Computer Assignment 3

Computer Assignment 3
Title:
Sampling with and without replacement the Binomial and Hypergeometric random variables
with
The Poisson random variable as a limiting case of the Binomial random variable.
Introduction
There are occasions when addressing a problem in elementary probability where it is not clear to the beginner whether
a problem is an instance of the binomial random variable or the hypergeometric random variable (RV). Some of this can
be attributed to learning new and unfamiliar concepts. These are two distinct RV; hence, there should be no confusion.
Further investigation will reveal that there is an overlap or common ground between the two probability mass functions
(pmf) for the RVs that when understood will help to eliminate uncertainty as to which RV applies in a given problem.
The reader has been exposed to the theory of both the binomial and hypergeometric RV. Also, the reader is aware that
the mean and variance of a RV can be obtained by several different approaches: application of the definition of
expectation, moment generating functions, and etc. The formulas describing the preceding are well established. What
follows then is an attempt to show that the two RV are more closely related than the reader may have supposed.
Arguably this relationship is best seen with the assistance of computer programs and graphics. So, in addition to the
mathematical exposition below computer programs are provided. The reader is encouraged to improve upon these
programs. Running these programs to visualize the behavior of the random variable’s (pmf) will provide the reader with
both visual and numerical evidence of how the two RVs behave in similar ways under certain circumstances.
Theory
Suppose that a bin contains 𝑁 blocks, 𝐺 of which are green and 𝑅 of which are red. So 𝐺 + 𝑅 = 𝑁. Consider the
random experiment of selecting 𝑛 blocks. There are two procedures for doing this. First, the blocks can be sampled
with replacement; this means that after a block is selected, its color is noted, the block is replaced into the bin, and the
blocks are mixed. Thus the composition of the bin remains the same from selection to selection. The second way to
choose them is to sample without replacement. This means that blocks are not replaced after they are selected.
Let the random variable 𝑌 be the number of green blocks chosen in 𝑛 drawings. In sampling with replacement the
selections are independent from one to next and
𝑝 = 𝑃(green block chosen) =
𝐺
𝑁
At any one selection trial. Consequently, 𝑌 is binomially distributed in sampling with replacement.
In sampling without replacement successive selections are not independent; if a green block is chosen at the first draw,
the composition of the bin is altered, so the probability of choosing green at the next draw is changed. Using
combinatorial ideas the event {𝑋 = 𝑗} means that among n selections, 𝑗 are green and 𝑛 − 𝑗 are red. Thus 𝑗 selections
must be made from the 𝐺 green blocks and 𝑛 − 𝑗 must be made from the 𝑅 red blocks. There are 𝑁𝐶𝑛 samples of size
𝑛 from all 𝑁 blocks. The number of ways to select 𝑗 green blocks is 𝐺𝐶𝑗
; the number of ways to choose the remaining
𝑛 − 𝑗 blocks from the set of 𝑅 red ones is 𝑅𝐶𝑛−𝑗
. Consequently, the number of ways to choose 𝑗 green and 𝑛 − 𝑗 red
blocks is the product
𝐺𝐶𝑗
∙ 𝑅𝐶𝑛−𝑗
.
The set of values that 𝑗 can assume is restricted, however. Not only must 0 ≤ 𝑗 ≤ 𝑛, but the number of green blocks 𝑗
cannot exceed the total number 𝐺 of green bocks; similarly, the number of red blocks 𝑛 − 𝑗 cannot exceed the total
number 𝑅 of red blocks. Therefore,
𝑗 ≤ 𝐺 and 𝑛 − 𝑗 ≤ 𝑅.
Probability and Stochastic Processes by Fredrick Solomon pages 98 and 99
Instructions: Do the two exercises below by hand similarly as was asked of the reader in computer assignment 2.
Exercises
A box of 100 ornamental light bulbs contains 40 green and 60 red bulbs. Four are selected at random. Find the
probability that three are red, assuming that the sampling is done (a) with replacement and (b) without replacement.
In a Klingon prison there are 40 Vulcans and 60 Humans. The Klingons select 4 at random to be executed. Let the
random variable be the number of Vulcans in the group of four to be executed. With the random variable taking on
values from zero to four construct two probability tables one using the hypergeometric model and the other using the
binomial model. Compare the results of the two approaches.
Instructions: Two programs have been provided. One of the programs outputs the pmf for a binomial RV. The other
program outputs the pmf for a hypergeometric RV. Combine these two programs into a single program and include a
numerical error. The numerical error is the difference between the valued of the pmf. This error will tend towards zero
as the two pmfs become more alike. Run it with at least the values stated. Visually compare the outputs. In which
case(s) do the two distributions appear to be equivalent? In these case(s) what is the disparity in their numerical
output?
Computer Programs
Binomial vs Hypergeometric
Run your binomial program with 25 and 100 trials. Make bar plots for probabilities of success equal to 0.1, 0.15, 0.25,
and 0.5.
Run your hypergeometric program with at least the following suggested amounts in the table.
We are viewing the number of greens as the number of success. The size samples will play a role
in making the pmf more or less like that for the binomial pmf.
Poisson
A computer program that plots the Poisson probability mass function has been provided. The instructions of what to do
with it in this computer assignment are provided in the comments in the file.
Deliverables
The two exercises solved by hand.
A copy of your Python program of the combined binomial and hypergeometric programs. The output from the program
showing when the pmfs diverge and when the pmfs merge. The output includes the numerical error between the two
pmfs.
The run of the Poisson program with the outputs and the questions answered. (Put the answers in the PDF not in the
Python file.)
Rubric
Name and I.D. #
Name of Assignment
Submission Date
Not Satisfactory:
Data requested absent.
Satisfactory:
Has all data requested.
Exercise 1. Not Satisfactory:
Missing steps in solution.
Answer missing.
Satisfactory:
Shows steps in solution towards
final answer.
Exercise 2. Not Satisfactory:
Missing steps in solution.
Answer missing.
Satisfactory:
Shows steps in solution towards
final answer.
A block comment at beginning
of the combined program
summarizing it. Line comments
throughout the program
explaining its operation.
Not Satisfactory:
Conveying incomplete thoughts.
Satisfactory:
A brief two or three sentence
explanation at the beginning.
Sufficient line comments
explaining the code.
List references. Not Satisfactory:
No references listed.
Satisfactory:
References – assignment
handout, internet, students,
and etc. listed
Green Red
3 17
5 15
10 30
25 75
50 50
In the PDF submitted solutions
to exercises 1 & 2, a copy of the
.py file of the combined
programs that compare the
pmfs, the outputs, also, the
Poisson outputs with the
answers to questions to
dropbox and a .py of the
combined program to dropbox.
Not Satisfactory:
Absent solutions and answers to
exercises 1 & 2.
Absent outputs.
Absent copy of program.
Satisfactory:
Solutions to exercises 1 & 2, a
copy of the combined Python
program file, the output all in
the PDF, also, the Poisson
output with questions answered
and the .py file. These two
complete files are submitted to
dropbox.
Appendix 1
Random Variables and Distributions
Binomial
A finite number of trials
Independent trials
Two outcomes
The probability of success 𝑝 is constant (this implies sampling with replacement)
The random variable 𝑌 is the number of successes
𝑃({𝑌 = 𝑦}) = 𝑛𝐶𝑦 𝑝
𝑦𝑞
(𝑛−𝑦)
 where 𝑞 = 1 − 𝑝 and 𝜇 = 𝑛𝑝 𝜎 = √𝑛𝑝𝑞
Hypergeometric
A finite number of trials
Two outcomes
The probability of success 𝑝 is not constant (this implies sampling without replacement)
Population size is known
The random variable 𝑋 is the number of successes
𝑃({𝑋 = 𝑗}) =
𝐺𝐶𝑗 𝑅𝐶𝑛−𝑗
𝑁𝐶𝑛
where G+R = N
Poisson
The occurrences must be random,
The occurrences must be independent of each other,
The occurrences must be uniformly distributed over the interval being used.
𝑃({𝑋 = 𝑥}) = 𝑒
−𝜆
𝜆
𝑥
𝑥!
 where 𝑥 = 0, 1, 2, …
Appendix 2
Mean and variance relationship between binomial and hypergeometric distributions.
As the reader is aware a random variable 𝑋 is said to have a hypergeometric probability distribution if and only if
𝑝({𝑋 = 𝑗}) =
(
𝑟
𝑗
) (
𝑁−𝑟
𝑛−𝑗
)
(
𝑁
𝑛
)
where 𝑗 is an integer 0, 1, 2, … , 𝑛, subject to the restrictions 𝑗 ≤ 𝑟 and 𝑛 − 𝑗 ≤ 𝑁 − 𝑟.
Further there is a relationship between the binomial and hypergeometric that may not be familiar to the reader with
respect to the mean and standard deviation.
If 𝑋 is a random variable with a hypergeometric distribution.
𝜇 = 𝐸(𝑋) =
𝑛𝑟
𝑁
 and 𝜎
2 = 𝑉(𝑋) = 𝑛 (
𝑟
𝑁
) (
𝑁−𝑟
𝑁
) (
𝑁−𝑛
𝑁−1
).
Although the mean and the variance of the hypergeometric random variable seem to be rather complicated, they bear a
striking resemblance to mean and variance of a binomial random variable. Indeed, if we define 𝑝 =
𝑟
𝑁
and 𝑞 = 1 − 𝑝 =
𝑁−𝑟
𝑁
,we can then express the mean the mean and variance of the hypergeometric as 𝜇 = 𝑛𝑝 and
𝜎
2 = 𝑛𝑝𝑞 (
𝑁 − 𝑛
𝑁 − 1
).
You can view the factor
𝑁 − 𝑛
𝑁 − 1
in 𝑉(𝑌) as an adjustment that is appropriate when 𝑛 is large relative to 𝑁. For fixed 𝑛, as 𝑁 → ∞,
𝑁 − 𝑛
𝑁 − 1
→ 1.
Mathematical Statistics with Applications by Wackerly, Mendenhall, and Scheaffer 5th Ed.