Ronald C. Serlin and Joel R. Levin

University of Wisconsin

Journal of Statistics Education v.4, n.2 (1996)

Copyright (c) 1996 by Ronald C. Serlin and Joel R. Levin, all rights reserved. This text may be freely shared among individuals, but it may not be republished in any medium without express written consent from the author and advance notification of the editor.

**Key Words:** Teaching statistics; Probability.

The common development of the hypergeometric probability formula is typically confusing to students in introductory statistics courses. Two alternative developments that appear to be more intuitive and conceptually consistent are presented.

1 In most elementary statistics texts, the hypergeometric probability formula is developed within a combinatorial framework. Unfortunately, this development can be confusing to the beginning student, especially if a combination has been previously defined as a unique grouping of distinct objects in which order is unimportant. This teaching tip presents two alternative developments, one focusing on permutations and the other on modifying the binomial formula. Both avoid certain conceptual pitfalls of the traditional development.

2 The source of the potential confusion can best be described in the context of the usual hypergeometric development. As generally stated, a population of N objects is given, in which A of the objects are all of one kind and the remaining N-A objects are all of a different kind. From this population, a sample of n objects is selected at random without replacement. The goal is to calculate the probability that the sample contains exactly a objects of the first kind and n-a objects of the second kind.

3 To help students set up and solve problems such as these, a table like that shown in Table 1 is formed, and the rationale is given as follows: The total number of samples of size n selected from a population of size N is given by the binomial coefficient, which will be designated here as (N,n). There are (A,a) ways of selecting the objects of the first kind and (N-A,n-a) ways of selecting the objects of the second kind. Hence, the desired probability is (A,a)(N-A,n-a)/(N,n). This probability can be easily computed using the entries in the top and bottom rows of Table 1.

**Table 1**. A Usual Hypergeometric Table

Type A Type B Total In Sample a n-a n Not in Sample A-a N-A-(n-a) N-n Population A N-A N

4 Conceptualizing the combinations involved in the above rationale often creates confusion. The beginning student has just learned, for example, that with identical objects (A's) the two permutations A(1)A(2) and A(2)A(1) are indistinguishable. Is it not the case, then, that there is only one combination of a identical objects taken from a total of A in the population? In particular, does not this combination consist simply of a objects of Type A? A similar case can be made for the one combination possible in choosing the n-a objects of Type B. Indeed, the confusion is compounded in the calculation of the denominator, because the beginning student might feel that there are only n+1 combinations possible: (1) all objects of Type A; (2) all but 1 of the objects of Type A; (3) all but 2 of the objects of Type A; ...; (n+1) no objects of Type A.

5 The two present authors have adopted different approaches in their graduate-level introductory educational statistics classes to motivate the development of the hypergeometric formula from a more conceptually consistent standpoint. In both cases, however, the same resounding conclusion has been reached: In comparison to the traditional development -- as already described and as represented by Table 1 -- the tables definitely should be turned! Let us now consider the respective approaches to illustrate what we mean. The first author's approach will be referred to as the identical permutation rationale, whereas the second author's approach will be termed the binomial derivative rationale. Both of these approaches represent extensions of Conover's (1980, pp. 9-10) argument that the number of combinations of a objects selected from n distinct objects can be found using the formula for the number of permutations of a and n-a indistinguishable objects. After more than ten years of presenting the hypergeometric formula in the traditional manner, the present authors now teach the hypergeometric formula exclusively in the more conceptually consistent forms that follow.

6 The previous combinatorial confusion can be avoided by using the formula for the number of permutations of M objects when I(1) and I(2) of these are identical -- a formula that has already been derived in the course of studying permutations. This is simply M!/I(1)!I(2)! = (M,I(1)). The desired result can then be developed in the following manner. The entire population can be ordered in N!/A!(N-A)! = (N,A) ways. The n objects in the sample can be ordered in n!/a!(n-a)! = (n,a) ways, while the N-n objects not in the sample can be ordered in (N-n)!/(A-a)![(N-n)-(A-a)]! = (N-n,A-a) ways. The desired probability, then, is given by (n,a)(N-n,A-a)/(N,A), which can be seen immediately if Table 1 is turned on its side to form the entries of Table 2. (Alternatively, this can be accomplished by asking the students to view Table 1 from a sideways perspective.)

** Table 2**. An Unusual Hypergeometric Table

In Sample Not in Sample Population Type A a A-a A Type B n-a N-n-(A-a) N-A Total n N-n N

7 An alternative development, with the same end result, follows on the heels of the presumably already developed binomial formula rationale. The approach is best illustrated through example. Assume a population contains 75 red balls and 25 black balls, and it is desired to find the probability of selecting a sample of size 5 that contains 3 reds and 2 blacks.

8 The probability of selecting such a sample with replacement in any given order is: (3/4)^3(1/4)^2, or concretely for this example: (75/100)(75/100)(75/100)(25/100)(25/100). However, because there are (5,3) possible ways of ordering 5 objects of which 3 are identical and 2 are identical, the desired probability is the familiar result: P(binomial) = (5,3)(75/100)(75/100)(75/100)(25/100)(25/100).

9 If we now stipulate sampling without replacement, the above result can be modified to become: P(hypergeometric) = (5,3)(75/100)(74/99)(73/98)(25/97)(24/96). Putting this in terms of factorials, we obtain: P(hypergeometric) = (5!/3!2!)(75!/72!)(25!/23!)(95!/100!), which, upon rearrangement of the probability terms, becomes: P(hypergeometric) = (5!/3!2!)(95!/72!23!)/(100!/75!25!) = (n,a)(N-n,A-a)/(N,A). This is the same result that was obtained according to the table-turning identical permutation rationale.

10 Of course, if desired, the terms can be rearranged further to lead to the traditional hypergeometric formulation: P(hypergeometric) = (75!/3!72!)(25!/2!23!)/(100!/5!95!) = (A,a)(N-A,n-a)/(N,n).

11 It is our perception (based on many years of informal student feedback, both verbal and nonverbal) that the two present alternatives to the traditional development of the hypergeometric formula are more compatible with the beginning student's newly acquired conceptual knowledge. We do not claim to have systematically evaluated the extent to which turning the tables on the hypergeometric formula has also turned around students' course achievement and attitudes. We do, however, have some anecdotal (and clearly unscientific) evidence that might potentially relate the new approach to student learning outcomes. As was noted at the outset, both authors have at one time taught the hypergeometric formula in the traditional manner. More recently, we have used the approach advocated here, wherein the hypergeometric formula is conceptualized as the number of permutations of indistinguishable objects. In glancing back over class records, one of us found that before changing to the new derivation, students' average performance on the first midterm (covering basic statistics) and the second midterm (covering probability) were quite comparable, both about 80 out of 100 points. After changing to the new derivation, the first midterm average has continued to be about 80, but the average on the second midterm has consistently been somewhat higher -- about 85. Both the nature of the midterms and student characteristics have remained relatively constant over the last 19 years.

12 As a final comment, note that the preceding notions and approaches can be easily extended to the multinomial and multiple-category hypergeometric cases, again giving familiar results.

Conover, W. J. (1980), Practical Nonparametric Statistics (2nd ed.), New York: Wiley.

Ronald C. Serlin SERLIN@MACC.WISC.EDU Joel R. Levin LEVIN@MACC.WISC.EDU Department of Educational Psychology

1025 W. Johnson St.

University of Wisconsin

Madison, WI 53706

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