Some Instructional Issues in Hypergeometric Distribution

A brief introduction to sampling without replacement is presented. We represent the probability of a sample outcome in sampling without replacement from a finite population by three equivalent forms involving permutation and combination. Then it is used to calculate the probability of any number of successes in a given sample. The resulting forms are equivalent to the well known mass function of the hypergeometric distribution. Vandermonde’s identity readily justifies different forms of the mass function. One of the new form of the mass function embodies binomial coefficient showing much resemblance to that of binomial distribution. It also yields some interesting identities. Some other related issues are discussed.

section_content: Usually hypergeometric probability distribution is introduced without really introducing sampling scheme without replacement. In this paper, we want to introduce the related issues of sampling without replacement to provide clarity in the understanding of the hypergeometric probability distribution. We represent the probability of a sample outcome in sampling without replacement from a finite population by three equivalent forms involving permutation and combination. Then it is used to calculate the probability of any number of successes in a given sample. The resulting forms are equivalent to the well known mass function of the hypergeometric distribution. Some related instructional issues are presented.
section_name: Introduction

section_content: Consider a population of three doctors and two nurses denoted by,, A B C and, D E respectively. Notice that the individuals are distinctly identified. The sample space of a sample of 3 persons selected without replacement is given by
section_name: Sampling Without Replacement

section_content: Let ( 1, 2, 3) i A i  be the event that Doctor A is included in the -th i selection. Then the probability that A is included in the 1 st selection is
section_name: ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE i) Probability That a Person Is Included in a Particular Draw

section_content: which is   

1 (1/ 5) 1 (1/ 4) (1/ 3) 1/ 5.
section_name: ( ) ( ) ( | ) ( | ) P A A A P A P A A P A A A   

section_content: Obviously, the probability that unit j of the population of N units is included in the -th i selection is given by

.
section_name:   

section_content: The probability that Doctor Ais included in the sample is

) (1/ 5) (1/ 5) (1/ 5) 3 / 5.
section_name: ii) Probability That a Person Is Included in a Sample

section_content: Thus each of the 5 persons have the same chance (3 / 5) of being selected in a without replacement sample of size 3.
section_name: P A P A A P

section_content: The number of samples of size n that contains unit j of the population of N units is 1 (1). 1

the probability that unit j of the population of N units is included in the sample is given by 1
section_name: Let

section_content: The probability of selecting a sampleof size 3 n  members from a population of 5 people discussed in Section 1. 1 is given by

In general, at the first draw the probability that one of the n specified units is selected is / n N. At the second draw the probability that one of the remaining ( 1) n  specified units is drawn is ( 1) /( 1) n N  , and so on. Hence the probability that all n specified units are selected in n draws is, 1977). An alternative argument is provided now. Since the number of samples that includes unit j and k of the population in the sample is given by 2 (2), 2

 the probability that unit j and k of the population will be included in the sample is given by 2 (2)

( 1). 2 (0) ( 1 )

Again the number of samples that includes unit, j k and l of the population is given by 3 (3), 3

and hence, the probability that unit, j k and l of the population will be included in the sample is given by

Since the number of samples that include specific n units of the population is

the probability that specified n units of the population of N units is included in a sample of size n is given by ( ) (0)

which is the same as (1. 3).
section_name: Probability of Selecting a Sample (Equiprobable Case)

section_content: There are two major ways of calculating hypergeometric probabilities. One assumes that the items in the population are distinguishable, or can be labeled to make distinguishable. This will be discussed in Section 3. 

In the other case, it is immaterial whether the items in the population are distinguishable or indistinguishable. This will yield a sample space  where outcomes are based on dichotomous nature of the population. This will be discussed in Section 2. 

Suppose that a population containing K items are of one kind (say defective) and N K  items are of different kind (say non-defective). Let n items be drawn at random in succession, without replacement, and X denote the number of defective items selected. 

The quantity The paper is organized as follows. In Section 2, we will clearly demonstrate unequally likely sample space  and related representations of the hypergeometric probabilities. In Section 3, we will discuss the equally likely sample space and the wellknow hypergeometric mass function. We compare them by putting the two sample spaces based on Example 2. 1 and Example 3. 1 side by side. In Section 4, we show by an example how exact hypergeometric probabilities can be calculated. In Section 6, we present Vandermonde's identities related to hypergeometric distribution. Some other related issues are discussed. In Section 7, we prove that the probability of a particular sequence of outcomes with exactly x defective items and n x  nondefective items in sampling without replacement, with unequally likely sample space, converges to that of a similar sequence in binomial distribution.
section_name: Hypergeometric Probabilities

section_content: Let the population items be divided into units of two exhaustive kinds and denote the unequally likely sample space of at most 2 n outcomes. It is usually done by a tree diagram for the case of indistinguishable items. But in this method, it is really immaterial whether the items are distinguishable or not. Then we have the following lemma. Lemma 2. 1 Suppose that an urn contains K items of one kind (say defective) and
section_name: Conditional Probability Method (Unequally Likely Sample Space)

section_content:  items are of a different kind (say non-defective). Let n items be drawn at random in succession, without replacement, and X denote the number of defective items selected. The probability of x successive successes in n trials is given by any of the following three mass functions:

Proof. The left hand side of (2. 1a) is given by

which is the same as (2. 1a). The above can be written as non-defectives out of at most 2 n outcomes. The elements in the sample space are not equally likely. The motivation that led to the following theorem is also implicit in Kendal and Stuart (1969, 133), and Joarder and Al-Sabah (2007). 

Theorem 2. 1 Suppose that an urn contains K items of one kind (say defective) and N K  items are of a different kind (say non-defective). The items may be distinguishable or indistinguishable in each of the two categories. Let n items be drawn at random in succession, without replacement, and X denote the number of defective items selected. The probability of x successes in n trials is equivalently given by any of the two equivalent forms:

Proof. Any sample outcome of n items that have exactly x defectives and n x  non-defective items is given by (2. 1). Since there are n x       such outcomes, out of a maximum of 2 n outcomes in the sample space, we have

Hence by (2. 1b) or (2. 1c), the mass function is given by (2. 2a) or (2. 2b). 

The name hypergeometric is derived from a series introduced by the Swiss mathematician and physicist, Leonard Euler, in 1769. The probabilities in (3) are the successive terms of

where 2 1 1 2 (, ; ; ) F a a b x is the generalized hypergeometric function (Johnson, Kotz and Kemp, 1993, 237). 

Example 2. 1 A large chain retailer purchases a certain kind of electronic device from a manufacturer. The manufacturer indicates that 3 out of 5 devices are defective. The inspector of the retailer randomly picks up 3 devices from a 5 devices. Determine the probability that the sample will have two defective devices.
section_name: N K

section_content: is the event that in the -th i draw we have a defective item and f

is the event that in the -th i draw we have a non-defective item. Then the sample space (use tree diagram) is given by

Since each of the element in the event of interest

It can also be done by (2. 2b) as follows:

2) 0. 60. 2

The number 60 in the denominator of probabilities is explained at the end of theorem 3. 1. The probability mass function is explicitly given by the following table:

3 /10 6 /10 1/10
section_name: Solution:

section_content: In this section, it is required that the populations units be distinguishable. In case, they are indistinguishable, one may label them to make them distinguishable. 

Theorem 3. 1. Suppose that an urn contains K items of one kind (say defective) and
section_name: Equiprobable Method

section_content:  items are of a different kind (say non-defective). The items may be distinguishable or indistinguishable in each of the two categories. Let n items be drawn at random in succession, without replacement, and X denote the number of defective items selected. The probability of x successes in n trials is given by
section_name: N K

section_content: Proof. There will be a total of N n       equally likely elements in the sample space. The combinatorial proof of this theorem is available in most textbooks on statistics and discrete mathematics (e. g. Barnett, 1998). There are

ways of choosing x of the K items (say defective items) and

non-defective items, and hence there are

which is the same as (2. 2b). 

The method in this section, also guarantees that sample outcomes are equally likely or equiprobable. Thus this method produces a Simple Random Sampling where "simple" refers to the equally likely outcomes. 

Example 3. 1 A random committee of size 3 is selected from 3 doctors and 2 nurses. Suppose that the doctors and members can be identified well making the individuals distinguishable. What is the probability that there will be 2 doctors in the committee?

Solution: Suppose the doctors are labeled as 1 2, D D and 3, D while the nurses are labeled as 1 N and 2 N to make the items in the population distinguishable. The sample space of outcomes is given by  
section_name: 1)

section_content: The event of having 2 doctors in the committee is given by

which has a probability of 6 /10. This can be directly done (3. 1) as the following

) 0. 6. P X   That is there exactly 6 ways of selecting 2 doctors and 1 nurse in a sample of size 3. 

There are 5 10 3
section_name: D D D D D N D D N D D N D D N D N N D D N D D N D N N D N N

section_content: Then the probability will be

The following table will provide insight into the number of sample points:

Unequally Likely Sample Space Equally Likely Sample Space Mass function ( ),

We have taught both the methods in some service courses in statistics and found that students get better insight by the Conditional Probability Method. The elements of the sample space of the Conditional Probability Method are not equally likely or equiprobable. One element the sample space in the Conditional Probability Method maps on to some elements of the sample space of the Equiprobable Method. 

Example 3. 2 Suppose that a shipment of 9 ( ) N digital voice recorders contains 4 ( ) K that are defective. If n voice recorders are randomly chosen without replacement for inspection, what is the probability that a. the first two of 3 n  checked will be defective but the third one will be non-defective? b. 2 of the 3 n  recorders will be defective?
section_name:        

section_content: a. The probability is

which can be directly solved by (2. 1c) as 3), the probability that 2 of the 3 voice recorders will be defective is given by

2 4 5 3 5 2 5 14

  which, by (2. 2b), can also be written as

Note that the maximum number of sample points is 3 2 8.  Alternatively, by using the equiprobable method (3. 1), we have

which is 30/84, where the number of sample points is 9 84.
section_name: Solution:

section_content: We remark that 84 sample points in formula (3. 1) has been inflated to 3!(84) 504  in the formula in (2. 2b). 

The students erroneously tend to use the hypergeometric probability function (3. 1) based on Equally Likely Sample Space to find the probability of a simple event in part (a). Hence we recommend to use Unequally Likely Sample Space which provides probability of a simple event (say, part a by (2. 1c)) or a compound event (say, part b by (2. 2b)). 

Example 3. 3 There are N devices in a box. A total of K of them are faulty and N K  are sound. One takes three devices in succession at random without replacement. The probability that first two are faulty but the third one is not faulty is 12 / 60. what is the probability that in a sample of 3 devices, two are faulty and one is sound?
section_name:       

section_content: The probability that the first two devices selected are faulty and the third one is sound is given by

Hence by the given condition, we have,

which can be solved by preparing the following table: Notice that there are two solutions 3, 2,

2) 0. 60. 5 P X    It will be difficult to solve the problem by (2. 2a) or by (3. 1), but easy by using (2. 2b).
section_name: Solution:

section_content: Since the hypergeometric parameters are integers, it is possible to calculate the exact probability of any event. The following lemmas are in Hua (1982). 

Lemma 4. 1 Let p be a prime. Then the exact exponents of p that divides ! n is given by

where x     is the largest integer less than or equal to. 

x Lemma 4. 2 For any positive integer 2, n  the quantity ! n can be written as a product of prime numbers in the following manner:

To decompose 9! by Lemma 4. 1, the exponent of 2 will be

which can be expressed as Trong (1993) developed an algorithm to calculate accurate Cumulative Distribution Function of Hypergeometric Distribution.
section_name: Computational Accuracy

section_content: Acceptance sampling is an important field of statistical quality control that was popularized by Dodge and Romig and originally applied by the U. S. military to the testing of bullets during World War II. If every bullet was tested in advance, no bullets would be left to ship. If, on the other hand, none were tested, malfunctions might occur in the field of battle, with potentially disastrous results. Dodge and Romig reasoned that a sample should be picked at random from the lot, and on the basis of information that was yielded by the sample, a decision should be made regarding the disposition of the lot. In general, the decision is either to accept or reject the lot. This process is called Lot Acceptance Sampling or just Acceptance Sampling.
section_name: Acceptance Sampling Plan

section_content: The identity

which is equivalent to part (b). Since (2. 2a) is a probability mass function, part (c) follows from (2. 2a). Similarly, part (d) follows from (3. 1) and part (e) follows from (2. 2b).
section_name: Vandermond'e Identity

section_content: Suppose that an urn contains K items of one kind (say defective) and N K  items are of a different kind (say non-defective). Let n items be drawn at random, with replacement in succession, and X denote the number of defective items selected. The probability that any item is defective at any draw is / p K N  (say). Then with arguments similar to section 2, the probability of having x successive defectives and ( ) n x  successive nondefectives is given by

which equals,,

x n x p q  so that For any other sequence having x successes and n x  failures, the probability, in the limit, will be the same as above. 

This shows the equivalence of binomial and hypergeometric distribution in the limit. Though the fact is available in most textbooks on statistics, the result in (7. 2) and the factor n       in the hypergeometric mass function (2. 2) will be insightful to the students and instructors.
section_name: Binomial and Hypergeometric Probabilities

section_name: Acknowledgements

section_content: The author gratefully acknowledges the excellent research support provided by King Fahd University of Petroleum and Minerals, Saudi Arabia. The author is also grateful to Dr. A. Laradji, Dr. M. H. Omar, Dr. Walid S. Sabah, Dr. R. M. Latif, Dr M. Riaz and Dr. I. Rahimov for many constructive suggestions.
section_name: Acknowledgements