24 Teaching Statistics. Volume 24, Number 1, Spring 2002

Probability With Less Pain

KEYWORDS:Teaching;Probability teaching;Two-way tables.

Maxine Pfannkuch, George A. F. Seberand Chris J. WildThe University of Auckland, New Zealand.e-mail: pfannkuc@math.aukland.nz

seber@stat.auckland.ac.nzwild@stat.auckland.ac.nz

SummaryThe teaching of probability theory has been stead-ily declining in introductory statistics courses asstudents have difficulty with handling the rules ofprobability. In this article, we give a data-drivenapproach, based on two-way tables, which helpsstudents to become familiar with using the usualrules but without the formal structure.

INTRODUCTION

The teaching of probability theory is steadilydeclining in introductory statistics, partly be-cause of an increasing emphasis on data analysis,and partly because it is seen as too hard. This isparticularly sad with regard to Bayes Theorem be-cause of the wealth of socially important problems,especially concerning screening programmes, wherethe answers can be surprisingly counter-intuitive.As far as manipulations are concerned, there areno essential differences between probabilities,proportions and percentages. We simply use thedevice pick an individual from this population atrandom the probability that an individual hasa particular characteristic is the population pro-portion with that characteristic. This article is notconcerned with notions of randomness, but withthe manipulations. When we apply those mani-pulations to proportions of a sample or population,what we are doing is not probability theory butdata analysis, a legitimate and arguably import-ant part of a course which makes data analysis itsfocus.

The authors have been teaching at a universitywhere about 50% of each incoming cohort (ornearly 3000 students each year) take elementarystatistics. Our courses therefore have to cater fora wide variety of backgrounds and mathematicalabilities. Probability theory was the crunch pointin the class, and our help facilities became over-loaded. Our approach, like many other coursestaught elsewhere, tended to be formulae-driven and

incorporated three types of representation: two-waytables, Venn diagrams, and tree diagrams. BayesTheorem was the killer! There was an exodus ofthose mathematically unprepared and math-phobicstudents who were free to leave the course, supple-mented by panic and agonized discussions withthose who were trapped by their course require-ments. Students needed to recognize not only whattype of representation would be useful for solvinga problem, but also what formula to apply. Earlyon we realized we were not making connectionsbetween the types of representation. While we wereworking to address this, we came to the realizationthat tables formed a single, unifying representationthat could be used to solve all of the problems wewere setting students. With this shift of emphasis,the panic and overload on help facilities have nowlargely disappeared. One of our colleagues likes totalk about statistics for poets. This is an approachwith which even formula-phobic poets can solvethe usual problems addressed in elementary statis-tics up to and including Partition Theorem andBayes Theorem problems.

In this article we present a unified approach toteaching probability which uses just two-way tables;Venn and tree diagrams can be avoided. Of courseusing two-way tables for calculating proportionsor probabilities for unions, intersections and com-plements of events, and calculating conditionalprobabilities, is fairly well known, and features ina number of statistics texts (e.g. Siegel and Morgan,1998; Weiss and Hassett, 1991). However, we foundthat two-way tables can be extended to handle prob-lems of Bayes Theorem type where conditional

Teaching Statistics. Volume 24, Number 1, Spring 2002 25

probabilities are reversed. We were pleased re-cently to come across the excellent article byRossman and Short (1995) demonstrating manyaspects of the method. Here the emphasis is onusing the data given to actually construct and com-plete the two-way table which can then be used tofind the probabilities in question. This approachalso extends to (i) problems which require prob-abilities involving unions of events, (ii) tables whichcontain an unknown which has to be solved foralgebraically, and (iii) tables involving just inde-pendent events. As this material does not seem tobe widely known we have brought it together inthis article.

In what follows, the formulae given are there forthe reader. At one extreme, we can teach this mat-erial without the student seeing any formulae. Atthe other extreme, theory and formulae can bedeveloped in parallel and used to formally justifythe operations we perform on the tables.

Students can cope quite easily with the followingthree rules for probabilities or proportions:

(1) The probabilities of all outcomes add to one.[P(S) = 1]

(2) P(A does not occur) = 1 P(A does occur).[P(i) = 1 P(A)]

(3) (Addition Rule) Probabilities of mutually ex-clusive events are added:P(A or B or C or ) = P(A) + P(B) + P(C) +

Here or is used in the inclusive sense of and/or.However, difficulties arise for many students whenthey meet more complicated rules like

(4) P(A or B)= P(A) + P(B) P(A and B)

and various partition rules. As these rules causeproblems, they are often omitted so that teachersthen avoid examples which apparently depend onthem. These rules, however, are applied implicitlywhen we perform simple manipulations on two-waytables.

A useful rule for dealing with information that ap-plies only to a subset (conditional information) is

(5) A proportion of a proportion is obtained bymultiplying.

It can be pointed out that this is just another incar-nation of a fraction of a fraction being obtained by

multiplying in ordinary arithmetic. Algebraically,the above rule includes the multiplication rule forconditional probabilities

P(A and B) = P(A|B)P(B) = P(B|A)P(A)

and also applies to the special case of independentevents

if A and B are independent,P(A and B) = P(A)P(B)

The fraction that one proportion is of anothercan be obtained by dividing. This is much moreobvious when looking at a particular table. Withnotation, this is the conditional probability definedas P(A|B) = P(A and B)/P(B).

With a formal statement of the multiplication rule,we have a choice of order, either P(B |A)P(A) orP(A)P(B |A). Both are useful. The former is mostnatural when we have data on subsets of a popula-tion and are taking a proportion of a proportion.The latter is useful when there is a time order. Forexample, an urn contains 6 black balls and 4 whiteballs. Two balls are selected, one at a time. Whatis the probability that both are black? This treehas unconditional information (first fan) and con-ditional information (second fan). The advantageof the tree diagram approach is that it can beextended to a sequence of more than two events. Adisadvantage is that it becomes awkward to con-struct when at least one of the fans has a largenumber of alternatives (as in Example 3 below).Also, from in-depth interviews of a few students,and from our experience, many students learn tomultiply the probabilities along the branches of thetree diagrams without understanding the logic oftaking a proportion of a proportion. Furthermore,determining an order is less critical for the tablemethod than for trees. We therefore prefer the tableapproach.

Example 1

Table 1(a), constructed from data in the WorldAlmanac (1993, p. 157), cross-classifies job lossesin the US over a three-year period. Here 5,584,000job losses are broken down by the sex of the per-son who lost the job and the reason given for los-ing it. After presenting the table to the students,we consider choosing a lost job at random so thatthe sample space consists of 5,584,000 lost jobs,which are equally likely. A necessary foundation

26 Teaching Statistics. Volume 24, Number 1, Spring 2002

P(male or pos. abol.) = 0.3050 + 0.2142 + 0.0981+ 0.0650 = 0.6823

as before.

Conditional probability can readily be motivatedby using the table and comparing a cell frequency(or proportion) with the relevant row or columntotal. For example, to obtain the probability that amale job loss was due to the position being abol-ished, we note that there are 3,447,000 job lossesto males (in the male row), and of these 548,000were due to the position being abolished. HenceP( pos. abol.|male) = 548/3447. If we wanted to useproportions instead of frequencies, we have

P(pos. abol.|male) =

5485584

34475584

0 09810 6173

..

=

This conditional probability can be contrastedwith the reverse probability P(male | pos. abol.)= 548/911.

The above approach applies to problems that canbe conceived in terms of the operation of twofactors, for example two events that may or maynot occur. We

(1) decide on the factors involved and set up atwo-way table

(2) put the available information in the table(3) complete the table using the fact that rows and

columns add(4) use the completed table to obtain the desired

probabilities.

Example 2

Suppose that, between the hours of 9.00 am and5.30 pm, Maxine is available for student help 70%of the time, George is available 60% of the timeand both are available (simultaneously) 40% of thetime. We begin by deciding that the two factors areMaxines availability and Georges availability. Weconstruct a two-way table and insert what we know,as in figure 1. It is then a simple matter to fill in therest of the table and produce the bracketed num-bers. Whether or not the algebraic table is seen bystudents depends upon the mathematical level atwhich the teacher is pitching the instruction.

We can now find various probabilities. For ex-ample, the probability that Maxine is in but notGeorge is read directly as 0.3 [giving P(A and j) ].

for our approach is the ability to read such tablesand understand their structure including theidea that interior cells correspond to both row andcolumn conditions applying simultaneously. Afterdiscussing the significance of the row and columntotals in relation to events (for example, there are1,760,000 people who give rise to the event of slackwork), the table is then converted to a table ofproportions, given in table 1(b), which is also ob-served to add across and down. (The unwaryteacher should note that in some examples theproportions may not add up correctly because ofrounding error: a disadvantage of using propor-tions.) We note that, under the pick a lost job atrandom scenario, the proportions in the cellsare now probabilities and that the cell events aremutually exclusive.

Without recourse to formulae, the students are thenintroduced to the notion of finding probabilities(compare the rules above) of events expressed inwords visually from the table. For example, sup-pose that we want to find the probability of theevent that a randomly selected person is male orhas had a position abolished, namely P(male orpos. abol.). People satisfying that event are all thosein row 1 and column 3, a total of (1703 + 1196 +548 + 363) 1000 = 3,810,000 people. (Switched-on students will see quicker ways of obtainingthis total.) Since these are all equally likely to bechosen, the probability of the event is P(male orpos. abol.) = 3810/5584 = 0.6823.

Probabilities involving more than one cell are foundby adding cell proportions. For instance, in theprevious example

(a) Numbers of job losses (in thousands)

Reason for job loss TotalWorkplace Slack work Position

moved/closed abolished

Male 1703 1196 548 3447Female 1210 564 363 2137Total 2913 1760 911 5584

TWO-WAY TABLES

(b) Proportions

Reason for job loss TotalWorkplace Slack work Position

moved/closed abolished

Male 0.3050 0.2142 0.0981 0.6173Female 0.2167 0.1010 0.0650 0.3827Total 0.5217 0.3152 0.1631 1.0000

Table 1. Job losses in the US for 1987 to 1991

Teaching Statistics. Volume 24, Number 1, Spring 2002 27

Fig 1. Building the table for Example 2.

0.6

0.4

(0.2)

George

Maxine

Total

In

Out

(0.4)

(0.3)

(0.1)

1.00

0.7

(0.3)

TotalIn Out

Descriptive tableP(George in and

Maxine in)

P(George in) P(Maxine in)

P(B)

P(A and B)

P(A and B)

Total

A

A

P(B)

P(A and B)

P(A and B)

1.00

P(A)

P(A)

TotalB B

Algebraic table

Fig 2. Contraceptive data.

0.38

0 0.38= 0

(0.3800)

Method

Outcome

Total

Failed

Didnt

1.00

(0.0592)

(0.9408)

TotalSteril.

P(Failed and Oral) =P(Failed|Oral) P(Oral)

[ = 5% of 32%]

P(Steril.) = 0.38

0.32

0.05 0.32= 0.0160

(0.3040)

Oral

0.24

0.14 0.24= 0.0336

(0.2064)

Barrier

0.03

0.06 0.03= 0.0018

(0.0282)

IUD

0.03

0.26 0.03= 0.0078

(0.0222)

Sperm.

P(Barrier) = 0.24 P(IUD) = 0.03

P(Failed and IUD) =P(Failed|IUD) P(IUD)

[ = 6% of 3%]

The probability that just one of them is in is foundby adding the two appropriate cells, i.e. 0.2 + 0.3 =0.5 [giving P( (A and j) or (i and B) ) ]. The prob-ability that at least one of them is available is 0.4 +0.2 + 0.3 = 0.9 [giving P(A or B) ]. Note that we donot need to use rule 4. The probability that Maxineis in on an occasion when George is out is given by0.3/0.4 [giving P(A|j ) ]. The complementary eventsidea becomes reasonably obvious in a table with alarge number of cells. The example we chose hereis unexciting, but uncluttered by context. The re-maining examples use real and interesting data.

CONDITIONAL PROBABILITY

Example 3

Of American women using contraception (Elmer-DeWitt, 1990), 38% are sterilized, 32% use oral con-traceptives, 24% use barrier methods (diaphragm,condom, cervical caps), 3% use IUDs and 3% relyon spermicides (foams, creams, jellies). If we definethe failure rates of a method as the percentage ofwomen who become pregnant during a year of useof the method, then the failure rates for each ofthese methods are approximately as follows: steril-ization 0%, the oral contraceptive pill 5%, barriermethods 14%, IUDs 6% and spermicides 26%. Onequestion we may like to answer is what percentage

of women using contraception have an unwantedpregnancy over the course of a year. Another is,if we look only at women who experienced con-traceptive failure, what proportions were using eachtype of contraceptive method.

Once again we have two factors. One is Method the type of contraceptive method the woman wasusing. The other is Outcome whether or not thewoman experienced contraceptive failure. This sug-gests constructing a two-way table with dimen-sions Method and Outcome. We insert informa-tion about the proportions of women using eachmethod directly into the bottom row. Our otherinformation is conditional and we fill in the rest ofthe first row using rule 5. In the case of P(Oral andFailed ), for example, the 5% failure rate appliesto only 32% of the women, giving a proportion0.05 0.32 = 0.016 which both used oral contra-ceptives and experienced failure. This completes theunbracketed entries in figure 2. The rest of the tableis then filled in.

From the completed table, we can read off thatapproximately 6% of the women sampled had ex-perienced contraceptive failure [P(Failed ) = 0.0592].Conditional probabilities are easily found fromthe information in the table. For example,P(Barrier |Failed ) is 0.0336/0.0592 = 0.568. This tellsus that, of the women experiencing contraceptive

28 Teaching Statistics. Volume 24, Number 1, Spring 2002

failure, 57% were using IUDs. We note that this sortof information is often quoted in the media, leadingto the thought that IUDs must be a particularlyunreliable form of contraception. We can pointout to students that P(IUD|Failed ) is the wrongprobability for deciding which method to use. Therelevant probability is P(Failed |IUD), i.e. the failurerate among those using the method. For IUDs thisis 6%, which is nearly as good as the 5% failurerate for oral contraceptives, and much better thanbarrier methods or spermicides. In this example weimplicitly solve a fairly complicated Bayes The-orem problem without Bayes Theorem. A tree dia-gram here would end up with 10 possible paths.

Example 4

This example can also incorporate the Normal dis-tribution. Diabetes is a major disease in modernsociety. A standard test for diabetes is based onglucose levels in the blood after fasting for a pre-scribed period. For healthy people, the mean fast-ing glucose level is 5.31 mmol/L and the standarddeviation is 0.58 mmol/L. For untreated diabetesthe mean is 11.74 and the standard deviation is3.50. In both groups the levels appear approxi-mately Normal. At this point the two overlapping

Normal distributions can be drawn (as in figure 3),and the class can discuss where they might set thecut-off point C using this simple diagnostic test. Ifa patients fasting glucose level is above C theytest positive and we say that they have diabetes,while if it is below C they test negative and we saythat they do not have diabetes. We can point outthat one can be wrong with both types of diagno-sis. The probabilities of a correct diagnosis whenthe patient has or has not got diabetes are calledthe sensitivity and the specificity of the test, respect-ively. For example, if we use C = 6.5 then, fromNormal distribution calculations, the sensitivity isP(Positive|Diabetes) = 0.933 and the specificity isP(Negative|Not Diabetes) = 0.98.In New Zealand the proportion of people with dia-betes (the base rate) is estimated to be between 2and 5%. Suppose we screened the whole New Zea-land population for diabetes, took the base rate as3% and took the cut-off point as 6.5 mmol/L forthe diagnostic test. Through constructing a two-way table (figure 4) we can find out the proportionof New Zealanders testing positive and the pro-portion of those testing positively who would actu-ally have diabetes. From the table we can read thatapproximately 4.7% of New Zealanders would test

Fig 4. Proportions by disease status and test result.

(0.04739)

0.933 0.03= 0.02799

(0.01940)

DiseaseStatus

Total

Diabetes

Not Diabetes

1.00

0.03

0.97

TotalPositive

P(Diabetes and Positive) =P(Positive|Diabetes) P(Diabetes)

[ = 93.3% of 3%]

(0.95261)

(0.00201)

0.98 0.97= 0.95060

Negative

P(Not Diabetes and Negative) =P(Negative |Not Diabetes) P(Not Diabetes)

[ = 98% of 97%]

P(Diabetes) = 0.03

P(Not Diabetes) = 0.97

Test Result

Fig 3. Glucose test for diabetes.

5.31 11.74

Distribution of glucose levelsfor healthy people

98% of healthyhave a levelbelow 6.5

(Specificity)

93.3% of diabeticshave a levelabove 6.5

(Sensitivity)

C = 6.5

Distribution of glucoselevels for untreated diabetics

98%

93.3%

Teaching Statistics. Volume 24, Number 1, Spring 2002 29

positively. From the calculation P(Diabetes|Positive)= 0.02799/0.04739 = 0.5906 we can say that, forthose 4.7% who tested positive, approximately 59%would actually have diabetes. Such results are notintuitive. For instance, in New Zealand the baserate for diabetes varies for different ethnic groupsand ranges between 1 and 10%. If the ethnic groupbase rate is 1% then P(Diabetes|Positive) = 0.320or 32%, whereas a base rate of 9% gives 82.2%.Following some recent high-profile cases in NewZealand about false negatives in cancer screeningand diagnosis, which then go untreated, we cansimilarly obtain that P(Diabetes|Negative) = 0.0007if the base rate is 1%, 0.002 if the base rate is 3%and 0.007 if the base rate is 9%. Fortunately, thesenumbers are all fairly small.

The question now arises as to how we might estim-ate the base rate for diabetes; that is, we want toestimate the unknown p = P(Diabetes). To find agood estimate of the proportion of people in thecommunity who test positively for diabetes, wecollect data from a random sample of people. Sup-pose that 10.3% of such a sample test positive. Wecan now obtain table 2, and from the first column,the equation 0.933p + 0.02(1 p) = 0.103, whichgives p = 0.091. Thus if 10.3% of the sample testpositively then we estimate that 9.1% actually havediabetes. This example may be outside the comfortzone of many students but it is an illustration ofthe power of the two-way table.

INDEPENDENT EVENTS

Example 5

There are a large number of genetic-based bloodgroup systems that are used for typing blood. Twoof these are the Rh system (with blood types Rh+and Rh) and the Kell system (with blood typesK+ and K). For any person, their blood type inany one system is independent of their blood typein any other. For Europeans in New Zealand, about81% are Rh+ and about 8% are K+. If a EuropeanNew Zealander is chosen at random, we can con-struct a two-way table to find the probability thatthey are either (Rh+, K+) or (Rh, K).

Disease status Test result Total

Positive Negative

Diabetes 0.933 p 0.067 p pNot diabetes 0.02 (1 p) 0.98 (1 p) 1 pTotal 0.103 0.897 1

Table 2. Proportions by disease status and test result

From the above information we have P(Rh+) =0.81, P(Rh) = 0.19, P(K+) = 0.08 and P(K) =0.92. These probabilities belong in the row andcolumn totals respectively in table 3.

How do we obtain the probabilities in the interiorof the table? We use a multiplication rule, exceptthat this time we use it for independent events andnot for conditional probabilities. For example,

P(Rh+ and K+) = P(Rh+)P(K+)= 0.81 0.08 = 0.0648

Or, intuitively, the message of independence is thatno matter whether we look at the whole popula-tion, the K+ group or the K group, 81% of thegroup we look at will be Rh+. Thus, P(Rh+ andK+) is 81% of 8%. In a similar manner we cancalculate all the probabilities in table 3, or simplyfill them in using additivity.

To find the answer to our original question, wesimply add the probabilities for the (Rh+, K+) and(Rh, K) cells to get 0.0648 + 0.1748 = 0.2396. Indeveloping the above blood example we can usemore well-known blood groups such as the ABOsystem. However, the use of the Kell system withone fairly rare marker (K) enables us to take theproblem one step further into forensics. Supposethat a murder victim has blood type (Rh+, K) buthas a blood stain on him with type (Rh, K+),presumably from the assailant. What is the prob-ability that a randomly selected person matchesthis type? The answer is read from table 3, namely0.0152, which is quite small. Suppose further thatone of the suspects has this blood type. Either thesuspect is innocent and a random event with prob-ability 0.0152 has occurred, or the suspect is guilty.The jury is out! For further comments on this kindof forensic problem, see Rossman and Short (1995).

The above is a typical example in which the as-sumption of independence is made right from thebeginning. This is somewhat artificial and back tofront. Usually, independence has to be deducedfrom data, not the other way round. For example,suppose by chance that a survey of a large numberof European New Zealanders yielded the relative

Rh system Kell system Total

K+ K

Rh+ 0.0648 0.7452 0.81Rh 0.0152 0.1748 0.19Total 0.08 0.92 1.00

Table 3. Blood type data

30 Teaching Statistics. Volume 24, Number 1, Spring 2002

frequencies in table 3. The question now is, Arethe two blood types independent?. The answer isyes, as each of the four cell proportions (e.g. 0.0648)is exactly the product of the corresponding rowand column proportions (e.g. 0.81 0.08). In prac-tice, however, because the sample of people is finite,the cell relative frequencies will only approximatethe true genetic probabilities. Even if the two bloodtypes are independent, exact equalities now becomeonly approximate equalities and table 3 will notgenerally have the nice product structure; this leadsto the subject of testing for independence in con-tingency tables.

DISCUSSION

Two-way tables have allowed many of our studentsto obtain the underlying ideas behind probabilitycalculations without the use of formulae. In thisapproach, the manipulations are entirely straight-forward and are based on common sense ratherthan half-remembered formulae. This allows atten-tion to be focused on the essentially hard problemof abstracting relevant information from context.For the more advanced students, two-way tablescan support an algebraic approach.

Two-way tables are not a complete panacea. Wehave noted the lack of extensibility to chains ofmore than two events. On the balance side, how-ever, the approach is closely tied to primaryintuitions, facilitates the posing and solution of in-teresting problems, has a data-analysis focus and

fosters a greater confidence in handling probabilityproblems. It has a different sort of extensibility tocontingency table data, relative risks, odds ratios,chi-square tests and so on. And it does reduce thelevel of fear. It really is probability with less pain.

We conclude with a final recommendation for stu-dents who cannot cope even with proportions intables. Tell students to apply the given proportionsto a large group of individuals, say 10,000. Theresulting tables, which are completed to make rowsand columns add as before, are tables of counts.The students can then think simply in terms ofnumbers of individuals. The calculation of desiredproportions from the table of counts is much moreobvious.

ReferencesElmer-DeWitt, P. (1990). TIME, 26 February,

p. 44.Rossman, A.J. and Short, T.H. (1995). Con-

ditional probability and education reform:are they compatible? Journal of StatisticsEducation, 3 (http://www.amstat.org/publi-cations/jse/ )

Siegel, A.F. and Morgan, C.J. (1998). Statis-tics and Data Analysis: An Introduction (2ndedn). New York: Wiley.

Weiss, N.A. and Hassett, M.J. (1991). Intro-ductory Statistics (3rd edn). New York:Addison-Wesley.

World Almanac and Book of Facts (1993).Mahwah, NJ: PRIMEDIA Reference Inc.

LOOK AHEADLOOK AHEAD

Look Ahead and I hope that what you can see isICOTS-6, the Sixth International Conference onTeaching Statistics, to be held in Durban from 7 to12 July. The ICOTS events are of course one of themajor highlights in statistical education world-wide.You can see the conference announcement on theinternet at http://www.beeri.org.il/icots6. If you cango, please do so you are unlikely to regret it.

Teaching Statistics is hoping to have a fewextra articles in the next issue that will be ofspecial reference to ICOTS and the internationaldimension.

Meantime, as usual we whet your appetites withthe titles of a few of the regular articles that will beappearing soon:

Using consulting for teaching elementarystatistics

Voting methods matterDucks and greenThe language of statistics