyes she may get.b) What’s the probability that

i) she never misses?ii) there are no more than 8 bull’s-eyes?

iii) there are exactly 8 bull’s-eyes?iv) she hits the bull’s-eye more often than she misses?

23. Vision. It is generally believed that nearsightedness affects about 12% of all children. A school district tests thevision of 169 incoming kindergarten children. How manywould you expect to be nearsighted? With what standarddeviation?



24. International students. At a certain college, 6% of allstudents come from outside the United States. Incomingstudents there are assigned at random to freshman dorms,where students live in residential clusters of 40 freshmensharing a common lounge area. How many internationalstudents would you expect to find in a typical cluster?With what standard deviation?

25. Tennis, anyone? A certain tennis player makes a suc-cessful first serve 70% of the time. Assume that each serveis independent of the others. If she serves 6 times, what’sthe probability she getsa) all 6 serves in?b) exactly 4 serves in?c) at least 4 serves in?d) no more than 4 serves in?

26. Frogs. A wildlife biologist examines frogs for a genetictrait he suspects may be linked to sensitivity to industrialtoxins in the environment. Previous research had estab-lished that this trait is usually found in 1 of every 8 frogs.He collects and examines a dozen frogs. If the frequencyof the trait has not changed, what’s the probability hefinds the trait ina) none of the 12 frogs?b) at least 2 frogs?c) 3 or 4 frogs?d) no more than 4 frogs?

27. And more tennis. Suppose the tennis player in Exer-cise 25 serves 80 times in a match.a) What are the mean and standard deviation of the

number of good first serves expected?b) Verify that you can use a Normal model to approxi-

mate the distribution of the number of good firstserves.

c) Use the 68–95–99.7 Rule to describe this distribution.d) What’s the probability she makes at least 65 first

serves?

28. More arrows. The archer in Exercise 18 will be shoot-ing 200 arrows in a large competition.a) What are the mean and standard deviation of the

number of bull’s-eyes she might get?b) Is a Normal model appropriate here? Explain.c) Use the 68–95–99.7 Rule to describe the distribution of

the number of bull’s-eyes she may get.d) Would you be surprised if she made only 140 bull’s-

eyes? Explain.

29. Apples. An orchard owner knows that he’ll have to useabout 6% of the apples he harvests for cider because theywill have bruises or blemishes. He expects a tree to pro-duce about 300 apples.a) Describe an appropriate model for the number of cider

apples that may come from that tree. Justify your model.b) Find the probability there will be no more than a

dozen cider apples.c) Is it likely there will be more than 50 cider apples?

Explain.

30. Frogs, part II. Based on concerns raised by his prelimi-nary research, the biologist in Exercise 26 decides to col-lect and examine 150 frogs.

a) Assuming the frequency of the trait is still 1 in 8, deter-mine the mean and standard deviation of the numberof frogs with the trait he should expect to find in hissample.

b) Verify that he can use a Normal model to approximatethe distribution of the number of frogs with the trait.

c) He found the trait in 22 of his frogs. Do you think thisproves that the trait has become more common? Explain.

31. Lefties again. A lecture hall has 200 seats with foldingarm tablets, 30 of which are designed for left-handers.The typical size of classes that meet there is 188, and wecan assume that about 13% of students are left-handed.What’s the probability that a right-handed student in oneof these classes is forced to use a lefty arm tablet?

32. No-shows. An airline, believing that 5% of passengersfail to show up for flights, overbooks (sells more ticketsthan there are seats). Suppose a plane will hold 265 pas-sengers, and the airline sells 275 tickets. What’s the prob-ability the airline will not have enough seats, so someonegets bumped?

33. Annoying phone calls. A newly hired telemarketer istold he will probably make a sale on about 12% of hisphone calls. The first week he called 200 people, but onlymade 10 sales. Should he suspect he was misled about thetrue success rate? Explain.

34. The euro. Shortly after the introduction of the eurocoin in Belgium, newspapers around the world publishedarticles claiming the coin is biased. The stories were basedon reports that someone had spun the coin 250 times andgotten 140 heads—that’s 56% heads. Do you think this isevidence that spinning a euro is unfair? Explain.

35. Seatbelts II. Police estimate that 80% of drivers nowwear their seatbelts. They set up a safety roadblock, stopping cars to check for seatbelt use.a) How many cars do they expect to stop before finding

a driver whose seatbelt is not buckled?b) What’s the probability that the first unbelted driver is

in the 6th car stopped?c) What’s the probability that the first 10 drivers are all

wearing their seatbelts?d) If they stop 30 cars during the first hour, find the

mean and standard deviation of the number of driversexpected to be wearing seatbelts.

e) If they stop 120 cars during this safety check, what’sthe probability they find at least 20 drivers not wear-ing their seatbelts?

36. Rickets. Vitamin D is essential for strong, healthybones. Our bodies produce vitamin D naturally whensunlight falls upon the skin, or it can be taken as a dietarysupplement. Although the bone disease rickets waslargely eliminated in England during the 1950s, somepeople there are concerned that this generation of chil-dren is at increased risk because they are more likely towatch TV or play computer games than spend time out-doors. Recent research indicated that about 20% of Britishchildren are deficient in vitamin D. Suppose doctors test agroup of elementary school children.

Exercises 403


a) What’s the probability that the first vitamin D– deficient child is the 8th one tested?

b) What’s the probability that the first 10 children testedare all okay?

c) How many kids do they expect to test before findingone who has this vitamin deficiency?

d) They will test 50 students at the third-grade level.Find the mean and standard deviation of the numberwho may be deficient in vitamin D.

e) If they test 320 children at this school, what’s the prob-ability that no more than 50 of them have the vitamindeficiency?

37. ESP. Scientists wish to test the mind-reading ability of aperson who claims to “have ESP.” They use five cardswith different and distinctive symbols (square, circle, tri-angle, line, squiggle). Someone picks a card at randomand thinks about the symbol. The “mind reader” mustcorrectly identify which symbol was on the card. If thetest consists of 100 trials, how many would this personneed to get right in order to convince you that ESP mayactually exist? Explain.

38. True-False. A true-false test consists of 50 questions.How many does a student have to get right to convinceyou that he is not merely guessing? Explain.

39. Hot hand. A basketball player who ordinarily makesabout 55% of his free throw shots has made 4 in a row. Isthis evidence that he has a “hot hand” tonight? That is, isthis streak so unusual that it means the probability hemakes a shot must have changed? Explain.

40. New bow. Our archer in Exercise 18 purchases a newbow, hoping that it will improve her success rate to morethan 80% bull’s-eyes. She is delighted when she first tests

her new bow and hits 6 consecutive bull’s-eyes. Do youthink this is compelling evidence that the new bow is better? In other words, is a streak like this unusual forher? Explain.

41. Hotter hand. Our basketball player in Exercise 39 hasnew sneakers, which he thinks improve his game. Overhis past 40 shots, he’s made 32—much better than the55% he usually shoots. Do you think his chances of mak-ing a shot really increased? In other words, is making atleast 32 of 40 shots really unusual for him? (Do you thinkit’s his sneakers?)

42. New bow, again. The archer in Exercise 40 continuesshooting arrows, ending up with 45 bull’s-eyes in 50 shots.Now are you convinced that the new bow is better? Explain.


JUST CHECKINGAnswers

1. There are two outcomes (contact, no contact), theprobability of contact is 0.76, and random callsshould be independent.

2. Binomial, with and . For actual calculations, we could approximate using a Normalmodel with and

3. Geometric, with .p = 0.29

s = 1npq = 1100010.76210.242 L 13.5.

m = np = 100010.762 = 760

p = 0.76n = 1000


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