Discrete Probability Distributions
Chapter 6
6-1
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Learning Objectives
LO6-1 Identify the characteristics of a probability
distribution
LO6-2 Distinguish between discrete and continuous
random variables
LO6-3 Compute the mean, variance, and standard
deviation of a discrete probability distribution
LO6-4 Explain the assumptions of the binomial
distribution and apply it to calculate
probabilities
LO6-5 Explain the assumptions of the Poisson
distribution and apply it to calculate
probabilities
6-2
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What is a Probability Distribution?
PROBABILITY DISTRIBUTION A listing of all the outcomes of an
experiment and the probability associated with each outcome.
CHARACTERISTICS OF A PROBABILITY DISTRIBUTION
1. The probability of a particular outcome is between 0 and 1
inclusive.
2. The outcomes are mutually exclusive.
3. The list of outcomes is exhaustive. So the sum of the probabilities of
the outcomes is equal to 1.

Example: A drug manufacturer may claim a treatment will
cause weight loss for 80% of the population. This claim could
be tested by a consumer protection agency using a sample and
statistical inference.
6-3
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Probability Distribution Example


Suppose we are interested in the number of heads
showing face up with 3 tosses of a coin
The possible outcomes are 0 heads, 1 head, 2 heads, and 3
heads
6-4
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Probability Distribution Table

Probability distribution table and chart for the events of
zero, one, two, and three heads
6-5
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Example
6-6
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Example
6-7
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Random Variables

In any experiment of chance, the outcomes occur
randomly, and so are called random variables

Examples


6-8
The number of employees absent from the day shift on
Monday: the number might be 0, 1, 2, 3, …The number absent
is the random variable
The grade level (Freshman, Sophomore, Junior, or Senior) of
the members of the St. James High School Varsity girls’
basketball team. Grade level is the random variable (and notice
that it is a qualitative variable).
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Two Types of Random Variables


One type of random variable is the discrete random
variable
Discrete variables are usually the result of counting

6-9
Example: Tossing a coin three times and counting the number
of heads
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Discrete Random Variable


For example, the Bank of the Carolinas counts the
number of credit cards carried by a group of customers
The number of cards carried is the discrete random
variable
6-10
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Continuous Random Variables

Continuous variables are usually the result of measuring

Examples


6-11
The time between flights between Atlanta and LA are 4.67
hours, 5.13 hours, and so on
The annual snowfall in Minneapolis, MN measured in inches
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Mean and Variance of a Discrete Probability
Distribution


The mean is a typical value used to represent the central
location of the data
The mean is also referred to as the expected value

The amount of spread (or variation) in the data is
described by the variance

The standard deviation of the probability distribution is
the positive square root of the variance
6-12
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Discrete Probability Distribution Mean
Example
John Ragsdale sells new
cars for Pelican Ford. John
usually sells the most cars
on Saturday. He has
developed a probability
distribution for the
number of cars he expects
to sell on Saturday.
6-13
1. What type of distribution is this?
2. How many cars does John expect
to sell on a typical Saturday?
3. What is the variance?
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Discrete Probability Distribution Variance
Example
The computational steps for variance:
 Subtract the mean from each value of x and square
 Multiply each squared difference by its probability
 Sum the resulting products to arrive at the variance
6-14
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Example
6-15
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Example
6-16
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Binomial Probability Distribution


The binomial probability distribution is a widely occurring
discrete probability distribution.
There are four requirements of a binomial probability
distribution
1.
2.
3.
4.

There are only two possible outcomes and the outcomes are
mutually exclusive, as either a success or a failure
The number of trials is fixed and known
The probability of a success is the same for each trial
Each trial is independent of any other trial
Example

6-17
A young family has two children, both boys. The probability of the
third birth being a boy is still .50. The gender of the third child is
independent of the gender of the other two.
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Binomial Probability Experiment

Use the number of trials, n, and the probability of a
success, π, to compute binomial probability
BINOMIAL PROBABILITY EXPERIMENT
1. An outcome on each trial of an experiment is classified into one of
two mutually exclusive categories—a success or a failure.
2. The random variable is the number of successes in a fixed number
of trials.
3. The probability of success is the same for each trial.
4. The trials are independent, meaning that the outcome of one trial
does not affect the outcome of any other trial.

Note: Do not confuse the symbol π, with the
mathematical constant 3.1416
6-18
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How is a Binomial Probability Computed?
6-19
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Binomial Probability Distribution
Recently, www.creditcards.com reported that 28% of purchases at coffee shops were
made with a debit card. For 10 randomly selected purchases at the Starbucks on the
corner of 12th Street and Main, what is the probability that exactly one of the
purchases was made with a debit card? What is the probability distribution for the
random variable, number of purchases made with a debit card? What is the
probability that six or more purchases out of 10 are made with a debit card? What is
the probability that five or less purchases out of 10 are made with a debit card?
6-20
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How is a Binomial Probability Computed?
What is the
probability that no
purchases were made
with a debit card?
Π = 0.28
n = 10
x=0
What is the
probability that
exactly one was made
with a debit card?
Π = 0.28
n = 10
x=1
6-21
P(x) = nCr(π)r 1 − π n − r
P(0) = 10C0(.28)0 1 − .28 10 − 0
= (1)(1)(.0374) = .0374
P(x) = nCr(π)r 1 − π n − r
P(1) = 10C1(.28)1 1 − .28 10 − 1
= (10)(25)(.0520) = .1456
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Shortcut Formulas

Using the preceding example of debit card purchases;
n=10 and π = .28 and the shortcut formulas
μ=n∗π
μ = 10 .28 = 2.8
σ2 = nπ 1 − π
σ2 = 10 .28 1 − .28 = 2.016
6-22
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Shortcut Formulas
6-23
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Binomial Probability Tables

Tables are already constructed for use as well
In the Southwest, 5% of all cell phone calls are dropped. What is the
probability that out of six randomly selected calls, none was dropped?
Exactly one? Exactly two? Exactly three? Exactly four? Exactly five?
Exactly six out of six? See the table below for the answers.
6-24
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Cumulative Binomial Probability
Distributions
A study by the Illinois Department of Transportation concluded that 76.2% of
front seat occupants wore seat belts. That is, both occupants of the front seat
were using their seat belts. Suppose we decide to compare that information with
current usage.We select a sample of 12 vehicles.
1. What is the probability that the front seat occupants in exactly 7 of the 12
vehicles are wearing seat belts?
P(x) = nCr(π)r 1 − π
n−r
P(x=7) = 12C7(.762)7 1 − .762 12 − 7
= 792(.149171)(.000764) = .0902
2. What is the probability that at least 7 of the 12 front seat occupants are
wearing seat belts?
P(x≥7) = P(x=7) + P(x=8) + P(x=9) + P(x=10) + P(x=11) + P(x=12)
=.0902 + .1805 + .2569 + .2467 + .1436 + .0383
=.9562
6-25
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Example
6-26
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Example
6-27
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Example
6-28
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rights reserved. No reproduction or
Example
P(x) = nCx(π)x 1 − π
6-29
n−x
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Poisson Probability Distribution




This describes the number of times some event occurs
during a specified interval
The interval can be time, distance, area, or volume
Two assumptions
 The probability is proportional to the length of the
interval
 The intervals are independent
The Poisson has many applications, like describing:
 The distribution of errors in data entry
 The number of accidents on I-75 during a three-month
period
6-30
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Poisson Distribution
6-31
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Poisson Distribution
6-32
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Poisson Distribution Example
Budget Airlines is a seasonal airline that operates flights from Myrtle Beach,
South Carolina, to various cities in the northeast. Recently Budget has been
concerned about the number of lost bags. Ann Poston from the Analytics
Department was asked to study the issue. She randomly selected a sample of
500 flights and found that a total of twenty bags were lost on the sampled flights.
The mean number of bags lost, μ, is found by 20/500 = .04
The probability that no bags are lost is found using formula 6-7.
P 0 =
μxe−μ
x!
=
.040𝑒−0.04
0!
= .9608
Then calculate the probability that one or more bags is lost.
P(x≥1) =1-P 0 = 1 −
6-33
μxe−μ
x!
=1-
.040𝑒−0.04
0!
= 1- .9608 = .0392
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Poisson Probability Distribution Tables
NewYork-LA Trucking company finds the mean number of breakdowns on
the New York to Los Angeles route is 0.30. From the table, we can locate
the probability of no breakdowns on a particular run. Find the column 0.3,
then read down that column to the row labeled 0; the value is .7408. The
probability of 1 breakdown is .2222
6-34
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Example
6-35
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Example
6-36
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Example
6-37
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Exercise
6-38
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Exercise
6-39
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Chapter 6 Practice Problems
6-40
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distribution without the prior written consent of McGraw-Hill Education.
Question 5
LO6-2,3
The information below is the number of daily emergency service calls made
by the volunteer ambulance service of Walterboro, South Carolina, for the last
50 days. To explain, there were 22 days when there were two emergency calls,
and 9 days when there were three emergency calls.
Convert this information on the number of calls to a probability
distribution.
Is this an example of a discrete or continuous probability distribution?
What is the probability that 3 or more calls are made in a day?
What is the mean number of emergency calls per day?
What is the standard deviation of the number of calls made daily?
a.
b.
c.
d.
e.
6-41
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Question 5
LO6-2,3
a. Calls,x




Frequency P(x) xP(x)
(x − μ)2 P(x)
0
8
.16
0
.4624
1
10
.20
.20
.0980
2
22
.44
.88
.0396
3
9
.18
.54
.3042
4
1
.02
.08
.1058
Total
50
1.70
1.0100
b. Discrete distribution, because only certain outcomes are
possible.
c. 0.20 found by P(x = 3) + P(x = 4) = 0.18 + 0.02 = 0.20
d. μ = Σx · P(x) = 1.70
e. σ = √1.01 = 1.005
6-42
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Question 13
LO6-4
An American Society of Investors survey found 30% of
individual investors have used a discount broker. In a
random sample of nine individuals, what is the probability:
a. Exactly two of the sampled individuals have used a
discount broker?
b. Exactly four of them have used a discount broker?
c. None of them has used a discount broker?

6-43
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Question 13



LO6-4
a. .2668, found by P(2) = 9!/((9 − 2)!2!) * (.3)^2(.7)^7
b. .1715, found by P(4) = 9!/((9 − 4)!4!) * (.3)^4(.7)^5
c. .0404, found by P(0) = 9!/((9 − 0)!0!) * (.3)^0(.7)^9
6-44
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Question 21
LO6-4
In a recent study, 90% of the homes in the United States
were found to have large-screen TVs. In a sample of nine
homes, what is the probability that:
a. All nine have large-screen TVs?
b. Less than five have large-screen TVs?
c. More than five have large-screen TVs?
d. At least seven homes have large-screen TVs?

6-45
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Question 21




LO6-4
a. 0.387, found from Appendix B.1 with n of 9, π of
0.90, and x of 9
b. P(x < 5) = 0.001
c. 0.992, found by 1 − 0.008
d. 0.947, found by 1 − 0.053
6-46
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Question 27
LO6-6

Ms. Bergen is a loan officer at Coast Bank and Trust. From
her years of experience, she estimates that the probability
is .025 that an applicant will not be able to repay his or
her installment loan. Last month she made 40 loans.
a.
What is the probability that three loans will be
defaulted?
What is the probability that at least three loans will be
defaulted?
b.
6-47
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Question 27


LO6-6
a. .0613
b. .0803
6-48
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