Clear-Sighted Statistics
Chapter 8: Discrete Probability Distributions
I. Introduction
Building on the basic concepts of probability, we now turn our attention to probability distributions. We will focus on discrete probability distributions in this chapter. In the chapters that follow, we will turn our attention to continuous probability distributions.
After completing this chapter, you will be able to:
• Define basic probability distribution terms.
• Distinguish between discrete and continuous probability distributions.
• Describe the characteristics of three major types of discrete probability distributions: Binomial, Hypergeometric, and Poisson.
• Calculate the probabilities of the Binomial, Hypergeometric, and Poisson distributions using tables, the discrete probability formulas, and Microsoft Excel.
Here is the list of files accompanying this chapter that you should download:
• Chapter08_Examples.xlsx, which shows the calculations for the examples shown in this chapter.
• Chapter08_Exercises.xlsx, which contains the data for the end-of-chapter exercises.
• Chapter08_BinomialTable.xlsx, which includes binomial probability tables for 1 trial up to 20 trials.
• Chapter08_Hypergeometric_Calculator.xlsx.
• Chapter08_PoissonTable.xlxs, which include Poisson Probability Tables for the mean, μ, from 0.01 through 10.0.
The Binomial, Hypergeometric, and Poisson tables are also included in Appendix 2: Statistical Tables.
II. Basic Probability Distribution Terms
Here are a few basic probability distribution terms with which you should become familiar. Some of these we have already covered in previous chapters.
1. Random Variables: Random variables are measures of objects of interest generated by experiments or observations. The values of these variables vary by chance. Whenever we mention variables from a sample, we are referring to random variables even when we do not include the word random.
2. Qualitative or Categorical Variables: These variables are not numerical: Colors, names, gender, and animal species. Yes or no survey questions like “do you have a pet?” yield qualitative variables. Counting the frequencies of the yes or no answers yields quantitative data.
3. Quantitative Variables: These variables are numerical: Height, weight, commute time, grade point average, bank balance, and hours in a day. There are two types of quantitative variables: Discrete variables and continuous variables.
4. Discrete Variables: These variables are often considered countable numbers or integers: 0, 1, 2, 3, 4…n. Discrete variables jump from one integer to the next leaving gaps in the measurement scale.
Examples of discrete variables include:
1. The number of students in a classroom.
2. The number of times you have been on a bus this month.
3. The number of times a day you brush your teeth.
4. The number of teeth in your mouth.
5. The number of pennies in your pocket or purse.
6. The number of spam phone calls you receive in a week.
5. Continuous Variables: These variables can take on an infinite number of values between integers. As a consequence, there are no gaps in a range of continuous variables. It is easy to count discrete variables like the number of fingers on your right hand: One, two, three, four, and five. But, it is impossible to count all possible random variables between one and two because there are an infinite number of fractions between integers.
Examples of continuous variables include:
1. The weight of a sumo wrestler.
2. The height of a wave in a tsunami.
3. The time to run a marathon.
4. The value of pi (π) in geometry (mathematicians have calculated its value to 200 million digits!).
5. The amount of sleep you got last night.
6. The distance between New York City and Paris.
6. Probability Distribution: A probability distribution provides the mathematical relationship of the probabilities of the different possible outcomes of an observation, experiment, or trial. Probability distributions can be expressed as a mathematical formula, table, or chart. As shown in Figure 1, probability distributions can be discrete or continuous.
Figure 1: Graphic Representations of Discrete and Continuous Probability Distributions
When the random variables are discrete, we have a discrete probability distribution. When the variables are continuous, we have a continuous probability distribution.
There are a wide variety of discrete and continuous probability distributions. Figure 2 shows the three discrete probability distributions we will cover in this chapter and the four continuous probability distributions we will cover in subsequent chapters:
Figure 2: Discrete and Continuous Probability distributions covered in Clear-Sighted Statistics
7. Key Probability and Probability Distribution Concepts: The probability for any outcome of an event is between zero and 100 percent (0.00 to 1.00). Outcomes are mutually exclusive and collectively exhaustive. To illustrate these points, let’s examine a discrete probability distribution of flipping a fair coin three times. The probability of getting a head on any one flip is 50 percent. Figure 3 shows the eight possible outcomes of flipping a coin three times:
Figure 3: Eight possible outcomes of flipping a coin three times
8. Probability Mass Functions (PMF) and Cumulative Probability Distributions (CPD): Each discrete probability function can be reported as a Probability Mass Function or a Cumulative Probability Distribution.
Probability Mass Functions provide the probability of the distribution of discrete random variables. The PMF function is also called a discrete density function. Figure 4 shows a chart that represents the PMF function for the outcomes of flipping a fair coin three times:
Figure 4: PMF for Three Flips of a Coin
The PMF shows that the experiment of flipping a coin three times will yield no heads 12.5 percent of the time, one head 37.5 percent of the time, two heads 37.5 percent of the time, and three heads 12.5 percent of the time.
A Cumulative Probability Distribution (CPD), which is also known as a Cumulative Distribution Function, provides the probability that a random variable, X, is less than or equal to the random variable. Figure 5 shows the CPD for flipping a coin three times:
Figure 5: CPD for Three Flips of a Coin
9. Expected Value, E(X) or Mean, μ: The expected value is the most important measure of location for a discrete probability distribution. The expected value is actually the mean, or more specifically, the weighted mean of the possible values of all random values. Here is the formula for the mean (expected value):
Equation 1: Mean or E(X) of a Discrete Probability Distribution
μ = E(X) = Σ[XP(X)]
Where: E(X) or μ: expected value
X: Random Variable
P(X): Probability of the Random Variable
10. The mean, μ, variance, σ2, and standard deviation, σ, of a discrete probability distribution: Figure 6 shows that the expected value or μ for the number of heads resulting from flipping a coin three times is 1.5.
Figure 6: Expected Value or Mean for Flipping a Coin Three Times = 1.5
Equations 2 and 3 show the formulas for calculating the variance and standard deviation of a discrete probability distribution:
Equation 2: Formula for Variance of a Discrete Probability Distribution
Equation 3: Formula for Standard Deviation of a Discrete Probability Distribution
Figure 7 shows the calculations for variance and standard deviation for the probability distribution of three flips of a coin.
Figure 7: Variance and Standard Deviation for Flipping a Coin 3 Times
11. Successes and Failures: Results of an experiment or trial are often dichotomous, which means that the outcomes fall into only one of two groups: Successes or failures. These labels do not carry the everyday meanings of the words success and failure. A success means that the outcome matches what we are looking for and a failure means that that outcome does not. Suppose, for example, we are interested in determining the proportion of start-up ventures that go bankrupt, which is to say, fail. But, given that we are looking for bankrupt start-ups, these firms would be considered successes because they meet the standard we seek to measure. Other dichotomous terms we might use are: Heads/tails as in the examples shown in Figure 7, voted/not voted, on/off, yes/no, open/closed, defective/not defective, or gender if we use a binary gender definition: female/male.
12. Sampling Techniques (With Replacement and Without Replacement): With discrete probability distributions there are two basic sampling techniques that we must consider. When the sampling is performed with replacement, the selected item is returned to the population after its selection. Doing so leaves the probability of selection unchanged from one event to another. Sampling without replacement means that a selected item is not returned to the population after it is selected. A consequence of this sampling technique is that the probability of a “success” changes after each trial or experiment. Most sampling is done without replacement.
III. A Quick Overview of Binomial, Hypergeometric, and Poisson Distributions
As shown in Figure 8, the three major discrete probability distributions have distinct uses. Binomial distributions are used for repeated trials with a constant probability of occurrence. Sampling for binomial distributions are performed with replacement. Hypergeometric distributions are used for repeated trials when the sampling is performed without replacement. Poisson distributions measures the probability of a given number of independent events occurring in a fixed interval of time or space at a constant rate.
Figure 8: Overview of Binomial, Hypergeometric, and Poisson Distributions
IV. Binomial Probability Distributions
A binomial probability distribution shows all the probabilities associated with the discrete random variables generated by Bernoulli trials. Bernoulli trials are named after the seventeenth century Swiss mathematician Jakob Bernoulli, who is sometimes called James or Jacques. The Bernoulli family produced eight famous mathematicians in the seventeenth and eighteenth centuries. Here are the key features of a Bernoulli trial:
1. The trial or experiment has only two possible outcomes: Success or failure.
2. The random variable is the number of successes in a fixed number of trials.
3. Sampling is done with replacement.
4. The probability of a success—P(Success)—does not change from the first trial through the last trial.
5. The trials are independent, which means that trials are unaffected by the outcome of previous trials. For example, the first flip of a coin does not affect the outcome of subsequent coin flips.
As with any discrete or continuous probability distribution, there is a family of binomial distributions. The family of binomial distributions is defined by:
1. The number of trials, symbolized by n,
2. The probability of a success, symbolized by π.
There are four ways to calculate the probabilities of a binomial distribution.
1. The binomial distribution probability table.
2. Pascal’s Triangle, named after the Blaise Pascal whose contributions to probability theory were examined in Chapter 7.
3. The binomial probability formula.
4. Microsoft Excel’s BINOM.DIST function.
We will review each method, but first we need a problem that can be solved using binomial probability. What are the probabilities associated with flipping a fair coin five times? There are six possible outcomes to this experiment: 0 heads, 1 head, 2 heads, 3 heads, 4 heads, and 5 heads. This is a perfect exercise for demonstrating the four methods for calculating binomial probabilities. Here is why: First, each trial has only two outcomes: Heads or tails. We will consider getting a head a success. Second, the trials are independent, which is another way of saying that the samples are performed with replacement and the probability of a success for each trial is 50 percent.
A. Finding the Binomial Probability Using a Binomial Distribution Probability Table
The Excel file Chapter08_BinomialTable.xlsx contains a collection of binomial probability tables. There are tables for sample sizes, or n, of 1 up to 20. The first column shows the number of successes from 0 up to the n. Then there are fourteen columns for the following probabilities of a success: 0.01, 0.05, 0.10, 0.15, 0.20, 0.25, 0.30, 0.35, 0.40, 0.45, 0.50, 0.55, 0.60, 0.65, 0.70, 0.75, 0.80, 0.85, 0.90, and 0.95. Of course, if the probability of success for your experiment is not one of these probabilities—48 percent, for example—you cannot use these tables.
Because our experiment has a sample size, n, of 5 based on five flips of a coin, we will focus on the table where n equals 5. Because the probability of a success is 50 percent, we will focus on the column labelled 0.50. Figure 9 shows the binomial probability table with the probability distribution for the possible outcomes of flipping a coin five times:
Figure 9: Binomial Probability Table for n = 5
Figure 10 shows a chart of this probability distribution. Please note: The probabilities are symmetrical because the probability of success, π, equals 0.50 or 50 percent.
Figure 10: Binomial Probability Distribution, n = 5, π = 50%
B. Finding the Binomial Probability Using Pascal’s Triangle
Figure 11 shows an abbreviated Pascal’s Triangle set up for a probability of 50 percent. We will focus on Row 5 because that row shows the number of successes for each of the six outcomes for flipping a coin five times:
Figure 11: Abbreviated Pascal’s Triangle
Table 1 shows the number of heads for all 32 unique outcomes. We can convert the number of successes for each of the six outcomes into probabilities by finding the relative frequencies of the number of successes.
Table 1: Converting Number of Successes into Probabilities
Constructing a Pascal’s Triangle can be laborious as the number of trials increase. There are better ways to calculate binomial probabilities. We will turn to these methods now.
C. Finding the Binomial Probability Using the Binomial Probability Formula
In this section, we will use the binomial probability formula along with formulas for calculating the Expected Frequency, E(X), or mean (μ), variance (σ2), and standard deviation (σ).
Figure 12 shows the binomial probability formula. It requires using combinations, which we covered in Chapter 7. This formula is not as scary as it looks.
Figure 12: Binomial Probability Formula
Where: P(X) = Probability of X successes in n trials, X/n
X = Number of successes
n = Number of trials
π = Probability of a success for each trial, (0.5 for this example)
nCX = The Combination of X things taken from n
Equation 4 shows the calculation of the binomial probability formula for zero successes:
Equation 4: Binomial Probability Formula
Now that we have calculated the probability for no success, we must calculate the probabilities for one, two, three, four, and five successes. This can be very time-consuming. In fact, I would argue it is a waste of time because we can use Excel, which will save us from this drudgery.
Before we turn to calculating binomial probabilities with Excel, let’s calculate three parameters of this distribution: μ, σ2, and σ.
Equation 5 shows the formula for finding the Expected Value, or mean, and the calculation for this problem:
Equation 5: Formula for E(X) or μ
Where: n = Sample size
π = Probability of a “success” for each trial
Equation 6 shows the formula for finding the variance and the calculation for this problem:
Equation 6: Formula for Variance
Where: n = Sample size
π = Probability of a “success” for each trial
1 - π = Probability of a “failure” for each trial
Equation 7 shows the formula for finding the standard deviation and the calculation for this problem:
Equation 7: Formula for Standard Deviation
D. Finding the Binomial Probability Using Microsoft Excel’s BINOM.DIST Function
Using Excel’s BINOM.DIST function will save time so you can focus on more important tasks for your analysis. Equation 8 shows the syntax of this function:
Equation 8: Syntax for BINOM.DIST
=BINON.DIST(#successes,#trials,Probability,Cumulative)
Where: #successes = the number of successes
#trials = the number of trials
Probability = The Probability of a success
Cumulative = True or False
Cumulative takes one of two values: True or False. If it is false, Excel returns the Probability Mass Function. If it is true, Excel returns the Cumulative Probability Distribution function.
Figure 13 shows the complete binomial probability distribution for an n of 5 and a π of 0.50.
Figure 13: Binomial Probability Distribution Prepared Using Microsoft Excel
You will notice that this worksheet also reports the three parameters: Mean or E(X), variance, and standard deviation. It took only a few minutes to construct this worksheet.
Another nice thing about using Excel is that it can quickly draw charts of the probability mass function and the cumulative probability distribution. Table 2 shows the charts of the Probability Mass Function and Cumulative Probability Distribution for this example:
Table 2: Charts of the Binomial Probability Distribution When n = 5 and π = 0.50
Please Note: When the probability of a success is 50 percent, the binomial probability distribution is symmetrical, as shown in the PMF chart in Table 2.
Binomial Probability Distribution Example When π is not 50 Percent
Let’s construct a binomial probability distributions for a problem that do not have a 50 percent probability of success. Our example deals with a deck of playing cards. As shown in Figure 14, the deck has 52 cards in four suits: Hearts, clubs, diamonds and spades. There are thirteen cards in each suit from two to ten, Jack, Queen, King, and Ace.
Figure 14: Deck of 52 Playing Cards
After we shuffle the cards, what is the probability that we will get a spade when there are 13 spades in a deck of 52 cards? The answer is π equals 13/52, which equals 0.25 or 25 percent. Here is the game we are going to play. We are going to pick a card and record whether the selected card is a spade or not a spade. We then return the selected card to the deck, shuffle, and select another card. We will do this a total of five times, n = 5. In our game, a success is selecting a spade. What are the probabilities that we will pick no spades, one spade, two spades, three spades, four spades, or five spades? Remember: There are four ways to do these calculations:
1. A binomial probability table,
2. Pascal’s triangle,
3. Binomial probability formula., and
4. Excel’s BINOM.DIST function.
We will go through all four methods. But, you should know that the binomial probability table and Pascal’s triangle were constructed using Excel’s BINOM.DIST function.
# 1: The Binomial Probability Table
We will use the table for an n of 5 because we will draw a card five times. Please note: After we select a card, we return it to the deck and shuffle the cards. See Figure 15. You can find this table in the workbook titled Chapter08_BinomialTables.xlsx on the Binomial Table worksheet. Notice that this table uses the BINOM.DIST function to calculate the probabilities.
Given that the probability of a success is 0.25, we will find our probabilities for the six possible outcomes in the column labelled 0.25.
Figure 15: Binomial Probability Table
# 2: Pascal’s Triangle
Figure 16 shows a Pascal’s triangle constructed in Excel. All you need to do is enter the probability for success in cell D2. This file is in the workbook titled Chapter08_DiscreteProbability.xlsx on the Pascal’s Triangle worksheet. This worksheet has advantages over using a paper table. Paper tables are inflexible. You are stuck with whatever probabilities for success are on the table. The Pascal’s Triangle worksheet allows you to enter the exact probability you need.
Figure 16: Pascal’s Triangle
When we compare the probabilities for the row with 5 trials in Pascal’s triangle to the table in Figure 16, we see that both report the same probabilities even though Pascal’s triangle rounds the probabilities off to four digits past the decimal point instead of three.
# 3: The Binomial Probability Formula
This is the most time-consuming method of calculating the probabilities of success. Here is the formula along with the calculation for no successes.
Equation 9: Binomial Probability Formula
P(X)=nCxπX(1-π)n-X=5C00.250(1-0.25)5-0=(1)(1)(0.2373)=0.2373
Where: P(X) = Probability of X successes in n trials, X/n
X = Number of successes
n = Number of trials
π = Probability of a success for each trial, (0.5 for this example)
nCX = The Combination of X things taken from n
Our answer of 0.2373 matches the probabilities of no spades in five trials, found using the binomial probability table and Pascal’s triangle. But, using the binomial probability formula requires more work because we have to calculate the probabilities for the other five outcomes.
# 4: Excel’s BINOM.DIST Function
Figure 17 shows the probability calculations performed using Excel. This file is in the workbook titled Chapter08_DiscreteProbability.xlsx on the Binomial_Spades worksheet. This worksheet calculates the mean, variance, and standard deviation as well as the probability mass function and cumulative probability distribution. The probabilities shown in the PMF match those found on the probability table and Pascal’s triangle.
Figure 17: Binomial Probabilities Found Using Excel for the P(Spade)
When the probability of success, π, is less than or more than fifty percent, the probability mass function is not symmetrical. This distribution is right-skewed as shown in Figure 18.
Figure 18: PMF for the P(Spade)
If we changed the definition of a success to the probability of not selecting a spade, π would be 0.75. You can calculate this probability two ways: 1) 39 non-spades/52 cards = 0.75, or 2) by using the complement rule, 1 – P(Spade) = 0.75. Figure 19 shows the binomial probabilities for no spades. These calculations can be found in Chapter08_DiscreteProbability.xlsx on the Binomial_No Spades worksheet.
Figure 19: Binomial Probabilities Found Using Excel for the P(No Spade)
When the probability of success, π, is greater than fifty percent, the probability mass function is skewed to the left. See Figure 20:
Figure 20: PMF for the P(No Spade)
V. Hypergeometric Probability Distributions
With binomial distributions the trials are independent. With hypergeometric distributions the trials are dependent because the sampling is done without replacement. We could, therefore, say that hypergeometric probability distributions are an extension of binomial distributions. In fact, hypergeometric distributions are also known as binomial distributions without replacement.
We use hypergeometrical distributions under the following circumstances:
1. The experiment has binomial outcomes, which means the outcomes are classified as either a success or failure.
2. The random samples are done without replacement.
3. The population is finite and the size of the population, N, is known.
4. The sample size, n, is more than 5 percent of the population, n/N ≥ 0.05.
Here is a problem that requires us to use the hypergeometric probability distribution to calculate the probabilities of success. The law firm of Dewey, Cheatem, & Howe is forming a committee to oversee the firm’s work for charities. Twenty senior partners are eligible to serve on this committee; five are women, fifteen are men. The committee will have five partners. What are the probabilities that this committee will have no women, one woman, two women, three women, four women, and five women?
This is a hypergeometric probability problem for the following reasons:
1. The outcomes are binomial, where success for a trial is the selection of a woman and failure is the selection of a man.
2. The sampling is done without replacement because a committee member cannot be selected twice.
3. The sample is selected from a finite population, N = 20 and the sample size, n = 5,
so n/N > 0.05, (5/20 = 0.25).
Equation 10 shows the hypergeometric probability formula and the calculation of the probability that no women will be on this committee:
Equation 10: Hypergeometric Probability Formula and Calculation
Where: P(X) = Random Variable
C = Combination
N = Population Size
n = Sample Size/Number of Trials
X = Number of successes in the sample
S = Number of success in the population
There is a 19.37 percent chance that no women will be serving on this committee. Calculating the remaining probabilities with all the combination calculations can be time-consuming. We can save time using Excel’s HYPGEOM.DIST function.
Here is the syntax for the HYPGEOM.DIST function:
Equation 11: HYPGEOM.DIST Function
=HYPGEOM(X,n,S,N,Cumulative)
Where: X = Number of successes in the sample
n = Sample Size/Number of Trials
S = Number of success in the population
N = Population Size
Cumulative = True or False
If Cumulate is False, Excel returns the PMF. If Cumulative is True, Excel returns the CPD.
Figure 21 shows the PMF and CPD hypergeometric calculations performed in Excel.
Figure 21: Hypergeometric Distribution
While the probability of no women on this committee is 19.37 percent, the probability of five women on this committee is extremely remote, 0.01 percent or one in a 1,000. Figure 22 shows a chart of the PMF distribution. Because the probability of success, π, is less than 50 percent, the distribution is skewed to the right.
Figure 22: Chart of the PMF Distribution
Based on the CPD, there is a 92.74 percent probability that the committee will have fewer than three women (0, 1, or 2 women). You can also calculate this figure by adding the probabilities for zero, one, and two successes.
Equation 12 shows the formulas and calculation for μ, σ2, σ:
Equation 12: μ, σ2, σ, for a Hypergeometric Distribution
VI. Poisson Probability Distributions
The third discrete probability distribution is the Poisson distribution, which is named after the French mathematician, Siméon Denis Poisson, who wrote a paper on this distribution in 1837. This distribution could have easily been named after another French mathematician, Abraham de Moivre, who developed a similar distribution in the eighteenth century. The Poisson distribution was popularized at the end of the nineteenth century by Ladislaus von Bortkiewicz, who used it to predict the number of Prussian soldiers who die annually from getting kicked by a horse.
The Poisson distribution is used to calculate the probability of a given number of independent outcomes when events occur at fixed intervals at a constant rate. Intervals may be time, area, distance, or volume. In addition, the intervals are independent and mutually exclusive. The random variable is the number of successes and the probability of a success, P(X), is proportional to the length of the interval. The longer the interval, the higher the probability of a success, π.
The mean of a Poisson distribution is calculated using the following formula: μ = nπ, where n is the number of trials. The variance, σ2, of a Poisson distribution is equal to the mean. Standard deviation is the square root of variance.
The Poisson probability distribution formula is shown in Equation 13:
Equation 13: Poisson Distribution Probability Formula
Where: P(X) = Probability of a specific number of successes
X = Number of successes
μ = Expected number of successes (Sometimes this is symbolized as λ)
e = Base of the logarithm system (2.71828)
Here is an example of the probability that a person’s credit card will be hacked is 0.0001 or 0.01 percent. Our sample size is 15,000.
First let’s calculate the mean, variance, and standard deviation:
Equation 14: Mean and Variance of a Poisson Distribution
μ and σ2 = nπ = ١٥,٠٠٠ * ٠.٠٠٠١ = ١.٥٠
Equation 15: Standard Deviation of a Poisson Distribution
We can use the Poisson distribution formula or Microsoft Excel to construct a probability distribution for the probability that a credit card will be hacked. A success is defined as a credit card being hacked. I know this sounds strange. After all, should not a hacked credit card be considered a failure? We consider it a success because it meets the condition we seek. Here is the calculation of the probability that none of the 15,000 credit cards will be hacked, P(0):
Equation 16: Probability of No Hacked Credit Cards
There is a 22.31 percent probability that none of the 15,000 credit cards will get hacked. To complete the probability distribution, we would have to calculate the probabilities of other outcomes: 1 card hacked, 2 cards hacked, etc. Just like the binomial and hypergeometric distributions, this can be time-consuming. Fortunately, we can save time using Excel’s POISSON.DIST function.
Equation 17: Excel’s POISSON.DIST Function
=POISSON.DIST(X,Mean,Cumulative).
Where: X = Number of successes
Mean = Expected value (Must be ≥ 0)
Cumulative = True or False
If Cumulative is False, Excel returns the PMF. If Cumulative is True, Excel returns the CPD.
Figure 23 shows the complete Poisson distribution calculation in Excel:
Figure 23: Poisson Distribution
Table 3 shows charts of the PMF and CPD distributions:
Figure 24: Charts of the Poisson PMF and CPD Distributions
We could also use the probability table to find these probabilities. Figure 24 shows the appropriate section of the Poisson Distribution.
Figure 25: Poisson Probability Distribution Table for μ = 1.50
The complete set of tables is in the file named Chapter08_PoissonTable.xlsx. Here is how this table works. Find the column header for μ, which is often called λ. The column labelled X shows the number of successes from 0 to 9. The probability of no successes is 0.2231, which matches the results found using the Poisson probability function and Excel. The Poisson probability table has a drawback. It does not provide a precise answer if the mean does not match one of the column headers.
VII. Summary
In this chapter, we examined the three most commonly used discrete probability distributions: Binomial, hypergeometric, and Poisson. Table 3 shows a comparison of these distributions.
Table 3: Comparison of the Three Discrete Probability Distributions
VIII. Exercises
Practice your discrete probability distribution skills by answering the following questions. Data for these exercises can be found in Chapter08_Exercises.xlsx.
Exercise 1:
You are touring the little country of Fredonia with your friends. At breakfast one morning you learn that the Fredonia national lottery’s grand prize is at an all-time high: 500 billion fireflys. One firefly is worth $0.001. The lottery requires a player to select six numbers from 46. A number can only be selected once.
Questions:
1. Which of the three discrete probability distributions must you use to compute the probability distribution? What is your rationale for this decision?
2. Consider a success picking a winning number. What are the probabilities that a ticket will have no winning numbers, 1 winning number, 2, 3, 4, 5, or 6 winning numbers?
Exercise 2:
You are still in Fredonia and having a great time. One evening, during an electrical blackout, the concierge of the hotel brings you candles. She tells you that on average there is an electric blackout once a week. You will be in Fredonia for a week. What is the probability of no blackouts, 1 blackout, two blackouts, and more than two blackouts in a seven-day period?
Questions:
1. Which of the three discrete probability distributions must you use to compute the probability distribution? What is your rationale for this decision?
2. Consider an electrical blackout a success. Construct the appropriate discrete probability distribution.
Exercise 3:
30 percent of students at the Nunya School of Business are freshmen. The dean of students selects four students at random to give high school seniors tours of the college’s campus. What is the probability that the dean will select no freshmen, one freshman, two freshmen, three freshmen, or four freshmen?
Questions:
1. Which of the three discrete probability distributions must you use to compute the probability distribution? What is your rationale?
2. Construct the appropriate discrete probability distribution.
Exercise 4:
You are working in the Quality Control department of Nile Deliveries. Your firm averages two lost packages per 1,000 deliveries. For the next 1,000 deliveries, what is the probability of no lost packages, one lost package, two lost packages, or more than two lost packages? The mean equals 2.0.
Questions:
1. Which of the three discrete probability distributions must you use to compute the probability distribution? What is your rationale?
2. Consider a lost package a success. Construct the appropriate discrete probability distribution.
Exercise 5:
Ten percent of the population is left-handed. This weekend, forty golfers play in a golf tournament. Ten foursomes will be drawn at random. What is the probability that a foursome will have no left-handed golfers, one left-handed golfer, two left-handed golfers, three left-handed golfers, or four left-handed golfers?
Questions:
1. Which of the three discrete probability distributions must you use to compute the probability distribution? What is your rationale?
2. Consider a success picking a winning number. Construct the appropriate discrete probability distribution.
Exercise 6:
You are taking a class in Ancient Greek philosophy. You know absolutely nothing about this subject. During the first class, your professor gives the class a surprise examination consisting of seven true/false questions. Given your lack of knowledge of Ancient Greek Philosophy, you guess the answers to all the questions. What is the probability that you will get no questions right, one question right, and two, three, four, five, six, and seven questions right?
Questions:
1. Which of the three discrete probability distributions must you use to compute the probability distribution? What is your rationale for this decision?
2. A success is considered getting a correct answer.
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