Stat02-Binomial.wxmx

TABLE OF CONTENTS

Preface 1

References 2

Discrete Distributions Defined 3

What Is a Discrete Random Variable? 4

Mean and Variance of Discrete Data Set 5

Fundamental Probability Laws 5

Permutations and Combinations 7

binomial (n, k) 9

The Binomial (n, p) Distribution 9

pdf_binomial (k, n, p), n>0, 0<p<1, k = 0,1,2,....,n 11

Two Tosses of a Balanced Coin 13

mean_binomial (n, p) 14

var_binomial (n, p) 14

std_binomial (n, p) 15

cdf_binomial (k, n, p) 15

Example 1 16

Example 2 18

quantile_binomial (q, n, p) 18

100 Flips of a Balanced Coin 18

Inverse Type of Problem 22

binomialCalc (k, n, p) 22

confidence (q, m, s) 23

Number of Side Effects from Medications 25

Number of Fraudulent Credit Card Transactions 28

Number of Spam Emails per Day 32

Number of River Overflows 32

Probability of Intercepting All Incoming Missiles 33

Number of Missiles Launched for Even Chance for Defense Penetration 35

random_binomial (n, p), random_binomial (n, p, m) 39

Random sample size m = 100 simulations, n = 100, p = 0.5 40

discrete_freq (data) 41

Random sample size m = 1000, n = 100, p = 0.5 42

Preface
In Stat02-Binomial.wxmx we discuss the discrete Binomial (n, p) probability distribution and its application, using Maxima tools and methods. This is the second worksheet in my Statistics with Maxima section.
Edwin L. (Ted) Woollett https://home.csulb.edu/~woollett/ April 9, 2024
References

In our series Statistics with Maxima we have used some examples and explanations (with much editing and additions) from:

Roger Barlow, Statistics: A Guide to the Use of Statistical Methods in the Physical Sciences, John Wiley, 1989,

Ch. 3 Reagle & Salvatore [RS], Statistics and Econometrics, 2nd ed, Schaum's Outlines, 2011, McGraw Hill,

Ch. 8 Fred Senese [FS], Symbolic Mathematics for Chemists: A Guide for Chemists, 2019, Wiley,

Louis Lyons, Statistics for Nuclear and Particle Physics, 1986, Cambridge Univ. Press, Luca Lista, 'Statistical Methods for Data Analysis in Particle Physics',

Lecture Notes in Physics 909, 2016, Springer-Verlag,

Frederick James, 'Statistical Methods in Experimental Physics', 2nd ed., 2006, World Scientific.

References specific to the Binomial distribution:

https://www.statology.org/binomial-distribution-real-life-examples/ https://www.statology.org/binomial-distribution-calculator/

https://online.stat.psu.edu/stat414/lesson/10

(%i5)

(%o1)

load (descriptive); load (distrib); fpprintprec : 6$ ratprint : false$ logexpand : all$

C:/maxima−5.43.2/share/maxima/5.43.2/share/descriptive/descriptive.mac

(%o2) C:/maxima−5.43.2/share/maxima/5.43.2/share/distrib/distrib.mac

Homemade functions fll, head, tail, Lsum are useful for looking at long lists.

(%i13)

3

fll ( aL) := [ first (aL), last (aL), length (aL) ]$ declare (fll, evfun)$

head(L) := if listp (L) then rest (L, - (length (L) - 3) ) else error("Input to 'head' must be a list of expressions ")$

declare(head,evfun)$

tail (L) := if listp (L) then rest (L, length (L) - 3 ) else error("Input to 'tail' must be a list of expressions ")$

declare(tail,evfun)$

Lsum (aList) := apply ("+", aList)$ declare (Lsum, evfun)$

Discrete Distributions Defined

From www.investopdedia.com:

"A discrete probability distribution counts occurrences that have countable or finite outcomes.

Discrete distributions contrast with continuous distributions, where outcomes can fall anywhere on a continuum.

Common examples of discrete distribution include the binomial, Poisson, and Bernoulli distributions.

These distributions often involve statistical analyses of "counts" or "how many times" an event occurs.

In finance, discrete distributions are used in options pricing and forecasting market shocks or recessions."

From http://www.stat.yale.edu/Courses/1997-98/101/ranvar.htm:

"If a random variable can take only a finite number of distinct values, then it must be discrete. Examples of discrete random variables include the number of children in a family, the Friday night attendance at a cinema, the number of patients in a doctor's surgery, the number of defective light bulbs in a box of ten."

What Is a Discrete Random Variable?

Quoting Luca Lista (Sec. 1.1):
"Many processes in nature have uncertain outcomes. This means that their result cannot be predicted before the process occurs. A random process is a process that can be reproduced, to some extent, within some given boundary and initial conditions, but whose outcome is uncertain. This situation may be due to insufficient information about the process intrinsic dynamics which prevents to predict its outcome, or lack of sufficient accuracy in reproducing the initial conditions in order to ensure its exact reproducibility. Some processes like
quantum mechanics phenomena have intrinsic randomness. This will lead to possibly different outcomes if the experiment is repeated several times, even if each time the initial conditions are exactly reproduced, within the possibility of control of the experimenter. Probability is a measurement of how favored one of the possible outcomes of such a random process is compared with any of the other possible outcomes."

A coin toss is "random" because we are ignorant of the 'initial conditions'. Repeated trials tell us something about how those initial conditions vary between trials

For the purposes of calculating things for experimental physics, we need physical probability. In particular we need 'frequentist probability':
"Probability is the frequency with which a particular outcome occurs in repeated trials."
P = (number of occasions on which that outcome occurs)/(total number of measurements).
Quoting L. Lyons, Sec. 2.1:
"In many situations we deal with experiments in which the essential circumstances are kept constant, and yet repititions of the experiment produce different results. Thus the result of an individual measurement or trial may be unpredictable, and yet the possible results of a series of such measurements have a well defined distribution."

What about events that can't be repeated? They don't have probabilities. Quoting [RS] Sec. 3.3:
"A random variable is a variable whose values are associated with some probability of being
observed. A discrete (as opposed to continuous) random variable is one that can assume only finite and distinct values. The set of all possible values of a random variable and its associated probabilities is called a probability distribution. The sum of all probabilites equals 1."

Quoting
https://saylordotorg.github.io/text_introductory-statistics/s08-discrete-random-variables.html,

"The probability distribution of a discrete random variable X is a listing of each possible value x taken by X along with the probability P(x) that X takes that value in one trial of the experiment.

The mean μ of a discrete random variable X is a number that indicates the average value of X over numerous trials of the experiment. It is computed using the formula μ=Σx P(x).

The variance σ^2 and standard deviation σ of a discrete random variable X are numbers that indicate the variability of X over numerous trials of the experiment. They may be computed using the formula σ^2 = (Σx^2 P(x) ) − μ^2, taking the square root to obtain σ."
Mean and Variance of Discrete Data Set

Consider a data set that contains M unique discrete values x_k, and assume the value x_k occurs with frequency f_k. Let N equal the sum of the frequencies.
N = sum (f_k, k, 1, M).
The mean
<x> = sum (f_k*x_k, k, 1, M) / N. The variance
Var(x) = sum ( f_k* (x_k - <x>)^2, k, 1, M )/ N.
The standard deviation is the square root of the variance.
Fundamental Probability Laws

We follow Matthews and Walker, Mathematical Methods of Physics, 2nd. Ed (1970) Ch. 14

"Let P(A) be the probability of something (called A) occurring when an experiment is performed. 0 <= P(A) <= 1. If A is certain to happen, then P(A) = 1. If A will certainly not happen,

then P(A)=0."

"To illustrate some more complicated possibilities consider an experiment with n equally likely outcomes, involving two events, A and B. Let

n1 = number of outcomes in which A occurs, but not B n2 = number of outcomes in which B occurs, but not A n3 = number of outcomes in which both A and B occur

n4 = number of outcomes in which neither A nor B occurs. Since we have exhausted all possibilities,

n1 + n2 + n3 + n4 = n .

The probabilities of A and of B are

P(A) = (n1 + n3)/n, P(B) = (n2 + n3)/n.

We can define more complicated probabilities.

The probability of either A or B (or both) occurring is P (A + B) = (n1 + n2 + n3)/n.

The probability of both A and B occurring is called the joint probability of A and B: P (A, B) = n3/n.

Finally, we may define "conditional probabilites"; the probability that A occurs, given that B occurs, is P(A | B) = n3/(n2 + n3).

Similarly, P(B | A) = n3 / (n1 + n3).

Two important rules may be extracted from this simple example, and in fact are easily seen to be true in general:

P (A + B) = P(A) + P(B) - P(A, B), (1)

since (n1+n3)/n + (n2+n3)/n - n3/n = (n1 + n2 + n3)/n.

Likewise we have the law, the probability of either A or B or C or any two, or all three occurring is P (A + B + C) = P(A) + P(B) + P(C) - P(A, B, C).

If events A, B, C are mutually exclusive events, then the joint probability P (A,B,C) = 0, and we are left with P (A + B + C) = P(A) + P(B) + P(C).

P (A, B) = P(B) P(A | B) = P(A) P(B | A), (2)

since n3/n = [(n2+n3)/n] * [ n3/(n2+n3)] = [(n1+n3)/n] * [n3/(n1+n3)]."

Consider the probability of drawing an ace from a deck of 52 cards. Since a deck contains 4 aces, P(1 ace) = 4/52 = 1/13. If event A is drawing an ace from deck 1 (of 2 decks) and

event B is drawing an ace from deck 2, the probability that when one card is drawn from each of two decks, at least one will be an ace is P(A + B) = P(A) + P(B) - P(A,B) = 1/13+1/13 - (1/13)^2

= (2*13 - 1)/13^2 = 25/169. In calculating P(A, B) = P(A)*P(B) we used the assumption that the result of drawing a card from deck 1 has no influence on the result when drawing a card from deck 2 - the two drawings are "independent".

Recall the probability of both A and B occurring is called the joint probability of A and B, P(A,B). In general, if events A and B are statistically independent, P(A, B) = P(A)*P(B). (3)

If events A and B are mutually exclusive (which means P(A, B) = 0) then P(A + B) = P(A) + P(B) as one can see from (1) above.

An example of P (A, B) = P(B) P(A | B) = P(A) P(B | A) is the probability of drawing two hearts, when two cards are drawn successively from the same deck of cards without replacement:

You have 13 hearts in a 52 card deck to start, so multiply the probabilities of each draw:

P(2 hearts) = P(draw 1 heart, draw2 heart) = P(1 heart)*P(1heart | 1heart) = (13/52)*(12/51) = 1/17

To find the probability of drawing a heart on the first draw and a club on the second draw from the same deck of cards, you follow a similar process. The probability of drawing a heart on the first draw is 13/52, and the probability of drawing a club on the second draw is 13/51 since there are 13 clubs remaining in the deck after removing one heart. The probability of drawing a heart

and then a club from the same deck is (13/52) x (13/51) = 169/2652.

(%i14)

(%o14)

7

(13/52)*(13/51);

204

Permutations and Combinations

Permutations

Quoting: https://en.wikipedia.org/wiki/Permutation
"In mathematics, a permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements. The word "permutation" also refers to the act or process of changing the linear order of an ordered set."

"For example, there are six permutations (orderings) of the set {1, 2, 3}: written as tuples, they are (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), and (3, 2, 1).
Anagrams of a word whose letters are all different are also permutations: the letters are already ordered in the original word, and the anagram reorders them. The study of permutations of finite sets is an important topic in combinatorics and group theory."

"The number of permutations of n distinct objects is n factorial, usually written as n!, which means the product of all positive integers less than or equal to n."

"The number of arrangements or 'permutations', of n objects is n!, since the first position can be occupied by any one of the n objects, the second by any one of the (n - 1) remaining objects, and so on."

"In elementary combinatorics, the k-permutations, or partial permutations, are the ordered arrangements of k distinct elements selected from a set. When k is equal to the size of the set, these are the permutations in the previous sense."

"In older literature and elementary textbooks, a k-permutation of n means an ordered arrangement (list) of a k-element subset of an n-set. The number of such k-permutations (k-arrangements) of n is denoted variously..." . We will use P (n, k) for that number.

P (n,k) = product of k factors = n*(n-1)*(n-2) ( n - (k - 1) )
= n*(n-1)*(n-2) ( n - k + 1) ) = n! / (n-k)!

since the first element of the sequence (permutation) can be chosen from any of n elements of the n-set, and then the second element of the sequence can be chosen from any of the n-1 remaining elements of the n-set, etc.

For example consider the example of a 3-set S = {1, 2, 3}, and form all 2-permutations: (1, 2), (1, 3), (2, 1), (2, 3), (3, 1), (3, 2).
The first element of each 2-permutation can be chosen in n = 3 ways, the second element in n-1 = 2 ways. So the total number of 2-permutations is n*(n-1) = 3*2 = 6. With k = 2,
n - k + 1 = 3 - 2 + 1 = 2. So we can write (using (n - k)! = (3 - 2)! = 1! = 1)
P(3, 2) = n*(n-1)....(n - k + 1) = 3*2 = 6 = n!/(n-k)! = 3! / 1! = (1*2*3)/ 1 = 6.
Combinations

Quoting: https://en.wikipedia.org/wiki/Combination
"In mathematics, a combination is a selection of items from a set that has distinct members, such that the order of selection does not matter (unlike permutations). For example, given three fruits, say an apple, an orange and a pear, there are three combinations of two that can be drawn from this set: an apple and a pear; an apple and an orange; or a pear and an orange. More formally, a k-combination of a set S is a subset of k distinct elements of S. [This means none of the members of a k-combination can be identical - 'no repititions'] So,
two combinations are identical if and only if each combination has the same members. [The arrangement of the members in each set does not matter.] If the set has n elements, the number of k-combinations, denoted by C(n,k), is equal to the binomial coefficient."

Given the set of 3 objects {1, 2, 3}, the number of 2-combinations of this set are: (1,2), (1, 3), (2, 3), so C (3, 2) = 3.
1. binomial (n, k)

The Maxima function binomial (n, k) can be used here:

(%i15) binomial (3, 2);

(%o15) 3

(%i16) 3! / (2! * (3 - 2)! );

(%o16) 3

In general each k-combination can be changed by changing the arrangement (order) of the k distinct elements, and there are k! ways of ordering those k elements. Hence the

number P(n,k) of k-permutations of n distinct elements equals the number of (unordered) k-combinations of n elements times the number of ways of ordering the k elements:

P(n, k) = C(n, k) * k!.

Solving for C(n, k) we get: C(n, k) = P(n, k) / k!, or

C(n, k) = n! / (n-k)! / k! == n! / ( k! * (n - k)! ) = binomial (n, k)

The Binomial (n, p) Distribution
1. Pb (k; n, p)
  The binomial distribution is used to find the probability of k 'successes' Pb (k; n, p) in n trials of the same experiment when (1) there are only two possible and
  mutually exclusive outcomes, (2) the n trials are independent, and (3) the probability of of 'success', p, remains constant in each trial. Then
  
  Pb (k; n, p) = ( n! / (k! * (n - k)! ) * p^k * (1 - p)^(n - k).
  
  where n! (read 'n factorial') = n*(n-1)*(n-2) *3*2*1, and 0! = 1 by definition.
  This assumes 0 < p < 1, n is a positive integer, and k = 1, 2, 3, , n.
  
  The [theoretical] mean of the binomial distribution is μ = n*p.
  The [theoretical] variance of the binomial distribution is n*p*(1 - p) = μ * (1 - p). The [theoretical] standard deviation is σ = sqrt (n*p*(1 - p) ) = sqrt (μ * (1 - p) ).
  
  If p < 0.5 the distribution is skewed to the right. If p > 0.5 the distribution is skewed to the left. If p = 0.5 the distribution is symmetrical.
  1. Origin of Pb (k; n,p) Formula
    Adapting the explanation in https://en.wikipedia.org/wiki/Binomial_distribution,
    
    The formula for Pb (k; n,p) can be understood as follows:
    
    p^k * (1-p)^(n-k) is the probability of obtaining the sequence of Bernoulli trials in which the first k trials are “successes“ and the remaining (last) (n - k) trials result in “failure“. Since the trials are independent with probabilities remaining constant between them, any arrangement of
    n trials with k successes (and (n - k) failures ) has the same probability of being achieved (regardless of positions of successes within the arrangement).
    
    How many distinguishable arrangements of n objects if n = n1 + n2, where n1 = k of the objects ar identical (type 'success') and the other n2 = (n - k) objects (type 'failure') are also identical
    (but different from the n1 type)? The total number of permutations of n objects is n! but each
    *distinguishable* permutation appears n1! * n2! times, so the number of *distinguishable* permutations is n!/(n1!*n2!) = n! / ( k! * (n-k)! ) = binomial (n, k).
    
    Here is a simple example: suppose n = 3, n! = 6, n = n1 + n2, n1 = 1 success, n1! = 1,
    n2 = 2, n2! = 2 failures, number of "distinguishable" permutations is n!/(n1!*n2!) = 6/ (1*2) = 3. Distinguishable permutations: {SFF, FSF, FFS}.
    
    Since the n trials are independent with probabilities remaining constant between them, any arrangement of n trials with n1 = k successes (and n2 = (n - k) failures ) has the same probability of being achieved (regardless of positions of successes within the arrangement). That probability is p^k * (1 - p)^(n-k).
    
    Recall from our section on basicd probability laws, we had the example:
    If events A, B, C are mutually exclusive events, then the joint probability P (A,B,C) = 0, and we are left with P (A + B + C) = P(A) + P(B) + P(C).
    
    Let N = number of distinguishable permutations = n! / ( k! * (n-k)! ) in the above example, and let {d1, d2, , dN} denote the distinguishable arrangements of the k successes each
    mutually exclusive events whose joint probability is equal to zero. Then
    P (d1 + d2 + d3 + ...+ dN) = P(d1) + P(d2) + P(d3) + + P(dN).
    In our case each term is equal to p^k * (1 - p)^(n-k), which is added n! / ( k! * (n-k)! ) times to produce Pb (k; n, p) = binomial (n, k) * p^k * (1 - p)^(n-k).
2. pdf_binomial (k, n, p), n>0, 0<p<1, k = 0,1,2, ,n
  What we have called Pb (k; n, p) is returned by the Maxima function pdf_binomial (k, n, p), with 0 < p < 1 and with n a positive integer and with k = 0, 1, 2, , n.
  ("pdf" standing for probability distribution function.)
  
  Assume p = 0.5 is the probability of success in each identical conditions trial. What is the probability of k = 0 success if we only have 1 such trial? 2 such trials? etc?
  Answer: 0.5, 0.25, 0.125 etc.
  Because p = 0.5, 1-p = 1 - 0.5 = 0.5 = p, and Pb (k; n, 0.5) =
  ( n! / (x! * (n - x)! ) * p^x * (1 - p)^(n - x) = (n!/(0!*n!)*(p^0)* (1-p)^n = (1-p)^n = p^n.
  
  (%i17) pdf_binomial (k, n, p);
  n
  (%o17)
  k
  ( 1 −p ) n − k pk
  
  (%i18)
  grind (%)$
  binomial(n,k)*(1-p)^(n-k)*p^k$
  
  (%i19) pdf_binomial (k, n, 0.5);
  (%o19) 0.5n n
  k
  
  (%i20)
  grind(%)$
  0.5^n*binomial(n,k)$
  
  (%i21) pdf_binomial (0, n, 0.5);
  (%o21) 1.0 0.5n
  
  The probability of zero success in one trial is 0.50 = p,
  
  (%i22) pdf_binomial (0, 1, 0.5);
  (%o22) 0.5
  
  The probability of zero successes in two trials is 0.25 since p^2 = (0.5)^2 = 0.25.
  
  (%i23) pdf_binomial (0 , 2, 0.5);
  (%o23) 0.25
  
  The probability of zero successes in ten trials is 0.0009765.
  
  (%i24) 0.5^10;
  (%o24) 9.76563 10−4
  Convert a probability (which must be in the interval [0, 1]) to % chance by multiplying by 100.
  
  (%i25) 100*%;
  (%o25) 0.0976563
  
  Watch out for the confusing notation - In Maxima the symbol % refers to the previous output, and has nothing to do with percent chance.
  
  We get the same probability using pdf_binomial (0, 10, 0.5).
  
  (%i26) pdf_binomial (0, 10, 0.5);
  (%o26) 9.76563 10−4
  
  What about the probability of getting zero successes in 100 trials (with p = 0.5 for each trial).
  
  (%i27) pdf_binomial (0, 100, 0.5);
  (%o27) 7.88861 10−31
3. [RS: Example 9]: Two Tosses of a Balanced Coin
  
  Sidebar:
  See the Oct. 11, 2023 article at: https://phys.org/news/2023-10-flipped-coins-fair-thought.html with the title: 'Flipped coins found not to be as fair as thought', by Bob Yirka , Phys.org,
  about practical conditions required for a 'fair' coin toss.
  
  Quoting [RS]:
  "The possible outcomes in two [successive] tosses [two trials] of a balanced coin are TT, TH, HT, and HH. Thus
  P(0 H) = 1/4, P(1 H) = 1/2, and P(2 H) = 1/4.
  [Note the sum of the probabilities is 1]
  The number of heads is therefore a discrete random variable, and the set of all possible outcomes [in two trials] is a 'discrete probability distribution'".
  
  Since "the binomial distribution is used to 'find the probability of k number of occurrences or successes of an event P(k), in n trials of the same experiment when (1) there are only two possible and mutually exclusive outcomes, (2) the n = 2 trials are independent, and (3) the probability of occurrence or success, p = 0.5 remains constant in each trial', the distribution of successes is described by a Binomial (2, 0.5) distribution.
  
  Applying the Maxima function pdf_binomial (k, n, p) with p = 0.5, n = 2, k = 0, 1, 2,
  
  (%i28) mypoints : makelist ([ k, pdf_binomial (k, 2, 0.5 ) ], k, 0, 2 );
  (mypoints) [ [ 0 , 0.25 ] , [ 1 , 0.5 ] , [ 2 , 0.25 ] ]
  (%i29)
  
  (%t29)
  wxdraw2d ( xrange = [-1, 3], yrange = [0, 0.6], points_joined = impulses, xlabel = "Number of Heads", ylabel = "Probability",
  title = "Distribution of Heads in Two Tosses of a Balanced Coin", background_color = light_gray, grid = true,
  line_width = 4, color = red, points (mypoints) )$
  1. mean_binomial (n, p)
    
    What is the 'mean number of heads' in two trials for which p = 1/2?
    
    The mean result (the average result) is advertised to be n*p = 2*(1/2) = 1, also given using sum (x*P(x), x, xmin, xmax) = 0*P(0) + 1*P(1) + 2*P(2) = 0 + 1*(1/2) + 2*(1/4) = 1/2 + 1/2 = 1.
    
    Using our basic definitions, we have a data set with M = 3 distinct values: x_1 = 0, x_2 = 1,
    and x_3 = 2 (heads), which occurs with frequencies f_1 = 1, f_2 = 2, f_3 = 1, and the sum of the frequencies sum (f_k, k, 1, M) = N = 4. Then the mean is given by our basic definition:
    mean = <x> = sum (f_k*x_k, k, 1, M) / N. = ( 1*0 + 2*1 + 1*2 )/ 4 = 4/4 = 1. Using Maxima's mean_binomial (n, p) function:
    (%i30) mean_binomial (2, 0.5);
    (%o30) 1.0
  2. var_binomial (n, p)
    
    The theoretical standard deviation is advertised to be σ = sqrt (n*p*(1 - p) ), and the variance is σ^2 = n*p*(1 - p) = μ * (1 - p) = 1*(1/2) = 1/2.
    The Maxima function: var_binomial (n, p) returns the variance of a Binomial (n, p) random variable, with 0 < p < 1 and n a positive integer.
    
    (%i31) var_binomial (2, 0.5);
    (%o31) 0.5
    
    Using our basic definition Var(x) = sum ( f_k* (x_k - <x>)^2, k, 1, M )/ N, we get (1*(0 - 1)^2 + 2*(1 - 1)^2 + 1*(2 - 1)^2) / 4 = 2/4 = 1/2.
  3. std_binomial (n, p)
    
    The standard deviation is the square root of the variance, so should be sqrt(1/2).
    
    (%i32)
    (%o32)
    sqrt (1/2);
    2
    1
    
    (%i33) %, numer;
    (%o33) 0.707107
    
    The theoretical standard deviation is σ = sqrt (n*p*(1 - p) ) = sqrt (μ * (1 - p) )
    = sqrt ( 1 * (1 - 1/2) ) = sqrt (1/2).
    
    The Maxima function std_binomial (n, p) returns the standard deviation σ of a Binomial (n, p) random variable, with 0 < p < 1 and n a positive integer.
    
    (%i34) std_binomial (2, 0.5);
    (%o34) 0.707107
  4. cdf_binomial (k, n, p)
    
    The 'cumulative distribution function' (cdf): cdf_geometric (k, p) returns the sum of the values P(k) for a Geometric (p) random variable, with 0 < p < 1, beginning at k=0 and ending at k.
    P (k <= n) = cdf_geometric (n, p)
    
    The 'cumulative distribution function' [cdf] cdf_binomial (k, n, p)
    returns the sum of the values P(k) for a Binomial (n, p) random variable, with 0 < p < 1 and n a positive integer, beginning at k = 0 and ending at k. If k is n, the returned answer should be 1, since the sum of the probabilities of all possible events is 1.
    P (k <= k1) = cdf_geometric (k1, n, p)
    (%i35)
    for j:0 thru 2 do print (j, cdf_binomial (j, 2, 0.5) )$
    0 0.25
    1 0.75
    2 1
    
    Here is a summary of the uses of pdf_binomial and cdf_binomial:
    
    P (k = k1) = pdf_binomial (k1, n, p), (probability of finding exact value k1),
    P (k < k1) = cdf_binomial (k1 - 1, n, p), (probability of finding a value less than k1),
    P (k <= k1) = cdf_binomial (k1, n, p), (probability of finding a value less than or equal to k1), P (k > k1) = 1 - cdf_binomial (k1, n, p), (probability of finding a value greater than k1),
    based on: P (k <= k1) + P (k > k1) = 1,
    P (k >= k1) = 1 - cdf_binomial (k1 - 1, n, p, (probability of finding a value greater than or equal to k1), based on: P (k < k1) + P (k >= k1) = 1.
  5. Example 1 of Use of cdf_binomial (k, n, p)
    
    This is Example 10.6 from https://online.stat.psu.edu/stat414/lesson/10/10.3
    
    By some estimates, twenty-percent (20%) of Americans have no health insurance. Randomly sample n = 15 Americans.
    1. Let k denote the number in the sample with no health insurance. What is the probability that exactly 3 of the 15 sampled have no health insurance?
      
      Solution:
      Since 15 is small relative to the population of N = 300,000,000 Americans, and all of the other criteria pass muster (two possible outcomes, independent trials, ), the random variable k can
      be assumed to follow a binomial distribution with n = 15 and p = 0.2.
      Using the probability mass function for a binomial random variable, the calculation is then relatively straightforward:
      P (k = 3) = binomial (15, 3)* (0.2)^3 * (0.8)^12 = 0.25.
      
      (%i37) binomial (15, 3)* (0.2)^3 * (0.8)^12; pdf_binomial (3, 15, 0.2);
      (%o36) 0.250139
      (%o37) 0.250139
      
      That is, there is a 25% chance, in sampling 15 random Americans, that we would find exactly 3 that had no health insurance.
    2. What is the probability that at most one of those sampled has no health insurance?
      
      Solution:
      "At most one" means either 0 or 1 of those sampled have no health insurance. That is, we need to find:
      P (k <= 1) = P (0) + P (1), which is given by cdf_binomial (1, n, p),
      
      (%i38) cdf_binomial (1, 15, 0.2);
      (%o38) 0.167126
      
      That is, we have a 16.7% chance, in sampling 15 random Americans, that we would find at most one that had no health insurance.
    3. What is the probability that more than seven have no health insurance? Solution:
      Yikes! "More than seven" in the sample means finding one of the values 8, 9, 10, 11, 12, 13, 14, 1
      P (k <= 7) + P (k > 7) = 1 implies P (k > 7) = 1 - P (k <= 7) = 1 - cdf_binomial (7, 15, 0.2).
      
      (%i39) 1 - cdf_binomial (7, 15, 0.2);
      (%o39) 0.00423975
      
      .... the probability that more than 7 in a random sample of 15 would have no health insurance is 0.0042 (the chance of finding that more than 7 in the random sample have no health insurance is about 0.4%).
    4. What is the probability that at least one of the 15 sampled has no health insurance?
      
      P (k < 1) + P ( k >= 1) = 1 implies that P (k >= 1) = 1 - P (k < 1) = 1 - cdf_binomial (0, 15, 0.2)
      
      (%i41) 1 - cdf_binomial (0, 15, 0.2);
      1 - pdf_binomial (0, 15, 0.2);
      (%o40) 0.964816
      (%o41) 0.964816
      
      That is, the probability that at least one person in a random sample of 15 would have no health insurance is 0.9648. (The chance of finding at least one person in a random sample of 15 with no health insurance is 96.5%.)
    5. What is the probability that fewer than 5 have no health insurance?
      
      This asks for the probability P (k < 5), ie., the number found is either zero, one, two, three, or four, which we can find using cdf_binomial (4, 15, 0.2), [P (k < 5) = P (k <= 4) ]:
      (%i42) cdf_binomial (4, 15, 0.2);
      (%o42) 0.835766
      
      .... the probability that fewer than 5 people in a random sample of 15 would have no health insurance is 0.8358.
  6. Example 2 of Use of cdf_binomial (k, n, p)
    
    This is Ex. 10.7 in https://online.stat.psu.edu/stat414/lesson/10/10.3
    
    Many utility companies promote energy conservation by offering discount rates to consumers who keep their energy usage below certain established subsidy standards. A recent EPA report notes that 70% of the island residents of Puerto Rico have reduced their electricity usage sufficiently to qualify for discounted rates. If ten residential subscribers are randomly selected
    from San Juan, Puerto Rico, what is the probability that at least four qualify for the favorable rates?
    
    The sample size is n = 10, the probability of a random island resident being qualified for the discount is p = 0.7. If k = the number in the sample of ten who qualify for the discount, this question asks for P (k >= 4). Since P (k < 4) + P (k >= 4) = 1, P (k >= 4) = 1 - P (k < 4)
    = 1 - cdf_binomial (3, 10, 0.7).
    
    (%i43) 1 - cdf_binomial (3, 10, 0.7);
    (%o43) 0.989408
    
    .... the probability that at least four people in a random sample of ten would qualify for favorable rates is 0.9894. (~ 99% chance)
  7. quantile_binomial (q, n, p)
    
    The Maxima function quantile_binomial (q,n,p) returns the q-quantile of a Binomial(n,p) random variable, with 0 <= p <= 1 and n a positive integer.This is the inverse
    of cdf_binomial. Argument q must be an element of [0,1] and q should be a value returned by cdf_binomial.
    
    (%i44)
4. for j:0 thru 2 do (
  qq : cdf_binomial (j, 2, 0.5),
  print (j, qq, quantile_binomial (qq, 2, 0.5) )) $
  0 0.25 1
  1 0.75 1
  2 1 2
  
  Number of Successes in 100 Flips of a Balanced Coin
  We continue to assume the probability of success in any one trial is p = 0.5 (50% probability).
  
  mean_binomial (n, p) produces the mean and std_binomial (n,p) calculates one standard deviation.
  
  (%i45) mean_binomial (100, 0.5);
  (%o45) 50.0
  
  The average (ie., the mean) number of heads out of 100 independent flips of a balanced coin is predicted to be 50. Here is std_binomial (100, 0.5) which measures the spread about the mean number of heads (50):
  
  (%i46) std_binomial (100, 0.5);
  (%o46) 5.0
  
  Let's calculate the probability of observing exactly k = 50 successes in n = 100 trials, if the probability of success on each single trial is p = 0.5.
  
  (%i47) pdf_binomial (50, 100, 0.5);
  (%o47) 0.0795892
  
  Thus about 8% chance for getting exactly 50 successes in 100 trials.
  
  Next we calculate the probability of observing k = 0 heads in 100 flips of a balanced coin.
  
  (%i48) pdf_binomial (0, 100, 0.5);
  (%o48) 7.88861 10−31
  
  (%i49)
  
  (%o49)/R/
  rat(%);
  1
  1267650600228229401496703205376
  
  (%i50) float(denom(%));
  (%o50) 1.26765 1030
  The probability of getting zero heads in the course of 100 flips of a balanced coin is one part in about 10 raised to the 30 power.
  
  See: https://en.wikipedia.org/wiki/Orders_of_magnitude_(numbers)
  10^30 is about one fifth of the estimated number of bacterial cells on planet earth, an incredibly large number.
  
  10^(-30) = (1,000)^(-10) = one quintillionth.
  
  To get the probability that we get somewhere in the range of 45 successes to 55 successes in 100 trials, (mean +/- std) we need to sum over these probabilities from x = 45 to x = 55.
  
  (%i51) sum (pdf_binomial (j, 100, 0.5), j, 45, 55);
  (%o51) 0.728747
  
  So about 73% chance of getting somewhere in the range of 45 - 55 successes in 100 trials. We can make a plot of the probability distribution function pdf_binomial (k,100,0.5).
  (%i55)
  mypoints : makelist ( [j, pdf_binomial (j, 100, 0.5) ], j, 20, 80)$ fll (mypoints);
  head (mypoints); tail (mypoints);
  (%o53) [ [ 20 , 4.22816 10−10 ] , [ 80 , 4.22816 10−10 ] , 61 ]
  (%o54) [ [ 20 , 4.22816 10−10 ] , [ 21 , 1.61073 10−9 ] , [ 22 , 5.78398 10−9 ] ]
  (%o55) [ [ 78 , 5.78398 10−9 ] , [ 79 , 1.61073 10−9 ] , [ 80 , 4.22816 10−10 ] ]
  
  The *continuous* normal distribution (see pdf_normal (x, m, s)) closely agrees with the discrete binomial distribution, provided m = mean_binomial(100, 0.5) = 50, and
  s = std_binomial (100, 0.5) = 5.
  (%i56)
  
  (%t56)
  wxdraw2d ( xrange = [30, 70], yrange = [0, 0.1], points_joined = impulses, xlabel = "number successes in 100 trials", ylabel = "probability", grid = true, title = "Binomial distribution n = 100, p = 0.5",
  background_color = light_gray,
  line_width = 2, color = red, points (mypoints), color = black, key_pos = top_left,
  key = "pdf normal (x, 50, 5)",
  explicit (pdf_normal (x, 50, 5), x, 30, 70) )$
  
  pdf_normal (x, m, s) produces the probability at location x for a nomal distribution having mean m and standard deviation s.
  
  The maximum of the normal distribution used is given by:
  
  (%i57) pdf_normal (50, 50, 5), numer;
  (%o57) 0.0797885
  
  which is also the maximum of our binomial distribution.
  
  The area (probability) in the x-interval [m - 3*s, m + 3*s], or within 3 standard deviations of the mean is almost equal to 1.
  
  (%i58) integrate (pdf_normal (x, 50, 5), x, 35, 65), numer;
  (%o58) 0.9973
  
  Summing the values of pdf_binomial (k, n, p) we get very close to the same answer:
  
  (%i59) sum (pdf_binomial (j, 100, 0.5), j, 35, 65);
  (%o59) 0.99821
  1. Inverse Type of Problem
    
    Question: There is a 95% chance that the number of heads (in 100 coin flips) will be in the range [0, x]. What is the approximate value of x?
    
    An *approximate* answer is provided by quantile_binomial (0.95, 100, 0.5), since as yet we don't know if 0.95 is a possible return value of cdf_binomial (k, 100, 0.5) with k equal to an integer.
    
    (%i60) quantile_binomial (0.95, 100, 0.5);
    (%o60) 58
    
    We can check this approximate value as follows:
    
    (%i61)
    for j: 57 thru 59 do print (j, cdf_binomial (j, 100, 0.5) )$
    57 0.933395
    58 0.955687
    59 0.971556
    
    There is about a 95% chance that the number of heads (in 100 coin flips) will lie in the range of 0 to between 57 and 58.
  2. binomialCalc (k, n, p)
    
    What is the probability (when n = 100 trials and p = 0.5) that k < 55?
    That is, what is the probability that the number of successes in 100 trials (flips) is less than 55?
    One way to calculate this is by summing the values of pdf_binomial (k, 100, 0.5) from k = 0 to k = 54.
    
    (%i62) sum (pdf_binomial (k, 100, 0.5), k, 0, 54);
    (%o62) 0.815899
    
    Another way is to use cdf_binomial (54, 100, 0.5):
    
    (%i63) cdf_binomial (54, 100, 0.5);
    (%o63) 0.815899
    
    What is the probability the number of successes in 100 trials (flips) is less than or equal to 55?
    
    (%i65) sum (pdf_binomial (k, 100, 0.5), k, 0, 55);
    cdf_binomial (55, 100, 0.5);
    (%o64) 0.864373
    (%o65) 0.864373
    What is the probability the number of successes in 100 trials (flips) is greater than 55?
    
    (%i67) 1 - sum (pdf_binomial (k, 100, 0.5), k, 0, 55);
    1 - cdf_binomial (55, 100, 0.5);
    (%o66) 0.135627
    (%o67) 0.135627
    
    What is the probability the number of successes in 100 trials is greater than or equal to 55?
    
    (%i69) 1 - sum (pdf_binomial (k, 100, 0.5), k, 0, 54);
    1 - cdf_binomial (54, 100, 0.5);
    (%o68) 0.184101
    (%o69) 0.184101
    
    We can incorporate these methods into a Maxima function binomialCalc(xx, n, p) which shows the probabilities P(k = xx), P (k < xx), P (k <= xx), P(k > xx), and P (k >= xx) as output.
    
    (%i70)
    binomialCalc (xx, nn, pp) := block (
    print (" "),
    print (sconcat (" For n = ", nn, ", p = ", pp, ", the probabilities are:")), print (" "),
    print (sconcat (" P ( k = ", xx, " ) : ", pdf_binomial (xx, nn, pp) )),
    print (sconcat (" P ( k < ", xx, " ) : ", cdf_binomial (xx - 1 , nn, pp) )),
    print (sconcat (" P ( k <= ", xx, " ) : ", cdf_binomial (xx, nn, pp) )),
    print (sconcat (" P ( k > ", xx, " ) : ", 1 - cdf_binomial (xx, nn, pp) )),
    print (sconcat (" P ( k >= ", xx, " ) : ", 1 - cdf_binomial (xx - 1, nn, pp) )), done )$
    
    (%i71) binomialCalc (55, 100, 0.5)$
    For n = 100, p = 0.5, the probabilities are: P ( k = 55 ) : 0.0484743
    P ( k < 55 ) : 0.815899 P ( k <= 55 ) : 0.864373 P ( k > 55 ) : 0.135627 P ( k >= 55 ) : 0.184101
    
    These answers agree with the results returned from https://www.statology.org/binomial-distribution-calculator/
  3. confidence (q, m, s)
    What is the q confidence interval about the mean m when we can approximate the distribution with a (continuous) Normal (m, s) distribution?
    
    Suppose q = 0.95 (in the case we are looking for the 95% confidence interval about the mean). The normal distribution is symmetrical, so we can say that there is 95% confidence that the resulting number of successes in 100 trials lies between x1 and x2, where x1 = m - dx,
    x2 = m + dx (for some dx); that is the area (probability) under the normal curve in the interval [m - dx, m + dx] is equal to 0.95. That means that the area left for the tails (left tail and right tail combined) is 0.05 (5%) and the area of the left tail (x < x1) is 0.025 (2.5%) and the area under the right tail (x > x2) is 0.025 (2.5%) as well.
    
    We then generalize to an arbitrary choice for q. We make use of the Maxima function quantile_normal (q, m, s) which we will cover in the worksheet Stat04-Normal.wxmx. Because the Normal (m,s) distribution is an example of a continous probability distribution function such as pdf_normal (x, m, s), where x can assume any real valued number, we can use quantile_normal (q, m, s) with any real number q.
    
    With q = 0.95, we find x2 using quantile_normal (0.975, m, s) since the combined left tail and the interval [x1,x2] area is 0.95 + 0.025 = 0.975. We find x1 using quantile_normal (0.025, m, s).
    We then find dx = (x2 - x1)/2
    
    (%i72)
    confidence (qq, mm, ss) := block ([ddx, xx1, xx2],
    float ( quantile_normal (qq + (1 - qq)/2, mm, ss) - quantile_normal ( (1 - qq)/2 , mm, ss) ),
    ddx : %% / 2, xx1 : mm - ddx, xx2 : mm + ddx,
    print ( "delx = ", ddx ),
    print ( " x1 = ", xx1, ", x2 = ", xx2 ), [xx1, xx2])$
    
    For our 100 flips of a balanced coin example, the mean m = 50, the standard deviation s = 5.
    
    (%i73)
    confidence (0.95, 50, 5)$
    delx = 9.79982
    x1 = 40.2002 , x2 = 59.7998
    
    (%i74)
    confidence (0.95, 50, 5);
    delx = 9.79982
    x1 = 40.2002 , x2 = 59.7998
    (%o74) [ 40.2002 , 59.7998 ]
    We can thus be 95% confident that the number of successes in 100 flips of a balanced coin will lie in the range (40 - 60).
    
    We can alter confidence (q,m,s) to confidence_integral (q,m,s) which rounds x1 and x2 to the closest integer.
    
    (%i75)
    confidence_integral (qq, mm, ss) := block ([ddx, xx1, xx2],
    float ( quantile_normal (qq + (1 - qq)/2, mm, ss) - quantile_normal ( (1 - qq)/2 , mm, ss) ),
    ddx : %% / 2, xx1 : mm - ddx, xx2 : mm + ddx,
    print ( "delx = ", ddx ),
    print ( " x1 = ", round (xx1), ", x2 = ", round (xx2) ), round ( [xx1, xx2] ))$
    
    (%i76)
    confidence_integral (0.95, 50, 5)$
    delx = 9.79982
    x1 = 40 , x2 = 60
    
    (%i77)
    confidence_integral (0.95, 50, 5);
    delx = 9.79982
    x1 = 40 , x2 = 60
    (%o77) [ 40 , 60 ]
    
    The following examples are from:
    https://www.statology.org/binomial-distribution-real-life-examples/ which offers the web based calculator referred to just above.
5. Statology 1: Number of Side Effects from Medications
  This example is from the webpage https://www.statology.org/binomial-distribution-real-life-examples/
  
  "Medical professionals use the binomial distribution to model the probability that a certain number of patients will experience side effects as a result of taking new medications."
  
  "For example, suppose it is known that 5% of adults who take a certain medication experience negative side effects. We can use a Binomial Distribution Calculator to find the probability that more than a certain number of patients in a random sample of 100 will experience negative side effects."
  
  "P(X > 5 patients experience side effects) = 0.38400 P(X > 10 patients experience side effects) = 0.01147
  P(X > 15 patients experience side effects) = 0.0004 [we will show this is too large by a factor of 10]. And so on."
  
  "This gives medical professionals an idea of how likely it is that more than a certain number of patients will experience negative side effects."
  
  In this example, each patient taking the medication either experiences negative side effects ('success') or does not experience negative side effects. The probability of 'success'
  is p = 0.05 (5%), and n = 100 patients take the same medication (100 trials), each with the same probability p of negative side effects. Then the average number of patients (out of 100) who experience negative side effects is 5 (out of 100) and the standard deviation is about 2.2.
  
  (%i79)
  mean_binomial (100, 0.05);
  std_binomial (100, 0.05);
  (%o78) 5.0
  (%o79) 2.17945
  
  Using our homemade Maxima function binomialCalc (x, n, p),
  
  (%i80) binomialCalc (5, 100, 0.05)$
  For n = 100, p = 0.05, the probabilities are: P ( k = 5 ) : 0.180018
  P ( k < 5 ) : 0.435981 P ( k <= 5 ) : 0.615999 P ( k > 5 ) : 0.384001 P ( k >= 5 ) : 0.564019
  
  The above output shows that the example assertion P(k > 5) = 0.384 is correct.
  (%i81) binomialCalc (10, 100, 0.05)$
  For n = 100, p = 0.05, the probabilities are: P ( k = 10 ) : 0.0167159
  P ( k < 10 ) : 0.971812 P ( k <= 10 ) : 0.988528 P ( k > 10 ) : 0.0114724 P ( k >= 10 ) : 0.0281883
  
  The output above shows that the example assertion P(k > 10) = 0.1147 is correct.
  
  (%i82) binomialCalc (15, 100, 0.05)$
  For n = 100, p = 0.05, the probabilities are: P ( k = 15 ) : 9.88002e−5
  P ( k < 15 ) : 0.999864 P ( k <= 15 ) : 0.999963
  P ( k > 15 ) : 3.70541e−5 P ( k >= 15 ) : 1.35854e−4
  
  This last calculation shows P(x > 15) = 3.7*(10^(-5)) = 3.7e-5 = 0.000037 ~ 0.00004, which is a factor of 10 smaller than the webpage asserts.
  
  There is a 95% chance that the number of patients out of 100 who experience negative side effects will be in the range [0, x]. What is the approximate value of x?
  
  An approximate answer is provided by quantile_binomial (0.95, 100, 0.05)
  
  (%i83) quantile_binomial (0.95, 100, 0.05);
  (%o83) 9
  
  We can check this estimate as follows:
  
  (%i85)
  cdf_binomial (8, 100, 0.05);
  cdf_binomial (9, 100, 0.05);
  (%o84) 0.93691
  (%o85) 0.971812
  
  There is a 95% chance the number of patients who experience negative side effects (out of 100) will lie in the range (0, 8 or 9).
  Here we try using confidence_integral (q, m, s).
  
  We can have 95% confidence that the number of patients who experience negative side effects (out of 100) will lie in the interval (m - dx, m + dx). We know the mean number is m = 5, and one standard deviation s = 2.2 approximately.
  
  (%i86)
  confidence_integral (0.95, 5, 2.2)$
  delx = 4.31192
  x1 = 1 , x2 = 9
  
  With dx about 4 and the mean equal to 5 we can be 95% confident that the number of patients experiencing negative side effects (out of 100) will be in the approximate range (1, 9).
  
  So both methods give the same approximate range.
6. Statology 2: Number of Fraudulent Credit Card Transactions
  
  This example is from the webpage https://www.statology.org/binomial-distribution-real-life-examples/
  
  "Banks use the binomial distribution to model the probability that a certain number of credit card transactions are fraudulent."
  
  "For example, suppose it is known that 2% of all credit card transactions in a certain region are fraudulent. If there are 50 transactions per day in a certain region, we can use a Binomial Distribution Calculator to find the probability that more than a certain number of fraudulent transactions occur in a given day:" [assuming a random sample of 50 transactions per day]
  
  "P(X > 1 fraudulent transaction) = 0.26423 P(X > 2 fraudulent transactions) = 0.07843 P(X > 3 fraudulent transactions) = 0.01776"
  
  "And so on."
  
  "This gives banks an idea of how likely it is that more than a certain number of fraudulent transactions will occur in a given day."
  
  In this example, each credit card transaction either is fraudulent ('success') or is not fraudulent.
  The probability of 'success' is p = 0.02 (2%), and n = 50 credit card transactions occur (50 trials), each with the same probability p of being fraudulent.
  (%i88) mean_binomial (50, 0.02);
  std_binomial (50, 0.02);
  (%o87) 1.0
  (%o88) 0.989949
  
  Let's show a plot of the probability distribution for this case Binomial (50, 0.02).
  
  (%i89) mypoints : makelist ( [j, pdf_binomial (j, 50, 0.02) ], j, 0, 5 );
  (mypoints) [ [ 0 , 0.36417 ] , [ 1 , 0.371602 ] , [ 2 , 0.185801 ] , [ 3 , 0.0606697 ] , [ 4 ,
  0.0145483 ] , [ 5 , 0.00273152 ] ]
  
  The *continuous* normal distribution (see pdf_normal (x, m, s)) closely agrees with the discrete binomial distribution, provided m = mean_binomial(50, 0.2) = 1, and
  s = std_binomial (50, 0.2) = 0.99, as we just calculated. Recall that pdf_binomial (k, n, p) is only nonzero for k = 0, 1, 2, ..., so pdf_binomial (0.5, n, p ) = 0 (see just below).
  
  (%i90) pdf_binomial (0, 50, 0.02);
  (%o90) 0.36417
  
  (%i91) pdf_binomial (0.5, 50, 0.02);
  (%o91) 0
  
  (%i92) pdf_binomial (1, 50, 0.02);
  (%o92) 0.371602
  (%i93)
  
  (%t93)
  wxdraw2d ( xrange = [0, 5], yrange = [0, 0.4], points_joined = impulses, xlabel = "number successes in 50 trials", ylabel = "probability", grid = true, title = "Binomial(50, 0.02), (red), Normal (1, 0.99), (black) ", background_color = light_gray,
  line_width = 2, color = red, points (mypoints),
  color = black, explicit (pdf_normal (x, 1, 0.99), x, 0, 5) )$
  
  Assertion: P(X > 1 credit card transactions [out of 50] are fraudulent) = 0.26423. Maxima check:
  
  (%i94) binomialCalc (1, 50, 0.02)$
  For n = 50, p = 0.02, the probabilities are: P ( k = 1 ) : 0.371602
  P ( k < 1 ) : 0.36417 P ( k <= 1 ) : 0.735771 P ( k > 1 ) : 0.264229 P ( k >= 1 ) : 0.63583
  
  The chance that more than 1 credit card transaction (out of 50) will be fraudulent is about 26%.
  
  Assertion: P(X > 2 credit card transactions [out of 50] are fraudulent) = 0.07843 Maxima check:
  (%i95) binomialCalc (2, 50, 0.02)$
  For n = 50, p = 0.02, the probabilities are: P ( k = 2 ) : 0.185801
  P ( k < 2 ) : 0.735771 P ( k <= 2 ) : 0.921572 P ( k > 2 ) : 0.0784277 P ( k >= 2 ) : 0.264229
  
  The chance that more than 2 credit card transactions (out of 50) will be fraudulent is about 8%.
  
  Assertion: P(X > 3 credit card transactions [out of 50] are fraudulent) = 0.01776 Maxima check:
  
  (%i96) binomialCalc (3, 50, 0.02)$
  For n = 50, p = 0.02, the probabilities are: P ( k = 3 ) : 0.0606697
  P ( k < 3 ) : 0.921572 P ( k <= 3 ) : 0.982242 P ( k > 3 ) : 0.0177581 P ( k >= 3 ) : 0.0784277
  
  The chance that more than 3 credit card transactions (out of 50) will be fraudulent is about 2%.
  
  There is a 95% chance that the number of fraudulent transactions will be in the range [0, x]. What is the approximate value of x?
  
  An approximate answer is provided by quantile_binomial (0.95, 50, 0.02).
  
  (%i97) quantile_binomial (0.95, 50, 0.02);
  (%o97) 3
  
  We can check this as follows:
  
  (%i98)
  for j:2 thru 4 do print (j, cdf_binomial (j, 50, 0.02) )$
  2 0.921572
  3 0.982242
  4 0.99679
  There is a 95% chance the number of fraudulent credit card transactions per day will lie in the interval [0, 2 or 3].
7. Statology 3: Number of Spam Emails per Day
  
  This example is from
  https://www.statology.org/binomial-distribution-real-life-examples/
  
  "Email companies use the binomial distribution to model the probability that a certain number of spam emails land in an inbox per day."
  
  "For example, suppose it is known that 4% of all emails are spam. If an account receives 20 emails in a given day, we can use a Binomial Distribution Calculator to find the probability that a certain number of those emails are spam:"
  
  "P(X = 0 spam emails) = 0.44200 P(X = 1 spam email) = 0.36834 P(X = 2 spam emails) = 0.14580 And so on."
  
  (%i99) makelist ([ j, pdf_binomial (j, 20, 0.04) ], j, 0, 2);
  (%o99) [ [ 0 , 0.442002 ] , [ 1 , 0.368335 ] , [ 2 , 0.145799 ] ]
  
  Out of 20 emails per day, (assuming a 4% spam rate) the chance of getting exactly no spam is 44%, the chance of getting one spam email is 36%, and the chance of getting two spam emails is about 15%.
8. Statology 4: Number of River Overflows
  This example is from
  https://www.statology.org/binomial-distribution-real-life-examples/
  
  "Park systems use the binomial distribution to model the probability that rivers overflow a certain number of times each year due to excessive rain."
  
  "For example, suppose it is known that a given river overflows during 5% of all storms. If there are 20 storms in a given year, we can use a Binomial Distribution Calculator to find the probability that the river overflows a certain number of times:"
  
  "P(X = 0 overflows) = 0.35849
  P(X = 1 overflow) = 0.37735
  P(X = 2 overflows) = 0.18868" "And so on."
  "This gives the parks departments an idea of how many times they may need to prepare for overflows throughout the year."
  
  (%i100) makelist ([j, pdf_binomial (j, 20, 0.05)], j, 0, 2);
  (%o100) [ [ 0 , 0.358486 ] , [ 1 , 0.377354 ] , [ 2 , 0.188677 ] ]
  
  Assuming a given river overflows during 5% of all storms, and assuming 20 storms/year, the chance there will be no overflow (X = 0) is about 36%, the chance there will be exactly one overflow during the 20 storms is about 38%, and the chance there will be exactly two river overflows during the 20 storms is about 19%.
  
  (%i101) binomialCalc (2, 20, 0.05)$
  For n = 20, p = 0.05, the probabilities are: P ( k = 2 ) : 0.188677
  P ( k < 2 ) : 0.73584 P ( k <= 2 ) : 0.924516
  P ( k > 2 ) : 0.0754837 P ( k >= 2 ) : 0.26416
  
  (%i102) mean_binomial (20, 0.05);
  (%o102) 1.0
9. Barlow Prob. 3.1
  A defence system is 99.5% efficient in intercepting ballistic missiles.
  1. What is the probability that it will intercept all of 100 missiles launched against it?
    
    The probability P(k,n) that the system will intercept exactly k missles out of n incoming if the defence system intercepts each missle with a probability p = 0.995, and we treat each missle as an (indepedent Bernoulli trial) is given by the binomial distribution.
    
    P_intercept (k,n) = binomial (n, k) * p^k * (1-p)^(n-k) = pdf_binomial (k,n,p) For n = 100 missiles and k = 100 intercepts,
    (%i103) pdf_binomial (100, 100, 0.995);
    (%o103) 0.60577
    
    So about 61% chance of intercepting all 100 out of 100 incoming.
    
    Let's make a plot of the values of the Binomial (n = 100, p = 0.995) for k <= 100.
    
    (%i104) mypoints : makelist ( [k, pdf_binomial (k, 100, 0.995) ], k, 97, 100 );
    (mypoints) [ [ 97 , 0.0124296 ] , [ 98 , 0.0757194 ] , [ 99 , 0.304407 ] , [ 100 , 0.60577 ] ]
    
    (%i105) wxdraw2d ( xrange = [97, 100], yrange = [0, 0.7], points_joined = impulses,
    xlabel = "number of interceptions in 100 trials", ylabel = "probability", grid = true, title = "Binomial(100, 0.995) (red), pdf binomial (x,100,0.995) (black) ", background_color = light_gray,
    line_width = 2, color = red, points (mypoints),
    color = black, explicit (pdf_binomial (x, 100, 0.995), x, 97, 100) )$
    
    (%t105)
  2. How many missiles must an aggressor launch to have a better than even chance of one or more penetrating the defences?
  'one or more' means at least one. Let x = number of missiles which evade the defense system. The probability of any given missile evading the defenses is pe = 1 - 0.995 = 0.005.
  
  This question is asking: for what number n of missiles launched do we have P_evade (x >= 1) = 0.5 ?
  
  Since P_evade (x < 1) + P_evade (x >= 1) = 1, we have the identity: P_evade (x >= 1) = 1 - P_evade (x < 1) = 1 - cdf_binomial (0, n, 0.005), or
  P_evade (x >= 1) = 1 - pdf_binomial (0, n, 0.005).
  1. Numerical Solution
    
    (%i106) eqn : 1 - pdf_binomial (0, n, 0.005) = 0.5;
    (eqn) 1 − 1.0 0.995n = 0.5
    
    (%i107) expr : lhs (eqn) - rhs(eqn); (expr) 0.5 − 1.0 0.995n
    (%i108) find_root (expr, n, 100, 200);
    (%o108) 138.283
  2. Graphical Solution
    
    We first survey the feasible solutions graphically. Define look (n1, n2) which plots the curve of 1 - pdf_binomial (0, n, 0.005) for n1 <= n <= n2.
    
    (%i109) look (n1, n2) := wxdraw2d (xlabel = "n", ylabel = "Pevade (x >= 1)", xrange = [n1, n2], title = " 1 - pdf binomial (0, n, 0.005)", grid = true, explicit (1 - pdf_binomial (0, n, 0.005), n, n1, n2) )$
    (%i110) look(100, 200)$
    
    (%t110)
    
    (%i111) look (130, 140)$
    
    (%t111)
    
    For 138 missiles launched we appear to reach P_evade = ~0.5. Print out the values near n = 138.
    
    (%i112) for n : 135 thru 140 do print (n, 1 - pdf_binomial (0, n, 0.005) )$
    135 0.491705
    136 0.494246
    137 0.496775
    138 0.499291
    139 0.501795
    140 0.504286
    Thus with 139 missiles launched, the agressor has an even chance of one or more (at least one) missiles evading the defence system.
  3. Analytic Solution
    
    For an analytic approach to this problem:
    
    (%i113) eqn : 1 - pdf_binomial (0, n, 0.005) = 0.5;
    (eqn) 1 − 1.0 0.995n = 0.5
    
    (%i114) eqn - 1;
    (%o114) − 1.0 0.995n =− 0.5
    
    (%i115) -1*%;
    (%o115) 1.0 0.995n = 0.5
    
    (%i116) log (%);
    (%o116) − 0.00501254 n=− 0.693147
    
    (%i117) solve (%, n), numer;
    (%o117) [ n= 138.283 ]
    
    The analytic solution thus says the agressor should launch 138 missiles to have an even chance of one or more evading the defense system.
10. Barlow Prob. 3.2
  
  In the previous question, how many missiles would be needed to ensure a better than even chance of more than two missiles evading the defences?
  
  P(x <= 2) + P (x > 2) = 1 implies: P(x > 2) = 1 - P ( x <= 2 ), or
  P(x > 2) = 1 - cdf_binomial (2, n, 0.005)
  1. numerical solution
    
    (%i118) eqn : expand (1 - cdf_binomial (2, n, 0.005) ) = 0.5;
    (eqn) − 1.25 10−5 n2 0.995n − 2 − 0.0049625 n 0.995n − 2 − 0.990025 0.995n − 2 +
    1 = 0.5
    (%i119) expr : lhs (eqn) - rhs (eqn);
    (expr) − 1.25 10−5 n2 0.995n − 2 − 0.0049625 n 0.995n − 2 − 0.990025 0.995n − 2 +
    0.5
    
    find_root (expr, variable, start, stop) looks for the value of the variable for which the expression expr passes through zero in the range [start, stop].
    
    (%i120) find_root (expr, n, 100, 800);
    (%o120) 534.475
  2. graphical solution
    
    Graphical exploration for various values of n.
    
    (%i121) look (n1, n2) := wxdraw2d (xlabel = "n", ylabel = "Pevade (x > 2)", xrange = [n1, n2], grid = true,
    title = " 1 - cdf binomial (2, n, 0.005) ",
    explicit (1 - cdf_binomial (2, n, 0.005), n, n1, n2) )$
    
    (%i122) look (100, 800)$
    
    (%t122)
    (%i123) look (520, 540)$
    
    (%t123)
    
    Print out values for n in the range [530, 540]:
    
    (%i124) for n : 530 thru 540 do print (n, 1 - cdf_binomial (2, n, 0.005) )$
    530 0.494453
    531 0.495695
    532 0.496936
    533 0.498175
    534 0.499413
    535 0.500649
    536 0.501883
    537 0.503116
    538 0.504347
    539 0.505577
    540 0.506805
    
    The agressor needs 535 missiles launched to have an even chance of more than two missiles (three or more) penetrating the defense system.
11. random_binomial (n, p), random_binomial (n, p, m)
  
  The Maxima function random_binomial (n, p) returns a Binomial (n, p) random variate, with 0 < p < 1 and n a positive integer. Each time you run the following command, you
  will get a different set of five numbers of successes (in general).
  Choosing n = 100, p = 0.5, reminds us of 100 flips of a balanced coin. With p = 0.5, the mean number of heads ("success") out of 100 flips is 50, so we expect random_binomial (100, 0.5) to return a number fairly close to 50 more often than not.
  
  (%i125) mean_binomial (100, 0.5);
  (%o125) 50.0
  
  (%i126) for j thru 5 do print (j, random_binomial (100, 0.5) )$
  1 52
  2 51
  3 46
  4 59
  5 46
  1. Random Sample of size m = 10, n = 100, p = 0.5
    
    The Maxima function random_binomial (n, p, m) returns a "random sample of size m", with
    each of the m samples being the result of a simulation of n trials of an identical experiment, each trial having the same probability of success p. Each trial is independent of the other trials, and in each trial there are only two possible and mutually exclusive outcomes.
    
    Thus random_binomial (100, 0.5, 10) calls random_binomial (100, 0.5) ten times and returns a list of ten values, each value being the number of successes found in 100 identical
    trials. For this example we know that the average (mean) number of successes in 100 identical trials is 50 (which is n*p).
    
    (%i130) rsample : random_binomial (100, 0.5, 10)$ fll (rsample);
    head (rsample); tail (rsample);
    (%o128) [ 45 , 49 , 10 ]
    (%o129) [ 45 , 49 , 54 ]
    (%o130) [ 45 , 57 , 49 ]
    
    Here is the whole list of 10 values.
    
    (%i131) rsample;
    (%o131) [ 45 , 49 , 54 , 55 , 48 , 49 , 49 , 45 , 57 , 49 ]
    
    With such a small sample there are no repeated integral values in this list.
  2. Random sample size m = 100 simulations, n = 100, p = 0.5
    We expect to get some integer repetitions in the returned list of values if we use random_binomial (100, 0.5, 100), which returns a list of 100 integers.
    
    (%i135) rsample : random_binomial (100, 0.5, 100)$ fll (rsample);
    head (rsample); tail (rsample);
    (%o133) [ 48 , 52 , 100 ]
    (%o134) [ 48 , 45 , 49 ]
    (%o135) [ 46 , 52 , 52 ]
  3. discrete_freq (data)
    
    The Maxima function discrete_freq( aList) counts the number of unique discrete "readings" of some instrument recorded in the list aList and returns a new list:
    [ list-of-unique-readings, list-of-frequency-of-each-unique-reading].
    
    The elements of the list 'frequencies' corresponds to the elements of the list 'uniqueData', in the following, element by element.
    
    (%i138) [uniqueData, frequencies] : discrete_freq (rsample)$ uniqueData;
    frequencies;
    (%o137) [ 39 , 40 , 41 , 42 , 43 , 44 , 45 , 46 , 47 , 48 , 49 , 50 , 51 , 52 , 53 , 54 , 55 , 56 , 57 ,
    58 , 59 , 60 ]
    (%o138) [ 1 , 1 , 1 , 2 , 7 , 3 , 2 , 11 , 5 , 9 , 6 , 2 , 4 , 8 , 10 , 6 , 7 , 5 , 5 , 2 , 2 , 1 ]
    
    If we add the integers in the list 'frequencies, we will get 100, since rsample has 100 elements, and each of the 100 elements (an integer) will contribute unity to one of the uniqueData elements.
    
    (%i139) Lsum (frequencies);
    (%o139) 100
    
    Let nfrequencies be a list of "normalized" frequencies (so the sum of the elements of nfrequencies equals 1 (we need the sum of the probabilities to equal 1).
    
    (%i140) nfrequencies : frequencies/100.0;
    (nfrequencies) [ 0.01 , 0.01 , 0.01 , 0.02 , 0.07 , 0.03 , 0.02 , 0.11 , 0.05 , 0.09 , 0.06 , 0.02 ,
    0.04 , 0.08 , 0.1 , 0.06 , 0.07 , 0.05 , 0.05 , 0.02 , 0.02 , 0.01 ]
    
    (%i141) Lsum (nfrequencies );
    (%o141) 1.0
    (%i142) lmax (nfrequencies);
    (%o142) 0.11
    
    To make a plot of probabilities (vertical axis) versus specific numbers of successes in
    100 trials (horizontal axis), we let mypoints be a list of 100 sublists, with each sublist containing [x, y] = [number successes in 100 trials, the corresponding probability]
    = [uniqueData[j], nfrequencies[j] ] based on the m = 100 simulations.
    
    (%i146) mypoints : makelist ([ uniqueData[j], nfrequencies [j] ], j, 1, length (uniqueData))$ fll (mypoints);
    head (mypoints); tail (mypoints);
    (%o144) [ [ 39 , 0.01 ] , [ 60 , 0.01 ] , 22 ]
    (%o145) [ [ 39 , 0.01 ] , [ 40 , 0.01 ] , [ 41 , 0.01 ] ]
    (%o146) [ [ 58 , 0.02 ] , [ 59 , 0.02 ] , [ 60 , 0.01 ] ]
    
    (%i147) wxdraw2d ( xrange = [35, 65], yrange = [0, 0.1], points_joined = impulses, xlabel = "number successes in 100 trials", ylabel = "probability", grid = true, title = "100 Random Binomial samples, each for n = 100, p = 0.5", background_color = light_gray,
    line_width = 2, color = red, points (mypoints) )$
    
    (%t147)
    
    When you run this script on your own, your probabilities (vertical axis) will be slightly different, and you can change the vertical range yrange in draw2d, if needed, to see all your points clearly.
  4. Random sample size m = 1000, n = 100, p = 0.5

(%i149) mean_binomial (100, 0.5);

std_binomial (100, 0.5);

(%o148) 50.0

(%o149) 5.0

(%i153) rsample : random_binomial (100, 0.5, 1000)$ fll (rsample);

head (rsample); tail (rsample);

(%o151) [ 53 , 48 , 1000 ]

(%o152) [ 53 , 52 , 39 ]

(%o153) [ 45 , 47 , 48 ]

(%i156) [uniqueData, frequencies] : discrete_freq (rsample)$ uniqueData;

frequencies;

(%o155) [ 34 , 35 , 36 , 37 , 38 , 39 , 40 , 41 , 42 , 43 , 44 , 45 , 46 , 47 , 48 , 49 , 50 , 51 , 52 ,

53 , 54 , 55 , 56 , 57 , 58 , 59 , 60 , 61 , 62 , 63 , 64 , 65 ]

(%o156) [ 2 , 2 , 1 , 2 , 8 , 11 , 8 , 9 , 23 , 29 , 35 , 52 , 54 , 65 , 94 , 77 , 64 , 81 , 66 , 74 , 57 ,

49 , 39 , 29 , 24 , 12 , 18 , 6 , 3 , 2 , 2 , 2 ]

Define a list of normalized frequencies, nfrequencies.

(%i157) nfrequencies : frequencies/Lsum (frequencies), numer;

(nfrequencies) [ 0.002 , 0.002 , 0.001 , 0.002 , 0.008 , 0.011 , 0.008 , 0.009 , 0.023 , 0.029 ,

0.035 , 0.052 , 0.054 , 0.065 , 0.094 , 0.077 , 0.064 , 0.081 , 0.066 , 0.074 , 0.057 ,

0.049 , 0.039 , 0.029 , 0.024 , 0.012 , 0.018 , 0.006 , 0.003 , 0.002 , 0.002 , 0.002 ]

(%i158) Lsum (frequencies);

(%o158) 1000

(%i159) Lsum (nfrequencies);

(%o159) 1.0

(%i160) lmax (nfrequencies);

(%o160) 0.094

(%i161) uniqueData;

(%o161) [ 34 , 35 , 36 , 37 , 38 , 39 , 40 , 41 , 42 , 43 , 44 , 45 , 46 , 47 , 48 , 49 , 50 , 51 , 52 ,

53 , 54 , 55 , 56 , 57 , 58 , 59 , 60 , 61 , 62 , 63 , 64 , 65 ]

(%i162) Lud : length (uniqueData); (Lud) 32

(%i166) mypoints : makelist ([ uniqueData[j], nfrequencies [j] ], j, 1, Lud)$ fll (mypoints);

head (mypoints); tail (mypoints);

(%o164) [ [ 34 , 0.002 ] , [ 65 , 0.002 ] , 32 ]

(%o165) [ [ 34 , 0.002 ] , [ 35 , 0.002 ] , [ 36 , 0.001 ] ]

(%o166) [ [ 63 , 0.002 ] , [ 64 , 0.002 ] , [ 65 , 0.002 ] ]

(%i171) print (" ")$

print (sconcat (Lud,

" normalized frequencies (red) from 1000 random simulations"))$ print (" pdf_binomial (k, 100, 0.5)", " (black) ")$

print (" ")$

wxdraw2d ( xrange = [35, 65], yrange = [0, 0.1], points_joined = impulses, xlabel = "number successes in 100 trials", ylabel = "probability", grid = true, title = sconcat (Lud, " normalized frequencies "),

background_color = light_gray,

line_width = 2, color = red, points (mypoints), color = black, line_width = 2, key_pos = top_left, key = "pdf binomial (k, 100, 0.5)",

explicit (pdf_binomial (kk, 100, 0.5), kk, 35, 65) )$

32 normalized frequencies (red) from 1000 random simulations pdf_binomial (k, 100, 0.5) (black)

(%t171)

Stat02-Binomial.wxmx

Preface

References

3

Discrete Distributions Defined

What Is a Discrete Random Variable?

Mean and Variance of Discrete Data Set

Fundamental Probability Laws

7

Permutations and Combinations

Permutations

Combinations

binomial (n, k)

The Binomial (n, p) Distribution

Pb (k; n, p)

Origin of Pb (k; n,p) Formula

pdf_binomial (k, n, p), n>0, 0<p<1, k = 0,1,2, ,n

( 1 −p ) n − k pk

binomial(n,k)*(1-p)^(n-k)*p^k$

0.5^n*binomial(n,k)$

[RS: Example 9]: Two Tosses of a Balanced Coin

(mypoints) [ [ 0 , 0.25 ] , [ 1 , 0.5 ] , [ 2 , 0.25 ] ]

mean_binomial (n, p)

var_binomial (n, p)

std_binomial (n, p)

cdf_binomial (k, n, p)

Example 1 of Use of cdf_binomial (k, n, p)

Example 2 of Use of cdf_binomial (k, n, p)

quantile_binomial (q, n, p)

Number of Successes in 100 Flips of a Balanced Coin

(%o53) [ [ 20 , 4.22816 10−10 ] , [ 80 , 4.22816 10−10 ] , 61 ]

(%o54) [ [ 20 , 4.22816 10−10 ] , [ 21 , 1.61073 10−9 ] , [ 22 , 5.78398 10−9 ] ]

(%o55) [ [ 78 , 5.78398 10−9 ] , [ 79 , 1.61073 10−9 ] , [ 80 , 4.22816 10−10 ] ]

Inverse Type of Problem

binomialCalc (k, n, p)

confidence (q, m, s)

(%o74) [ 40.2002 , 59.7998 ]

(%o77) [ 40 , 60 ]

Statology 1: Number of Side Effects from Medications

Statology 2: Number of Fraudulent Credit Card Transactions

(mypoints) [ [ 0 , 0.36417 ] , [ 1 , 0.371602 ] , [ 2 , 0.185801 ] , [ 3 , 0.0606697 ] , [ 4 ,

Statology 3: Number of Spam Emails per Day

(%o99) [ [ 0 , 0.442002 ] , [ 1 , 0.368335 ] , [ 2 , 0.145799 ] ]

Statology 4: Number of River Overflows

(%o100) [ [ 0 , 0.358486 ] , [ 1 , 0.377354 ] , [ 2 , 0.188677 ] ]

Barlow Prob. 3.1

(mypoints) [ [ 97 , 0.0124296 ] , [ 98 , 0.0757194 ] , [ 99 , 0.304407 ] , [ 100 , 0.60577 ] ]

Numerical Solution

Graphical Solution

Analytic Solution

(%o117) [ n= 138.283 ]

Barlow Prob. 3.2

numerical solution

graphical solution

random_binomial (n, p), random_binomial (n, p, m)

Random Sample of size m = 10, n = 100, p = 0.5

(%o128) [ 45 , 49 , 10 ]

(%o129) [ 45 , 49 , 54 ]

(%o130) [ 45 , 57 , 49 ]

(%o131) [ 45 , 49 , 54 , 55 , 48 , 49 , 49 , 45 , 57 , 49 ]

Random sample size m = 100 simulations, n = 100, p = 0.5

(%o133) [ 48 , 52 , 100 ]

(%o134) [ 48 , 45 , 49 ]

(%o135) [ 46 , 52 , 52 ]

discrete_freq (data)

(%o137) [ 39 , 40 , 41 , 42 , 43 , 44 , 45 , 46 , 47 , 48 , 49 , 50 , 51 , 52 , 53 , 54 , 55 , 56 , 57 ,

(%o138) [ 1 , 1 , 1 , 2 , 7 , 3 , 2 , 11 , 5 , 9 , 6 , 2 , 4 , 8 , 10 , 6 , 7 , 5 , 5 , 2 , 2 , 1 ]

(nfrequencies) [ 0.01 , 0.01 , 0.01 , 0.02 , 0.07 , 0.03 , 0.02 , 0.11 , 0.05 , 0.09 , 0.06 , 0.02 ,

(%o144) [ [ 39 , 0.01 ] , [ 60 , 0.01 ] , 22 ]

(%o145) [ [ 39 , 0.01 ] , [ 40 , 0.01 ] , [ 41 , 0.01 ] ]

(%o146) [ [ 58 , 0.02 ] , [ 59 , 0.02 ] , [ 60 , 0.01 ] ]

Random sample size m = 1000, n = 100, p = 0.5