Chapter 17: Probability Models

In previous chapters we have seen many different probability models. This chapter looks at three special types of models: the geometric model, the binomial model and the Poisson model.

The geometric model is sometimes useful and has the benefit that the associated probability formula is relatively easy to derive and understand.

The binomial model will be the most important of the three, as we will use it a great deal throughout the next part of the course; the associated probability formulas are not quite as easy to derive (and we won't do so in this class) but there are quick shortcuts available on the TI-84. For this reason, you should ignore the binomial probability formula in the book and use the calculator instead.

The Poisson model does have interesting applications, but we will rarely encounter it again in this course; for that reason you can consider the material and exercises about the Poisson model optional and it will not be covered on exams.

Bernoulli trials

A Bernoulli trial consists of situation in which we select one person or thing from a larger group, such as randomly selecting people from the general population for the purpose of a public opinion poll.

With each trial we must ask a yes-or-no question, e.g. "Do you plan to vote for Barack Obama for president in the 2008?" Notice that this question can be answered "yes" or "no"; if we had asked "Who are you going to vote for?" the answer might have been "Obama" or "Clinton" or "McCain" or "Giuliani" or "Hagel" or "I don't know" or "I don't care" or a whole host of other responses. By knowing the answer to the second question we also know the answer to the first, but for our purposes it is important to phrase the question in a yes-or-no fashion. We often call a "yes" answer a "success" and a "no" answer a "failure" � even when a "success" might be an affirmative response to a question like "Did the patient come down with the flu?"

In a Bernoulli trial we must also have a fixed probability of success. This is not true with small samples: if we have 20 M&Ms and 5 of them are green, the probability of choosing an M&M and getting a green one is 5/20 on the first trial, but on the second trial the probability of getting a green candy has changed to 4/19 (if we selected a green M&M the first time) or 5/19 (if not); this situation would not constitute a Bernoulli trial.

In the presidential poll example, we are most likely polling people selected from among all voters in the U.S. If 25% of 1,000,000 voters favor Obama, the probability of choosing an Obama supporter on the first trial (250,000/1,000,000 = 0.25) is essentially the same as choosing an Obama supporter on the second trial (249,999/999,999 � 0.25 or 250,000/999,999 � 0.25), so the "fixed probability of success" condition is satisfied in this case, even though the size of the population is finite.

Finally, the trials need to be independent. In the M&M example, the trials are not, since the probability of choosing a green M&M on the second trial depends on the outcome of the first trial. The trials in the Obama example are independent, as long as we use a simple random sample (calling only voters in Illinois, Obama's home state, or people who had visited Obama's Web site would violate this condition).

Once we have a Bernoulli trial we can compute probabilities associated with geometric and binomial models.

Geometric model

A geometric model counts the number of trials until the first success. According to the EdCC Web site, 59% of students enrolled at EdCC are female. Suppose we randomly select students from among all of those enrolled at EdCC until we find a female student. This constitutes a Bernoulli trial because:

There are two possible outcomes: female (success) and male (failure).

There is a constant probability of success (59%); this isn't precisely true, but since there are over 10,000 students enrolled at EdCC, removing a few of these students from contention won't significantly change the probability of finding a female student among those remaining.

The trials are independent: since we are randomly selecting the students, the gender of one selected student should have no effect on the gender of the other students selected.

Let the random variable $\normalsize X$ represent the number of students we need to contact before finding a female student. We note that:

$\normalsize P(X=1) \ = \ 0.59$

since the probability that the first student we select is female is 59%. We then compute:

$\normalsize \begin{eqnarray} P(X=2) \ &=& \ P(\mbox{male and then female}) \\vspace{5} \&=& \ P(\mbox{male} ) \times P(\mbox{female} ) \\vspace{5} \&=& \ (0.41)(0.59) \\vspace{5} \&=& \ 0.2419 \end{eqnarray}$

and then:

$\normalsize \begin{eqnarray} P(X=3) \ &=& \ P(\mbox{male and then male and then female}) \\vspace{5} \&=& \ P(\mbox{male} ) \times P(\mbox{male} ) \times P(\mbox{female} ) \\vspace{5} \&=& \ (0.41)(0.41)(0.59) \\vspace{5} \&=& \ 0.099179 \end{eqnarray}$

and then:

$\normalsize \begin{eqnarray} P(X=4) \ &=& \ P(\mbox{male and then male and then male and then female}) \\vspace{5} \&=& \ P(\mbox{male} ) \times P(\mbox{male} ) \times P(\mbox{male} ) \times P(\mbox{female} ) \\vspace{5} \&=& \ (0.41)(0.41)(0.41)(0.59) \\vspace{5} \&=& \ 0.04066339 \end{eqnarray}$

and so on. By this point you should notice a pattern:

$\normalsize P(X=x) \ = (0.41)^{x-1}(0.59)$

and more generally:

$\normalsize P(X=x) \ = p^{x-1}q$

where $\normalsize p$ is the probability of success for any geometric model and $\normalsize q = 1-p$ is the corresponding probability of failure.

Returning to our EdCC student example, we can collect our probabilities in a table that looks like this:

$\normalsize x$	$\normalsize P(X=x)$
1	0.5900
2	0.2419
3	0.0992
4	0.0407
$\normalsize \vdots$	$\normalsize \vdots$

Notice that this probability model continues forever. Sure, there is only a very small chance that we will need to contact more than 10 EdCC students to find a female student, but theoretically we could contact 20 or 30 or 100 students and not find a female.

The geometric probability formula is also built into the TI-84 (under the DISTR menu); for example:

$\normalsize P(X = 4)$ = geometpdf(0.59,4) � 0.0407

How do we find the mean and standard deviation of the geometric model? We could use the formulas from Chapter 16, but one problem is that the sums involved for the geometric model are infinite sums, which (unless you've taken three quarters of calculus) you have most likely never encountered before. However it is a fact that the standard formula for the mean reduces to a much simpler formula:

$\normalsize \begin{eqnarray} \mu \ &=& \ E(X) \\vspace{5} \&=& \sum \left[ x \, \cdot \, P(X = x) \right] \\vspace{5} \&=& \sum \left[x \, \cdot \, p^{x-1}q \right] \\vspace{5} \&=& 1/p \end{eqnarray}$

Why is it true that � = 1/p? You can find the ugly details in a separate document in the Chapter 17 materials, but the formula itself should be intuitive: If, for example, 20% of all M&Ms (or 1 in 5) are green, then on average we would need to randomly select 1/0.20 = 5 M&Ms before we found a green one. Sometimes, of course, we would find a green one sooner, and sometimes it might take more than 5 trials, but the average would be 5: if we assigned everyone in class to randomly draw M&Ms out of a large batch until they found a green one, there would be a variety of answers, but the average of the number drawn would be about 5.)

What is the standard deviation of the geometric model?

$\normalsize \begin{eqnarray} \sigma \ &=& \ SD(X) \\vspace{5} \&=& \sqrt{\sum \left[ (x-\mu)^2 \, \cdot \, p^{x-1}q \right]} \\vspace{5} \&=& \frac{\sqrt{q}}{p}\end{eqnarray}$

Again, the ugly details can be found elsewhere. While we will sometimes need to compute the mean of a geometric model, we will rarely need to compute the standard deviation.

In our EdCC student example, the mean would be 1/0.59 � 1.6949 (in other words, on average we would need to randomly contact about 1.7 students in order to reach a female).

The geometric probability formula is fairly straightforward, so in most cases the geometpdf feature on the TI-84 is not much easier than using the formula directly; however in certain cases a related feature on the TI-84 can be very useful. Suppose we want to know the probability of finding a female among the first three students we contact. This is given by:

$\normalsize \begin{eqnarray} P(X \leq 3) &=& P(X=1\mbox{ or }X=2\mbox{ or }X=3) \\vspace{5} \&=& P(X=1)+P(X=2)+P(X=3) \\vspace{5} \&\approx& 0.5900+0.2419+0.0992 \\vspace{5} \&=& 0.9311\end{eqnarray}$

but you can get the answer even faster using the TI-84:

$\normalsize P(X \leq 3)$ = geometcdf(0.59,3) = 0.931079

Note that we are using geometcdf here (the c stands for "cumulative"); it gives us the probability of achieving success using at most the specified number of trials.

Binomial model

Now let's consider a different, but related, question involving Bernoulli trials. Suppose we randomly select five EdCC students. What is the probability that none of them are female?

A binomial model is appropriate here: not only is a Bernoulli trial involved, there is a fixed number of trials (in this case, 5). Let $\normalsize Y$ be a random variable representing the number of females in our five-person sample. We then have:

$\normalsize \begin{eqnarray} P(Y = 0) \ &=& \ P(\mbox{no females} ) \ \vspace{5} \ &=& \ P(\mbox{male and male and male and male and male} ) \ \vspace{5} \ &=& \ P(\mbox{male} )\times P(\mbox{male} )\times P(\mbox{male} )\times P(\mbox{male} )\times P(\mbox{male} ) \ \vspace{5} \ &=& \ (0.41)(0.41)(0.41)(0.41)(0.41) \ \vspace{5} \ &=& \ (0.41)^5 \ \vspace{5} \ &\approx& \ 0.0116 \end{eqnarray}$

What is the probability that exactly one student is female? We could compute the probability that only the first student is female:

$\normalsize \begin{eqnarray} P(\mbox{only first student female} ) &=& \ P(\mbox{female and male and male and male and male} ) \ \vspace{5} \ &=& \ P(\mbox{female} )\times P(\mbox{male} )\times P(\mbox{male} )\times P(\mbox{male} )\times P(\mbox{male} ) \ \vspace{5} \ &=& \ (0.59)(0.41)(0.41)(0.41)(0.41) \ \vspace{5} \ &=& \ (0.59)(0.41)^4 \ \vspace{5} \ &\approx& \ 0.0167 \end{eqnarray}$

but the one female might be the second student instead:

$\normalsize \begin{eqnarray} P(\mbox{only second student female} ) &=& \ P(\mbox{male and female and male and male and male} ) \ \vspace{5} \ &=& \ P(\mbox{male} )\times P(\mbox{female} )\times P(\mbox{male} )\times P(\mbox{male} )\times P(\mbox{male} ) \ \vspace{5} \ &=& \ (0.41)(0.59)(0.41)(0.41)(0.41) \ \vspace{5} \ &=& \ (0.41)(0.59)(0.41)^3 \ \vspace{5} \ &=& \ (0.59)(0.41)^4 \ \vspace{5} \ &\approx& \ 0.0167 \end{eqnarray}$

or the third or the fourth or the fifth. These probabilities are all the same, so the probability that exactly one of the five students is female is given by:

$\normalsize \begin{eqnarray} P(Y=1) &=& \ P(\mbox{first only or second only or third only or fourth only or fifth only} ) \ \vspace{5} \ &=& \ P(\mbox{first only} )+ P(\mbox{second only} )+P(\mbox{third only} )+P(\mbox{fourth only} )+P(\mbox{fifth only} ) \ \vspace{5} \ &\approx& \ 0.0167+0.0167+0.0167+0.0167+0.0167 \ \vspace{5} \ &=& \ (5)(0.0167) \ \vspace{5} \ &\approx& \ 0.0834 \end{eqnarray}$

After this point it gets more complicated. For the probability that exactly two of the five students are female we need to determine the number of ways we can select two of the five students to be female; without too much difficulty we could derive a formula for this, and then derive a formula for the corresponding binomial probability. However, this formula isn't quite as intuitive as the geometric formula (and in practice we'll use the calculator anyway) so let's just jump directly to the calculator:

$\normalsize P(Y = 2)$ = binompdf(5,0.59,2) � 0.2399

Let's check our previous answers using the calculator:

$\normalsize P(Y = 0)$ = binompdf(5,0.59,0) � 0.0116

$\normalsize P(Y = 1)$ = binompdf(5,0.59,1) � 0.0834

and compute the remaining probabilities:

$\normalsize P(Y = 3)$ = binompdf(5,0.59,3) � 0.3452

$\normalsize P(Y = 4)$ = binompdf(5,0.59,4) � 0.2484

$\normalsize P(Y = 5)$ = binompdf(5,0.59,5) � 0.0715

You should check that the sum of these probabilities is 1. There are fairly simple formulas for the mean and standard deviation of a binomial distribution:

$\normalsize \mu = E(Y) = np$ and $\normalsize \sigma = SD(Y) = \sqrt{npq}$

The gory details of deriving these formulas can be found elsewhere. In our EdCC student example:

$\normalsize \mu \ = \ np \ = \ 5(0.59) = 2.95$

so, if we repeatedly selected 5 EdCC students at random and counted the number of female students in each 5-student group, we would expect to find an average of about 2.95 females per group, with a standard deviation of:

$\normalsize \sigma \ = \ \sqrt{npq} \ = \ \sqrt{5(0.59)(0.41)} \ = \ \sqrt{1.2095} \ \approx \ 1.10$ .

If we want to know the probability that we find no more than two females in our five-student sample, we can add the appropriate individual probabilities found above, or we can use the binomcdf feature on the TI-84:

$\normalsize P(Y \leq 2)$ = binomcdf(5,0.59,2) � 0.3349

Note that we are using the binomcdf feature here (where again c stands for "cumulative").

Binomial model revisited

Various polls conducted in late January 2007 report that about 35% of registered voters support President Bush's plan to send additional troops to Iraq. Suppose we want to ask these voters more detailed questions about the reasons they support the troop surge. We randomly select 1000 voters registered for our follow-up survey.

Let $\normalsize X$ represent the number of voters (out of the 1000 we contact) who approve of the troop surge. Let's check that a binomial model applies to this random variable:

Two outcomes: Either the voters answers "yes, I support the troop surge" or they do not; if they do not answer "yes" they might say "no, I do not support he troop surge" or "I don't know" or "I have no opinion" but we will classify all such responses as failures and we will consider a voter who answers "yes" as a success.

Constant probability of success: We are assuming that 35% of all registered voters support the troop surge. Since we are only sampling 1000 voters out of more than 100,000,000, the probability of success should not substantially change as we remove voters from the general population into our sample.

Independent trials: Since we are selecting the voters at random, their opinions about the troop surge should be independent of one another.

Suppose we want to get at least 400 voters who support the troop surge for our follow-up survey. What is the probability that our sample of 1000 randomly selected voters will contain at least 400 people who support the troop surge?

Since a binomial model applies, we can answer use the methods outlined above to find the answer. Note that:

$\normalsize P(X\geq 400) \ = \ 1 - P(X \leq 399)$ = 1-binomcdf(1000,0.35,399) � 0.00057 = 0.057%

so it is very unlikely that we will get enough voters in our sample who support the troop surge. We should plan to contact more than 1000 voters for our follow-up survey.

We've answered the question at hand, but there's another way to solve this problem that will be useful in future applications, so we employ it now to show that it gives (approximately) the same answer as our first method.

When $\normalsize np \geq 10$ and $\normalsize nq \geq 10$ we can approximate the binomial model with a Normal model. (Now is a good time to revisit the Normal model computations from Chapter 6 if necessary; to see why this condition is necessary, see the discussion on pages 386�387.)

In our present example we have $\normalsize n = 1000$ so:

$\normalsize np \ = \ 1000(0.35) \ = \ 350 \ \geq \ 10$

and:

$\normalsize nq \ = \ 1000(0.65) \ = \ 650 \ \geq \ 10$

so this condition is certainly satisfied.

Of course, in order to use a Normal model we need to specify the mean and standard deviation. For our present example, the mean is:

$\normalsize \mu \ = \ np \ = \ 1000(0.35) = 350$

and the standard deviation is:

$\normalsize \sigma \ = \ \sqrt{npq} \ = \ \sqrt{1000(0.35)(0.65)} \ = \ \sqrt{227.5} \ \approx \ 15.08$ .

Thus we want to use the Normal model N(350,15.08). To compute $\normalsize P(X\geq 400)$ we compute:

normalcdf(400,1E99,350,15.08) � 0.00046

Notice that this isn't equal to the exact answer we got with 1-binomcdf(1000,0.35,399), but it's fairly close (in fact, close enough to reach the same conclusion that we will need to survey more than 1000 voters).

We can sometimes get a more accurate answer if we realize that our random variable $\normalsize X$ is discrete: it might equal 398, or 399, or 400, or 401, but it can't equal 399.8; thus, any number between 399.5 and 400.5 would be interpreted as being equal to 400. So we adjust our computation slightly and use:

$\normalsize P(X \geq 400)$ � normalcdf(399.5,1E99,350,15.08) � 0.00051,

which is closer (although still not equal to the exact answer). This adjustment is called a continuity correction; while you're welcome to use it, our Normal model approximations will usually be good enough without it. Just keep in mind that the Normal model only gives us an approximation to the binomial model.

The Poisson Model*

The third model we study in this chapter is a bit different, as it does not involve Bernoulli trials. Instead, we seek to compute probabilities for a random variable that counts the number of occurrences of an event in a specified interval (often a time interval) where the mean number of occurrences per interval is fixed. We will use the Poisson model rarely in this class after Chapter 17, and very few exercises in Chapter 17 require it, so you may wish to skip this part of Chapter 17 unless you have time to spare.

Suppose that a math instructor tabulates the number of times per 50-minute lecture that a cell phone rings during class and finds that the mean number of cell-phone disruptions is 0.6 rings/lecture. If we assume that the cell phone disruptions are independent (they may not be, since once one phone rings other students are likely to make sure that their phones are turned off) and if we assume that the mean number of disruptions remains constant (it might not, as students may become more�or less�conscientious about turning off their phones as the quarter progresses, or at least on exam days) then we can apply the Poisson model.

Let $\normalsize Y$ be the random variable that counts the number of cell-phone disruptions during a randomly selected 50-minute lecture. According to the Poisson formula:

$\normalsize P(Y = 0) \ = \ \Large{\frac{e^{-0.6}0.6^0}{0!}} \normalsize \ = \ e^{-0.6} \ \approx \ 0.549$

so we would expect there to be no disruptions during about 55% of all lectures. Furthermore:

$\normalsize P(Y = 1) \ = \ \Large{\frac{e^{-0.6}0.6^1}{1!}} \normalsize \ = \ e^{-0.6} (0.6) \normalsize \ \approx \ 0.329$

so we would expect there to be exactly one disruption during about 33% of all lectures. And so on.

The formula is not difficult to use once we fine the e and ! keys on the calculator, but as with the geometric and binomial models there is a shortcut:

$\normalsize P(Y = 0)$ = poissonpdf(0.6,0) � 0.549

If we want to know the probability that no more than two disruptions occur during a 50-minute lecture, there is a shortcut for that, too:

$\normalsize P(Y = 2)$ = poissoncdf(0.6,2) � 0.967

Looking at this another way, we can see that there will be three or more disruptions in only about 2% of the lectures.

The Poisson Model Revisited*

While the Poisson model is interesting in its own right, we study it here primarily because it can be used as an approximation to the binomial model, even in some cases where the Normal model cannot (and when the calculator shortcuts might not work due to behind-the-scenes computations taxing the capabilities of the TI-84).

Suppose we wish to conduct a poll to determine support for a "fringe" presidential candidate who only has the support of 0.5% of all voters. Even if we poll 1000 voters, a Normal model does not apply since

$\normalsize np \ = \ 1000(0.005) \ = \ 5 \ < \ 10$ .

But we can use the Poisson model with mean:

$\normalsize \lambda \ = \ np \ = \ 1000(0.005) \ = \ 5$

Would it be unusual to poll 1000 voters and get three or fewer who support the fringe candidate? Let $\normalsize Y$ be the number of supporters of this fringe candidate among 1000 randomly selected voters:

$\normalsize P(Y = 3)$ = poissoncdf(5,3) � 0.265

This tells us that finding 3 or fewer supporters out of 1000 is not that surprising.

Homework

Work the following exercise in Chapter 17: 1�17 odd, 25, 27 and 33.

Errata

The binomial probability formula on page 384 (which you should ignore�use the TI-84 instead) is incorrect. It should read:

$\normalsize P(X=x) \ = \ _nC_x p^x q^{n-x}$ , where $\normalsize _nC_x \ = \ \frac{n!}{x!(n-x)!}$

The formula in the footnote also includes a typo; it should read:

$\normalsize \sum_{x=0}^{n} x \, \cdot \, _nC_x p^x q^{n-x} \ = \ np$

ActivStats

Work through the lessons on page 17-1 in the ActivStats lesson book, as time permits.

Additional Resources

Binomial Probability Histogram: This applet from the Univeristy of California at Berkeley Statistics Department offers useful insight into how the Normal model approximates the binomial model.
Binomial Distributions: Episode 17 from Against All Odds features a discussion of binomial probabilities, although some of the terminology may be different.
Sofia: Binomial Probability: Lesson 4.3 of the Sofia Open Content Initiative's Elementary Statistics course include a discussion of the binomial model and Lesson 4.4 discusses the Poisson model.