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Chapter 5

Discrete Probability Distributions

95

96 CHAPTER 5. DISCRETE PROBABILITY DISTRIBUTIONS

5.1 Random Variables

We have looked at some probability calculations in the previous chapter. In this chapter we willlook at some situations where the type of problem is the same but the only difference is the specificapplication. For example, if we are looking at probabilities of flipping a coin five times and lookingat the probability of getting three heads or tossing a die five times and looking at the probabilityof getting 3 even numbers, the problems are the same. The only difference is the setting of theproblem. We need to distinguish discrete and continuous random variables.

We now turn our attention to Random Variables. These are just a more formal way to talkabout variables. In this chapter, our focus will be on quantitative variables. We will be examiningtwo different types of random variables: continuous and discrete.

A variable whose value is determined by a random process is a Random Variable.

A Discrete Random Variable is a random variable that can only take on a countable numberof possible values.

A Continuous Random Variable is a random variable that can take on any value in an intervalor intervals.

We will discuss discrete random variables in this chapter and continuous random variables inthe next chapter.

Example 5.1.1.

Classify the following as continous or discrete random variables:

1. The number of DUI arrests at a randomly selected sobriety checkpoint2. The height of a randomly selected adult3. The number of strikeouts a pitcher has in a randomly selected game4. The time it takes a randomly selected taxpayer to prepare their taxes5. The speed of a randomly selected car on the freeway at noon

Solution.

The key to determining which it is is by figuring out what the possible values of the randomvariable are. If you can list them all (even if you use an ellipsis . . .) then it is discrete. If the randomvariable can be anything between values, it is a continuous random variable.

Notice that they all include ‘random’ or ‘randomly’ in their definition. It needs to be clearhow the value is obtained and what it represents. In the first one, we randomly select a sobrietycheckpoint and then count the number of DUI arrests. The random variable is not ‘DUI arrests’:it is not clear what we are talking about. Is this in the entire United States? Santa Cruz county?We don’t know. Our random variable would not be well defined.

5.1. RANDOM VARIABLES 97

1. The number of DUI arrests at a randomly selected sobriety checkpoint: The possible vauesare 0, 1, 2, 3, . . . Therefore this random variable is discrete.

2. The height of a randomly selected adult: A typical height might be 67 inches. What is next?There isn’t a ‘next’ possible value. If you think 68 inches, what about 68.5? or 68.1? Theheight can take on any value in an interval. This random variable is continous.

3. The number of strikeouts a pitcher has in a randomly selected game: How many strikeoutscan a pitcher have? 0, 1, 2, 3, . . . Therefore this is a discrete random variable.

4. The time it takes a randomly selected taxpayer to prepare their taxes: also continous. Wecan’t list the possible values. As soon as you list two possible values the variable can take onany value between the two values. That makes this random variable continuous.

5. The speed of a randomly selected car on the freeway at noon: Again, this random variable iscontinous. If it were discrete, instead of your car accelerating smoothly, your drive would bevery jerky. You’d be driving at 60 mph then, BAM, you’re driving 61 mph etc.

We have looked in the past at frequency distributions, percentage distributions, etc. In thissection we introduce another distribution: the probability distribution.

Just as the frequency distibution has a list of all possible values and the associated frequencies,a probability distribution will have a list of all possible values of the random variable and theassociated probabilities.

The following are examples of probabiltiy distributions

X P(X)0 0.261 0.232 0.183 0.174 0.125 0.04

X P(X)0 0.621 0.252 0.13

X P(X)2 1/163 3/164 4/165 4/166 3/167 1/16

Notice that there are two conditions that every probability distribution must meet: the probabilitiesmust add up to one and the probabitilities need to all be positive. You can check the abovedistributions to check this.

Once we have the distribution, we can also look at a graph of the distribtution similar to a barchart.

Following is the graph of the probabaility distribution.


0 1 2 3 4 5 60

.05

.10

.15

.20

.25

.30

XP(X

)

The ‘bars’ that we use in histograms and bar charts are replaced with line segments. If therewas any width to these ‘bars’ we would wonder what the width represented.

For this distribution we would describe it as skewed right.

5.1.1 The Mean and Standard Deviation of a Discrete Random Variable

For a random variable the mean, µ, is referred to as the Expected Value, E(X). It is what we‘expect’. If X represents the number of calls a tow truck recieves on a randomly selected day, themean would simply represent the number of tows we can ‘expect’ in a given day.

Let us assume that the probability distribution is based on a frequency distribution with a totalof 100. Our distribution becomes

X Frequency0 261 232 183 124 75 4

The mean, or expected value, would be simply obtained as

µ =0 + 0 + · · ·+ 0 + 1 + 1 + · · ·+ 1 + 2 + 2 + · · ·+ 2 + 3 + · · ·+ 5

100

There are 26 0’s, hence the · · · etc. This then becomes

µ =(0× 26) + (1× 23) + (2× 18) + (3× 17) + (4× 12) + (5× 4)

100

This can be written as

µ = 0× 26

100+ 1× 23

100+ 2× 18

100+ 3× 17

100+ 4× 12

100+ 5× 4

100

Notice that this is the sum of the random variable times the associated probability. We canwrite a formula:

µ = E(X) =∑

XP (X)


We can also do this on our calculatorInput values of X in to L1 and the probabilities in L2.STAT > CALC > 1:1-Var StatsSpecify lists, L1, L2 (you need to make sure you have the comma between lists)Our calculator gives usX=1.78σx = 1.49n = 1Notice that the Sx is blank. This is expected because n = 1.So we have E(X) = 1.78 and σ = 1.49Recall that the mean is where the graph ‘balances’. Look at the graph. It is reasonable that

the graph would balance at 1.78

Example 5.1.2.

In roulette, the wheel consists of 38 slots: 18 red slots, 18 black slots, and 2 green slots. You bet$1 on black. If black comes up, you win $1. If not, you lose. Construct a probability distributionand find the expected value where X is the amount of money won. Interpret the expected value.

Solution.

Each slot is equally likely so the probabilties are straighforward.There are only two things that can happen: win or lose. There are a total of 18 winning slots

and 20 losers.

X P (X)(win) 1 18/38

(lose) -1 20/38

To calculate the expected value we will use the formula µ = E(X) =∑XP (X)

X P (X) XP (X)(win) 1 18/38 18/38

(lose) -1 20/38 -20/38-2/38

The expected value is -$.0526 (=-2/38) or -5.26 ¢. What this means is that every time we betone dollar, we lose on average, about a nickel.

Example 5.1.3.

A car dealership has sent out an advertisement that promises people who come in will receive agift. Guests are guaranteed one of the following: a new car or $50,000, a TV worth $2,000, ticketsto a local theme park worth $200, or a $5 Target gift card. What are your expected winnings? Howmuch does it cost the dealership? The probabilities are stated as the car/$50,000, 1 in 20,000, theTV is 2 in 20,000, theme park tickets 5 in 20,000, and $5 gift card is 19,993 in 20,000.


Solution.

Since we want to know our ‘expected winnings’ we are looking for the mean of the winnings.This means our random variable will be the amount we win. Let us construct the probabiltiydistribution. And use it to find the mean.

X P(X) XP(X)50000 1/20000 50000/200002000 2/20000 4000/20000200 5/20000 1000/20000

5 19993/20000 99965/20000154965/20000

The probability distribution consists of just the first two columns above. The third is used tocalculate the mean which is $7.75 (=154965/20000). When we look at the probabilities, we assumethat the flier is being sent to 20,000 customers. Assuming this is the case, we find the total amountis $154,964 plus any additional costs: postage, printing, etc.

This example was inspired by an advertisement received by the author in the mail. If severaldealerships work together to put this on they can split the expenses and focus on only customersthat might be in the market for a car.

Example 5.1.4.

20% of students at a large university are juniors. A random sample of 2 students are to beselected. Let X be the number of juniors selected. Find the probability distribution.

Solution.

Let us construct a tree diagram with the information. In the diagram, J represents getting aJunior and J represents not getting a Junior.

J.20

.20

J

.80

J

.80.20

J

.80

.04 X = 2

.16 X = 1

.16 X = 1

.64 X = 0

J

J

To the right of the tree are the probabilites and the appropriate values of the random variable.Starting with the root of the tree, count the number of J ’s and we get the values of X.

Note that there are two different places in the tree where X = 1. We need to add these together.Summarizing, we get


X P (X)0 .641 .322 .04

Note that, as needed, the probabilties add up to one.

5.1.2 Exercises

1. Determine if the following are continous or discrete random variables.

(a) The number of fish a fisherman catches during a randomly selected fishing trip

(b) The number of foul balls hit into the stands during a randomly selected baseball game

(c) The weight of a randomly selected blue whale

(d) The time a randomly selected teenager spent online in the previous 24 hours

2. Determine if the following are continous or discrete random variables.

(a) The time it takes for a randomly selected flight to get to its destination

(b) How many dogs are rescued at a particular animal shelter in a randomly selected week

(c) The number of high school students that graduated with a 4.0 or higher last year at arandomly selcted high school

(d) The amount of gas a randomly selected car gets the next time they visit a gas station

3. For the following, determine if the given distribution is a possible probability distribution. Ifit is, find the mean and standard deviation of the random variable.

X P(X)0 .321 .212 .203 .124 .095 .06

X P(X)4 1/105 2/106 2/107 3/108 1/109 1/10

X P(X)0 .201 .362 .183 .194 .115 .10

X P(X)2 -.213 .354 .415 .186 .157 .12

4. For the following, determine if the given distribution is a possible probability distribution. Ifit is, find the mean and standard deviation of the random variable.

X P(X).25 .41.50 .26.75 .131.25 .111.50 .09

X P(X)2 .334 .266 .208 .4910 .11

X P(X)-5 .10-4 .23-3 .26-2 .39-1 .02

X P(X)10 .1820 .32200 .222000 .15-10 .13

5. One estimate gives the percentage of people that are left-handed at 15%. Assuming this istrue find probability distribution of X where X is the number of left-handed people selectedat random in a sample of 2.


6. In a large city, 30% of residents are over the age of 50. Let X be the random variable thatrepresents the number of residents that are over 50 in a random sample of 2 residents. Findthe probability distribution of X.

7. A pen holder contains 5 black pens and 3 blue pens. A person randomly grabs two pens. LetX be the number of blue pens selected in a random selection of two pens from the container.Find the probability distribution of X.

8. While getting ready for a long trip, a book lover randomly grabs 2 books off the shelf. Theshelf contains 5 romance novels and 6 mysteries. Let X be the number of romance novelsselected in a random sample of 2 books. Find the probability distribution of X.

9. 30% of a particular model of a car are painted white. If 3 cars are randomly selected, findthe probability distribution of X where X is the number of white cars in a random sample of3 cars.

10. It has been reported that 40% of teens text while they drive. Assuming this is true, find theprobability distribution for the random variable that describes the number of teens that textwhile they drive in a random sample of 3 teens

11. A farmer is looking ahead to the amount of rain during the growing season. If there is adrought, the farmer will lose $1,000. If there is marginal amount of rain, the farmer will earn$1,500. If there is sufficient rain, the farmer will earn $5,000. If there is too much rain, thefarmer will experience damaged fruit and will only make $3,000. The probability of a droughtis estimated to be .08, marginal rain is estimated to have a probability of .35, sufficient rainhas a probability of .53 and the probability of too much rain is .04. Find the expected amountthe farmer will earn.

12. An insurance company has a policy for a concert. If the headliner cancels, the insuancecompany will pay the concert promoters $1,000,000 to cover lost revenue from the concert.The concert promoter has paid $15,000 for this policy. The insurance company estimatesthere is a 2.5% chance the headliner will cancel. Find the expected value of the payout to thepromoter. Is this a good deal for the insurance company?

13. A company offers extended warranties to car owners. The policy costs $300 per year. Thecompany estimates the following amount will need to be paid out and the probabilites: $3,000,probability of .001 $1,500, probability of .02, $500, probality of .25. What is the expectedamount of money the company takes in for a policy?

14. In roulette, there are 38 slots, each numbered. If you bet $1 on a number and it comes up, youwin $35. Find the probability distribution and the expected winnings for this game. Comparewith betting on a color. See example in this section. Which bet is better: a number or color?

15. Two friends come up with a game. A person rolls a fair die. If a 1 comes up, player A givesplayer B $8. If a 2 or 3 comes up, player A gives player B $2. If a 4 comes up, player A givesplayer B $5. If a 5 or 6 comes up, player B gives player A $10. If the game is to be playedseveral times, who is expected to come out ahead: A or B? If they play the game 100 timeswho we expect to end up richer and by how much?


16. A fair game is a game where the expected value is 0. In a die game, a player pays a fee toplay. The die is rolled and the player recieves an amount, in dollars, equal to the numberrolled (For example, if a 3 comes up, the player gets $3). How much does the player need topay to play the game if it is going to be a fair game?


5.2 Binomial Distribution

Some of the probability problems we have seen before can be grouped together by type once westrip away the specifics of the problem. The binomial distribution is one such distribution. We canthink of it as a coin flipping problem where we want to know the probability of flipping a specifiednumber of heads in a given number of flips. In these ‘coin flipping problems’, the probability offlipping heads is not necessarily .5. We need to get some terminology down.

There are some specific properties of a coin flipping problem that, although perhaps obvious,we need to list.

1. We flip the same coin several times.2. Each time we flip the coin there are only two possibilities: heads or tails.3. The probability of getting heads doesn’t change.4. If we get heads, the probability of getting heads on the next flip doesn’t change.

If we look at rolling a die several times and we are only interested in, say, how many times aone is rolled, this follows the four items listed about a coin if we simply replace: ‘coin’ with ‘die’,‘heads’ with ‘one’, and ‘tails’ with ‘not a one’. Now let us generalize this.

A Binomial Experiment is an experiment in which the following conditions are met:

1. A trial is repeated n times.2. Each trial consists of two possible outcomes: success and failure.3. The probability of success does not change throughout the experiment.4. The trials are independent.

What we are interested in is the number of successes in a binomial experiment. This leads to thefollowing.

A Binomial Random Variable is a random variable that represents the number of successes ina binomial experiment. We write

X ∼ B(n, p)

n and p are called the parameters of the distribution.

This notation, and similar notation for other distributions, will be used throughout. The X isthe name of the random variable, B stands for Binomial, the parameters n and p give the numberof trials and the probability of success. The parameters are what we need to totally describe thedistribution.

If we flip a fair coin 10 times and are looking at how many times heads is flipped this will bea binomial experiment. In this case, n is 10, there are only two possible outcomes (heads andtails), the probability of heads and tails never change, and what happens on a flip does not affectwhat happens on subsequent flips. Therefore, this is a binomial experiment. We would writeX ∼ B(10, .5)

We can use a formula to calculate the formula or use technology to find the probability.

5.2. BINOMIAL DISTRIBUTION 105

If X ∼ B(n, p)

P (X = k) =

(n

k

)pnq1−n

n is the number of trialsp is the probability of successq is the probability of failure and q = 1− pYour calculator will also give this probability using a binomialPDF command: P (X =

k) =binomPDF(n,p,k).For cumulative probabilties P (X ≤ k), you would need a binomialCDF command: P (X ≤

k) =binomCDF(n,p,k). (C stands for cumulative.)

When we attack our probability problems in this chapter and beyond we will follow the following4 steps.

Four steps in solving probability problems:

1. Indicate what the random variable represents2. Indicate the appropriate distribution and parameter(s)3. Indicate what you are calculating, symbolically4. Answer the question

Example 5.2.1.

According to a report, 20% of all adults in the United States are smokers. Find the probabiltiythat in a random sample of 18 adults in the US, 5 are smokers.

Solution.

Let us start by determining what makes up a trial. In this case it would be randomly selectingone person and determining if they were a smoker or not. This will be done 18 times in total. Alsonote there are two possible outcomes: smoker or non-smoker. We are randomly choosing 18 fromthe population of US adults, a huge population compared to 18, so our probability of success willnot change enough to worry about. Also, we are randomly choosing these people so the trials willbe random. This meets all the criteria of a binomial experiment. Our ‘success’ is whatever we arecounting. In this case a ‘smoker’ is our ‘success’.

1. Our random variable X is the number of successes. In this example we getX=number of smokers in a random sample of 18 US adults.

2. We have decided that it is a binomial experiment and so X is a binomial random variable sowe haveX ∼ B(18, .2)

3. We want the probability that we get 5 smokers so we write P (X = 5)


4. Lastly, we can get the probability using the formula or our calculator. So we have.(185

)(.2)5(.8)13 = .1507, (or binomPDF(18,.2,5) on our calculator) so about 15% of the time

you select 18 adults, 5 will be smokers.

In the next example we will look at using our calculator to evaluate the probabilities. Wehave two options on our calculator: PDF and CDF corresponding to P (X = k) andP (X ≤ k),respectively. In order to use the calculator we will need to write all of our probabilities as acombination of these.

Example 5.2.2.

According to the National Institute of Mental Health, in 15% of births the mother suffers frompostpartum depression (PPD). Assume that this is true for all births. A random sample of 25women who recently gave birth is to be selected. What is the probability that the number of thesenew moms who suffer PPD is...

(a) Fewer than 5(b) At least 3(c) Exactly 6(d) 2 to 7

Solution.

We will proceed with our 4 steps as before. Note that this is a binomial experiment: there aretwo possible outcomes, we are repeating a trial (select one mom and determine if she had PPD),the probability is not expected to change, and the trials are independent. Also, we are countingthe number of women with PPD so that is what we consider a ‘success’.

1. Let X = The number of women with PPD in a random sample of 25 women who recentlygave birth.

2. X ∼ B(25, .15)3. (a) P (X < 5)

(b) P (X ≥ 3)

(c) P (X = 6)

(d) P (2 ≤ X ≤ 7)4. (a) For P (X < 5), we need to rewrite as P (X ≤ 4). This can easily be evaluated using

binomCDF(25,.15,5)=.8385

(b) For P (X ≥ 3), we must rewrite as 1− P (X ≤ 2) = 1−binomCDF(25,.15,2) = .7463

(c) P (X = 6) is ready to input as, binomPDF(25,.15,6)= .0920

(d) For P (2 ≤ X ≤ 7) we rewrite as P (X ≤ 7)− P (X ≤ 1) =

binomCDF(25,.15,7)−binomCDF(25,.15,1)=.8815.

A bit of common confusion amongst students is the 1 in the second inequality. Below isthe reasoning, with all the X’s removed for a better fit on the page.

5.2. BINOMIAL DISTRIBUTION 107

P (7 ≤ X ≤ 2)

= P (x ≤ 7)− P (X ≤ 1)

= P (7) +P (6) +P (5) +P (4) +P (3) +P (2)

= P (7) +P (6) +P (5) +P (4) +P (3) +P (2) +P (1) +P (0)

− (P (1) + P (0))

5.2.1 Exercises

Solve the following probability problems by addressing the 4 steps outlined in the section.

1. According to one site, 15% of Americans are left-handed. Assume this is true for the currentpopulation. What is the probability that in a random sample of 16 Americans, 4 are left-handed

2. According to breastcancer.org, 12.4% of women will develop invasive breast cancer in thecourse of their life. In a random sample of 17 women, what is the probability that 3 to 6women will develop breast cancer in the course of their life?

3. According the Red Cross, 53% of Latino-Americans have O+ blood. In a random sample of10 Latino-Americans at most 5 have O+ blood.

4. A professional baseball player has a batting average of .302. This means they get a hit 30.2%of the time. What is the probability that at 12 randomly selected trips to the plate, the playerwill get at most 5 hits?

5. An April 8, 2015 Pew Research Center survey indicated that 73% of teens have or have accessto a smartphone. Assume that is true for the current population of American teens. Findthe probability that of 18 randomly selected American teens more than 14 will have or haveaccess to smartphones.

6. The California Elections Board reported that as of October 24, 2016 24.27% of registeredvoters were registered as ‘No Party Preference’. Assume that is true for the current populationof registered voters in California. A random sample of 13 California registered voters are tobe selected. What is the probability of selecting at most 5 voters that are registered as ‘NoParty Preference’?

7. In the report ‘Sexual Activity and Contraceptive Use Among Teenagers in the United States:2011-2015’, the CDC reported that 42% of female teens aged 15-19 reported having ever hadsex. Assume that the report is an accurate depiction of the current population of female teensaged 15-19. What is the probabiltiy that in a random sample of 14 female teens aged 15-19,more than 8 will have ever had sex.

8. According to www.ptsd.va.gov, estimates that 30% of Vietnam Veterans have had Post Trau-matic Stress Disorder (PTSD) in their lifetime. Assume this is true. What is the probabilitythat in a random sample of 9 Vietnam Veterans, 3 to 6 will have had PTSD at some point intheir lifetime.


9. www.ptsd.va.gov goes on to report that 23% of women who use VA health care reported sexualassault when in the military. Find the probability that in a random sample of 8 women whouse VA health care, fewer than 3 will have reported sexual assault.

10. The National Alliance of Mental Illness (NAMI) reports that 18.1% of American adults livewith anxiety disorders. If this is true of the current population, find the probability that morethan 4 of 12 randomly selected American adults live with anxiety disorders.

11. The National Association of Anorexia Nervosa and Associated Disorders reports that 13% ofwomen over 50 engage in eating disorder behaviors. If this accurately describes the currentpopulation of women over 50, find the probability of selecting 4 women over 50 who engagein eating disorder behaviors from a random sample of 13

12. The American Veterinarian Medical Association (AVMA) reports that 36.5% of Americanhouseholds own at least one dog. A random sample of 9 households is to be selected. Findthe probability that fewer than 4 own at least one dog.

13. The American Veterinarian Medical Association (AVMA) reports that 30.4% of Americanhouseholds own at least one cat. A random sample of 12 households is to be selected. Findthe probability that fewer than 6 own at least one cat.

14. Your parents are watching you. According to an October 24th, 2018 report by Pew Research,68% of US adults use Facebook. 7 randomly selected US adults are asked about Facebook.What is the probability that at least 5 use Facebook?

15. Drunk drivers accounted for 29% of all fatalities on American roads in 2015 according tofinder.com. If the percentage is still the same, find the probability that in a random sampleof 17 fatalities on American roads, at most 4 are accounted to drunk drivers.

16. In the report ‘Parents in Prison and Their Minor Children’(US Department of Justice, 2008),it is reported that 54% of Americans in federal prisons were parents of minor children. Findthe probability that more than 4 of 10 randomly selected Americans in federal prisons areparents of minor children.

17. 61% of Americans wear visual aids occasionally according to a CBS report from September20, 2013. Assume the current population is the same. Determine the probability that 2 to 6of a random sample of 9 Americans wear visual aids occasionally.

18. An article from the Huffington Post from May 25, 2011 reports that 18% of Americans thinkthe Sun revolves around the Earth. Proceed as if this is true currently. What is the probabilitythat in a random sample of 15 Americans, 3 to 6 that think the Sun revolves around the Earth?

19. Go online and find a site that gives the percent of a large population that has some charac-teristic. Write up a binomial probability problem and solve.

5.3. THE HYPERGEOMETRIC DISTRIBUTION 109

5.3 The Hypergeometric Distribution

In the preceeding section we looked at the binomial distrubtion. In this section we look at a similardistribution, the hypergeometric distribution. The main distinction between the binomial and thehypergeometric experiment is that in a binomial distribution the probability of success remainsconstant whereas in a hypergeometric experiment the probabilties change. Let us begin with anexample.

Example 5.3.1.

A box contains 12 light bulbs. Four of the bulbs are burned out. You randomly select 3 bulbs.Let X be the number of burned out bulbs in a random sample of 3 bulbs. Find the probabiltiy ofgetting 1 burned out bulb.

Solution.

To find the required probability for this problem we need to count the number of ways to getthe different possible outcomes.

First off, how many ways can we pick 3 bulbs? We are picking 3 items out of 12, this meanseither a combination or permutation. Since order doesn’t matter, we want a combination. So wehave

(123

)= 220. Recall that to get the probabiltiy distribution we need to determine all possible

values the random variable can take on and the associated probabilities.When we say we have one defect (X = 1) what we really mean is: one bulb is defective and 2

are not defective. (remember, we are picking 3 bulbs). Now let’s count the number of ways to dothis. First, count the number of ways of picking the one defective. This is

(41

)= 4. The number

of ways of picking the 2 good bulbs is(

82

)= 28 ((if 4 are burned out then the rest, 8, are good).

For each way to pick 1 burned out bulb there are 28 ways to pick the remaining 2 good bulbs. Todetermine the total number of ways of picking 1 burned out and 2 good bulbs we multiply togetherand get 4× 28 = 112. Finally, the probability is 112

220 = .5091.

Let us now formally define a few things.

A hypergeometric experiment must meet the following conditions:

� Items are selected at random from a population, without replacement.� The population constists of N items.� The population consists of only ‘successes’ and ‘failures’.� There are r ‘successes’ in the population.

If we let X be the number of ‘successes’ in a hypergeometric experiment the we get a hyper-geometric random variable and we write

X ∼ H(N, r, n)

To calculate the probability we use the formula


P (X = k) =

(rk

)(N−rn−k

)(Nn

)

r is the total number of successes

k is the number of successes you want the probability of

n− k is the number of failures you want the probability of

N − r is the total number of failures

N is the population size

n is the sample size

Example 5.3.2.

A elementary school has 13 teachers. Seven of the teachers have CPR certifications. Fiveteachers from the school are to be randomly selected to attend a conference. Find the probabiltiythat three of the teachers have CPR certifications.

Solution.

Let’s begin by noting that the ‘population’ (the 13 teachers) fall into two categories: they haveCPR certification or they don’t. We are picking 5 of them without replacing. This sounds likeeither binomial or hypergeometric. To determine which one, we need to ask ourselves one question:we are picking five teachers out of how many? If the population is large then binomial would beappropriate, (13 isn’t large). Otherwise we use the hypergeometric distribution.

� Let X=The number of teachers with CPR certification in a random sample of 5 teachers.� X ∼ H(13, 7, 5)� P (X = 3)

� P (X = 3) =

(73

)(62

)(

135

) =35× 15

1287= .4079

The formula for the hypergeometric probabilities are for the probability that X equals a number.If we have an inequality we will need to apply the formula more than once. This is illustrated inthe following example.

Example 5.3.3.

21 chickens have just hatched. Twelve of the newly hatched chicks are female. Four chicks areto be selected. Find the probabiltiy that at most 2 are female


Solution.

We are picking 4 items without replacement out of 20 and we want the probability of at most2 ‘successess’. Hypergeometric is appropriate.

� Let X=The number of females selected in a random sample of 5 chicks.� X ∼ H(20, 12, 4)� P (X ≤ 2)� P (X ≤ 2) = P (X = 0) + P (X = 1) + P (X = 2)

=

(120

)(94

)(

214

) +

(121

)(93

)(

214

) +

(122

)(92

)(

214

)

=1× 126

5985+

12× 84

5985+

66× 36

5985= .5865

5.3.1 Exercises

1. Let X ∼ H(12, 5, 4), Find P (X = 3)

2. Let X ∼ H(20, 8, 5), Find P (X = 2)

3. Let X ∼ H(9, 4, 4), Find P (X ≤ 3)

4. Let X ∼ H(16, 12, 5), Find P (X ≥ 4)

For the following problems address all four steps used to solve probability problems.

5. A teacher has a box which contains 8 whiteboard markers. Three of them are dried out. Theteacher randomly selects 4 markers. What is the probability that 2 of the markers selectedare dried out?

6. At a party, the hosts have put out a bowl of cookies for their guests. Five of the cookieshave been made with organic ingredients and the remaining 8 have been made with non-organic ingredients. A partygoer randomly selects 5 cookies and to take home. What is theprobability that the guest gets 3 organic cookies.

7. A candy lover is also allergic to peanuts. Before our candy lover is a plate which contains 20pieces of fudge: Four of the pieces of fudge have traces of peanuts in them. Our candy fan isplanning on eating four pieces of fudge. What is the probability of the candy lover having anallergic reaction to peanuts?

8. At an NHL game, there are 25 pucks. Unfortunately, 9 of the pucks are not of the properweight. Seven pucks are to be randomly selected. Find the probability of getting 2 pucks thatare not of the proper weight.

9. A case of beer has a variety of types: 6 IPAs, 4 stouts, and 2 lagers. If a beer lover was torandomly select 5 beers, what is the probability of getting 2 IPAs?


10. An appliance store has 19 refrigerators. Six of the refrigerators have a stainless steel front.What is the probability that of 10 randomly selected refrigerators at most 2 will have stainlesssteel fronts?

11. A car dealer tells the assistant to park some of the new models in front of the dealership. Theassistant plans to randomly select 6 car keys to determine which cars to park in front. Thedealership has 12 new models from which the assistance can choose. Three of the cars areconvertibles. Find the probability that the assistant doesn’t park any convertibles in front.

12. A classroom has 12 students in it. All have strong preferences about which computer theyprefer: PC or Apple. Eight are Apple users and 4 are PC users. The teacher has no ideaabout this plans to randomly select 12 computers from the storage room. Students then willselect one of the computers. In the storage room there are 20 Apple computers and 9 PCs.What is the probability that the teacher will not have any students that are unhappy withwhat computer they have?

13. A box of light bulbs contains 8 halogen light bulbs and 9 incandescent bulbs. A homeownerwill randomly pick 5 of the bulbs to replace some burned out bulbs. Find the probability ofgetting at most 2 halogen bulbs.

14. While getting ready for an early morning flight a traveler is going to grab 8 loose socks froma drawer that contains 10 black socks and 6 dark blue socks. The blue socks all match andthe black socks all match. Unfortunately for the traveler, the electricity is out and the socksall look alike in the dark. Find the probability that the traveler will be able to match thesocks without any black/blue combinations?

15. At a traffic school for people that get tickets, there are 21 students: 16 got speeding ticketsand the remaining 5 got other traffic violations. For an in-class demonstration, 6 studentsare to be selected to perform a traffic safety skit. What is the probability that 4 of them gotspeeding tickets?

16. At a meeting of engineers, 7 are electrical engineers and 9 are mechanical engineers. If arandom sample of 3 engineers are to be selected, find the probability of getting at least 2mechanical engineers.

17. For an article on airline arrival times, a reporter is going to randomly select 11 flights thatarrived at an airport during the a given hour and report on how late the flights were. Duringthe hour in question, there were 23 flights and 5 were more than 30 minutes late. What is theprobability of the reporter selecting at least 2 flights that were more than 30 minutes late?

18. It is time for standardized testing for a student. The student was told to bring a number 2pencil for the test. The student prefers to have at least 2 pencils when taking a test. Thestudent has a cup on their desk with 12 pencils in it. Although the student thinks they areall number 2 pencils, 5 of them are not. The student is in a hurry and will just grab 4 fromthe cup. What is the probability that the student will have at least two number 2 pencils?

19. At a veteran’s reunion, there are 5 officers and 12 enlisted. The veterans are challenged toa pick-up game of basketball so they will randomly select 5 of the veterans. What is theprobability of selecting no more than 2 officers?


20. While planning a trip, a vacationer takes 15 DVD’s with them. Of those taken, 9 are comedies.The vacationer is planning to have a movie-fest and so is going to randomly pick 5 DVD’s towatch. Find the probability of selecting at least 1 comedy.

21. In the fall an avid gardener is planting bulbs for spring. The gardener has 5 white bulbs, and7 red bulbs. The gardener is planning on planting 4 bulbs. What is the probability that allof the bulbs will be the same color?

22. A photographer is applying to a program and needs to send in 5 photos. The photographernarrows down the best photos taken: it is down to 8 portraits and 6 panoramas. If the choiceis made at random, find the probability that there is at least one of each type.

23. At a costume party there are 15 people in costume and 4 that aren’t in costume. What is theprobability that in a random sample of 6 partiers at least 4 are in costumes.

24. The dog park in a park is a popular spot. On a specific day there are 12 dogs present, eachwith a different owner. Of the dogs present, 7 have been spayed or neutered. A representativefrom a local animal group is also at the park and has information about spaying/neuteringpets. The representative only has time to talk to 4 randomly selected pet owners. What isthe probability of selecting at most 1 owner whose dog has already been spayed/neutered?


5.4 The Poisson Probability Distribution

The final discrete distribution we discuss is the Poisson probabiltiy distribution. The distributionis named after Simeon Poisson, a French mathematician. The distribution deals with the numberof occurances in a time or space interval.

Let X be the random variable that represents the number of occurances in a time or spaceinterval where the number of occurances are random and independent. Then X is a Poissonrandom variable and we write X ∼ P(λ).

We will use the symbol P so we don’t confuse it with a P for probability. It would be a goodidea at this point to look at some examples.

Consider the number of pieces space debris that fall to Earth in a given week. The reader maybe unaware, but there are a great number of objects that are orbiting about the Earth. Each week,it is estimated that an average of 5 objects fall to the Earth each week.

It is reasonable that the debris falling are independent of one another. That is, an object fallingdoes not cause another object to fall or not fall. It will fall independently of the first object. That isnot to say that they cannot be dependent. It is just that most of the time they will be independent.Our model here, as all models, are not going to match reality 100%. But it will be close enough.As for randomness, when the objects hit the Earth could be any time. If X is the number of piecesof debris falling to Earth in a randomly selected week then X would be a Poisson random variableand we would write X ∼ P(5).

Emergency room visits are another example of a Poisson random variable. Suppose we have anaverage of 8.2 people come into the emergency room to be seen between 2 and 3 in the afternoon.Let X to be the number of people that come into an emergency room to be seen during that timeinterval. In this case, when people arrive at the emergency room is random: you go to the ERwhen you are in need of stiches, diagnosis for an injury, etc. Also, the events are indpendent of oneanother. (This could be violated if a car accident resulted in two people needing to go the ER, notvery likely so we won’t let it concern us.) We would write. X∼ P(8.2).

Don’t you just love junk mail? The amount of jumk mail you recieve in a randomly selectedday can be modeled as a Poisson random variable. Assume you receive 2.1 pieces of junk mail eachday, on average. It is reasonable that the junk mail you get are independent of one another andwhen they hit the mail stream prior to you getting them is random. In this case if X is the numberof pieces of junk mail in a randomly selected day then we would write X∼ P(2.1)

As with the binomial distribution we can either use a formula or our calculators to find aprobability.

If X ∼ P(λ)

P (X = k) =e−λλk

k!

λ is the average number of occurances

5.4. THE POISSON PROBABILITY DISTRIBUTION 115

Your calculator will also give this probability using a poissonpdf command:For single probabilities: P (x = k) = poissonpdf(λ, k).For cumulative probabilties: P (X ≤ k) = poissoncdf(λ, k).

Example 5.4.1.

The average number of battery fires for a popular cell phone is 2.3 fires per week. Find theprobability that there will be 4 battery fires for this phone in a randomly selected week.

Solution.

We are looking at the number of occurances (an occurance is a phone catches fire) in a timeinterval (one week). It is reasonable that the incidents are independent of one another and theycould occur anywhere. So the Poisson distribution is appropriate.

1. Our random variable X is the number of occurances in a time interval. In this example wegetX=number of cell phone batteried that catch fire in a randomly selected week.

2. We have decided that the Poisson distribution is appopriate so we haveX ∼ P(2.3)

3. We want the probability of 4 fires so we write P (X = 4)4. Lastly, we can get the probability using the formula or our calculator. So we have.

P (X = k) =e−2.32.34

4!= .1169 So about 12% of all weeks there will be 4 cell phones that

catch fire.

Example 5.4.2.

For each large truckload of earth that is excavated from a mine, an average of 1.4 diamonds ofgem quality are found. Find the probability that in a randomly selected large truckload of earththere we be at most 2 diamonds of gem quality found.

Solution.

Unlike the last example, in this example we are looking at the number of occurances in a spaceinterval. We still will apply the Poisson probability distribution. It is reasonable that the diamondsare randomly distributed and hence being mined will be independent of one another. They couldoccur anywhere in the truckload (randomness).

1. Let X=number of gem quality diamond found in a randomly selected large truckload.2. X ∼ P(1.4)3. P (X ≤ 2)4. Lastly, we can get the probability using our calculator. So we have poissoncdf(1.4,2) = .8335


5.4.1 Approximating the Binomial Distribution with the Poisson Distri-bution

As we look at the problems above and those in the excercises, we might notice that some of theproblems look a lot like the binomial problems we have seen in the past. This is because one of theuses of the Poisson dstribution is to approximate a binomial distribution. We can do so when theprobability of success, p, is very small and the number of trials, n, is very large. If you examing thecell phone fire from above, it is really a binomial problem: a randomly selected cell phone will catchfire or it won’t, etc. . . . . In this case, the probability of a cell phone catching fire, p, is extremelysmall(and unknown) but the number of cell phones used, n, is very large (also unknown). Themean of this binomial distribution would be np, a very large number times a very small numberyielding 2.3 fires per week. This is a common situation, n and p are unknown but we know whatnp is. As you proceed with the excercises, look for the problems which are really binomial.

5.4.2 Exercises

1. Let X ∼ P(1.4), Find P (X = 2)

2. Let X ∼ P(3.6), Find P (X = 3)

3. Let X ∼ P(3.2), Find P (X ≤ 3)

4. Let X ∼ P(5.1), Find P (X ≥ 4)

5. Let X ∼ P(2.7), Find P (2 ≤ X ≤ 5)

6. Let X ∼ P(4.3), Find P (3 ≤ X ≤ 7)

For the following problems address all four steps used to solve the problems.

7. On average, there are 5.2 broken bones seen at a local hospital’s emergency room in a givenweek. Find the probability that in a randomly selected week there will be 4 broken bonesseen in the emergency room.

8. At a large assembly line, production needs to be shut down unexpectedly an average of 2.6times per day. What is the probability that the line will be shut down 4 times in a randomlyselected day.

9. It is raining at a fairly steady rate when you notice that an average of 5.6 raindrops fall in abird bath each second. Find the probability that you will observe at most 8 raindrops in thebird bath in a randomly chosen second.

10. At a large orange orchard, a farmer grows seedless oranges. The variety isn’t truly seedless,a very rare few have seeds. The farmer notes that during harvest, an average of 5.9 orangesare found each day that have seeds. What is the probability that the farmer will find at most4 oranges with seeds in a randomly selected day?

11. A high energy particle detector detects an average of 9.4 particles per day. What is theprobability that there will be more than 8 particles detected on a randomly selected day?

5.4. THE POISSON PROBABILITY DISTRIBUTION 117

12. A rare tumor is diagnosed an average of 7.4 times per week. Find the probability that thetumor will be diagnosed more than 8 times in a randomly selected week.

13. An entomologist has set traps to collect bugs. On average, there are 2.1 uncommon flies thatare found in the traps each week. What is the probability that in a week there will be 2 to 6uncommon flies found.

14. A contractor specializes in concrete work. For each 1000 square feet of concrete laid, anaverage of 1.6 cracks form. What is the probability that in a randomly selected 1000 squarefeet poured, there will be 2 to 5 cracks formed?

15. A manufacture of glass extrudes the glass into a continuous sheet of glass. Although themanufacturer tries to avoid detectible air bubbles, an average of 2.6 bubbles are detected foreach 100 feet of glass. Determine the probability of finding more than 3 air bubbles in a 100foot long stretch of glass.

16. A nurse administers flu shots to patients. Although negative reactions to the flu shot is rare,the nurse has an average of 2.3 patients per month exhibit a fever after getting the shot.What is the probability that in a randomly selected month more than 4 patients will get afever after getting the shot.

17. The Empire State Building is hit by lightning and average of 23 times a year. What isthe probability that the building will be hit with lightening at least 25 times in a randomlyselected year.

18. California is hit by an average of 6.4 tornadoes each year. What is the probability that therewill be at least 5 tornadoes in a randomly selected year?

19. A copy machine that gets a lot of use jams an average of 1.2 times for each ream of paperused. What is the probability that there will be no jams if one ream of paper is used?

20. It has been estimated that about 5 people are fatally attacked by sharks each year. Assumingthis is true what is the probability that there will be at least 7 fatal shark attacks in arandomly selected year.

21. At 9 am each weekday, a maintenance worker comes to work and replaces any light bulbs ona large lighted display are burned out. On average, 1.3 light bulbs each day burn out. Themaintenance worker is the only one that replaces the bubs and only works Monday throughFriday. What is the probability that when the worker comes to work on Monday, there willbe at most five bulbs that are burned out? What is the probability that no bulbs will burnout when the worker is there if the worker leaves at 5 pm?

22. A secretary has an old computer. On average, if freezes up 1.8 times per day. The secretaryworks Monday through Friday. What is the probability that the computer will freeze at most5 times during a work week?

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