Chapter 5
Discrete Probability
Distributions
1
CHAPTER
Discrete Probability Distributions
Outline
5-1 Probability Distributions
5-2 Mean, Variance, Standard Deviation, and
Expectation
5-3 The Binomial Distribution
5
CHAPTER
Discrete Probability Distributions
Objectives
1
2
3
4
5
Construct a probability distribution for a random
variable.
Find the mean, variance, standard deviation, and
expected value for a discrete random variable.
Find the exact probability for X successes in n trials
of a binomial experiment.
Find the mean, variance, and standard deviation for
the variable of a binomial distribution.
5.1 Probability Distributions

A random variable is a variable whose values
are determined by chance.

A discrete probability distribution consists of
the values a random variable can assume and
the corresponding probabilities of the values.

The sum of the probabilities of all events in a
sample space add up to 1. Each probability is
between 0 and 1, inclusively.
Bluman, Chapter 5
4
Chapter 5
Discrete Probability Distributions
Section 5-1
Example 5-1
Page #262
Bluman, Chapter 5
5
Example 5-1: Rolling a Die
Construct a probability distribution for rolling a
single die.
Bluman, Chapter 5
6
Chapter 5
Discrete Probability Distributions
Section 5-1
Example 5-2
Page #262
Bluman, Chapter 5
7
Example 5-2: Tossing Coins
Represent graphically the probability distribution
for the sample space for tossing three coins.
.
Bluman, Chapter 5
8
Exercise #19
The probabilities that a patient will have 0, 1, 2, or 3 tests
6 5 3 1
performed on entering a hospital are 15 , 15 , 15 , 15
1
and 15 respectively. Construct a probability
distribution for the data and draw a graph
for the distribution.
X
0
1
2
3
P(X )
6
15
5
15
3
15
1
15
P(X)
The probabilities that a patient will have 0, 1, 2, or 3 tests
6 5 3 1
performed on entering a hospital are 15 , 15 , 15 , 15
respectively. Construct a probability
distribution for the data and draw a graph
for the distribution.
.50
.40
.30
.20
.10
0
0
1 Tests2
3
Exercise #27
A box contains two $1 bills, three $5 bills, one $10 bill, and
three $20 bills. Construct a probability distribution for
the data.
X
P( X )
$1
2
9
$5
1
3
$10
1
9
$20
1
3
Exercise #29
Construct a probability distribution for drawing a card
from a deck of 40 cards consisting of 10 cards numbered
1, 10 cards numbered 2, 15 cards numbered 3, and 5 cards
numbered 4.
X
1
2
3
4
P(X)
1
4
1
4
3
8
1
8
Example 2
5-2 Mean, Variance, Standard
Deviation, and Expectation
MEAN:    X  P  X 
VARIANCE:
    X  P  X    
2
2
Bluman, Chapter 5
2
14
Mean, Variance, Standard
Deviation, and Expectation
Rounding Rule
The mean, variance, and standard deviation
should be rounded to one more decimal place
than the outcome X.
When fractions are used, they should be reduced
to lowest terms.
Bluman, Chapter 5
15
Chapter 5
Discrete Probability Distributions
Section 5-2
Example 5-5
Page #268
Bluman, Chapter 5
16
Example 5-5: Rolling a Die
Find the mean of the number of spots that appear
when a die is tossed.
.
   X  P X 
 1 16  2  16  3  16  4  16  5  16  6  16

21
6
 3.5
Bluman, Chapter 5
17
Chapter 5
Discrete Probability Distributions
Section 5-2
Example 5-8
Page #269
Bluman, Chapter 5
18
Example 5-8: Trips of 5 Nights or More
The probability distribution shown represents the
number of trips of five nights or more that
American adults take per year. (That is, 6% do
not take any trips lasting five nights or more, 70%
take one trip lasting five nights or more per year,
etc.) Find the mean.
.
Bluman, Chapter 5
19
Example 5-8: Trips of 5 Nights or More
   X  P X 
 0  0.06   1 0.70   2  0.20 
 3  0.03  4  0.01
 1.2
Bluman, Chapter 5
20
Chapter 5
Discrete Probability Distributions
Section 5-2
Example 5-9
Page #270
Bluman, Chapter 5
21
Example 5-9: Rolling a Die
Compute the variance and standard deviation
for the probability distribution in Example 5–5.
.
 2    X 2  P  X     2
 2  12  16  22  16  32  16  42  16
 5   6    3.5 
2
 2  2.9 ,
1
6
2
1
6
2
  1.7
Bluman, Chapter 5
22
Chapter 5
Discrete Probability Distributions
Section 5-2
Example 5-11
Page #271
Bluman, Chapter 5
23
Example 5-11: On Hold for Talk Radio
A talk radio station has four telephone lines. If the
host is unable to talk (i.e., during a commercial) or
is talking to a person, the other callers are placed
on hold. When all lines are in use, others who are
trying to call in get a busy signal. The probability
that 0, 1, 2, 3, or 4 people will get through is
shown in the distribution. Find the variance and
standard deviation for the distribution.
Bluman, Chapter 5
24
Example 5-11: On Hold for Talk Radio
  0  0.18   1 0.34   2  0.23
 3  0.21  4  0.04   1.6
  0  0.18   1  0.34   2  0.23
2
2
2
2
 3  0.21  4  0.04   1.6 
2
 2  1.2 ,
2
2
  1.1
Bluman, Chapter 5
25
Example 5-11: On Hold for Talk Radio
A talk radio station has four telephone lines. If the
host is unable to talk (i.e., during a commercial) or
is talking to a person, the other callers are placed
on hold. When all lines are in use, others who are
trying to call in get a busy signal.
Should the station have considered getting more
phone lines installed?
Bluman, Chapter 5
26
Example 5-11: On Hold for Talk Radio
No, the four phone lines should be sufficient.
The mean number of people calling at any one
time is 1.6.
Since the standard deviation is 1.1, most callers
would be accommodated by having four phone
lines because µ + 2 would be
1.6 + 2(1.1) = 1.6 + 2.2 = 3.8.
Very few callers would get a busy signal since at
least 75% of the callers would either get through
or be put on hold. (See Chebyshev’s theorem in
Section 3–2.)
Bluman, Chapter 5
27
Expectation

The expected value, or expectation, of
a discrete random variable of a probability
distribution is the theoretical average of
the variable.

The expected value is, by definition, the
mean of the probability distribution.
EX    X PX
Bluman, Chapter 5
28
Chapter 5
Discrete Probability Distributions
Section 5-2
Example 5-13
Page #273
Bluman, Chapter 5
29
Example 5-13: Special Die
A special six-sided die is made in which 3 sides
have 6 spots, 2 sides have 4 spots, and 1 side
has 1 spot.
If the die is rolled, find the expected value of the
number of spots that will occur.
Bluman, Chapter 5
30
Exercise #3
A bank vice president feels that each savings account
customer has, on average, three credit cards. The
following distribution represents the number of credit
cards people own. Find the mean, variance,
and standard deviation. Is the vice
president correct?
Number of
cards X
0
1
2
3
4
Probability P(X) 0.18 0.44 0.27 0.08 0.03
 = 0.92 = 0.96 or 1
 =  X  P(X )
= 0(0.18) + 1(0.44) + 2(0.27)
No, on average each person
+ 3(0.08) + 4(0.03)
has
= 1.34 or 1.3
about one credit card.
 2 =  X 2  P(X ) –  2
= [02 (0.18) + 12 (0.44) + 22 (0.27)
+ 32 (0.08) + 4 2 (0.03)] –1.34 2
= 0.92 or 0.9
A bank vice president feels that each savings account
customer has, on average, three credit cards. The
following distribution represents the number of credit
cards people own. Find the mean, variance,
and standard deviation. Is the vice
president correct? = 1.34
= 2.72
X
P(X)
0
1
2
3
4
0.18
0.44
0.27
0.08
0.03
X • P(X) X 2 • P(X)
0
0.44
0.54
0.24
0.12
0
0.44
1.08
0.72
0.48
 =1.34
2.72
Exercise #15
A lottery offers one $1000 prize, one $500 prize, and five
$100 prizes. One thousand tickets are sold at $3 each.
Find the expectation if a person buys one ticket.
 1 
 1   5   993 
E ( X ) = 997 
–3
 + 497 
+ 97 
= – $1.00


1000
1000



  1000   1000 
Alternate solution:
E(X ) =  X  P(X )
 1 
 1 
= $1000 
 + $500 

 1000 
 1000 
 5 
+ $100 
 – $3.00 = – $1.00
 1000 
Exercise #17
For a daily lottery, a person selects a three-digit number.
If the person plays for $1, she can win $500.
Find the expectation.
E ( X ) =  X  P( X )
 1 
= $500 
 – $1.00= – $0.50
 1000 
Alternate solution :
 1 
 999 
E ( X ) = $499 
–1 
 = – $0.50
 1000 
 1000 
In the same daily lottery, if a person boxes a number, she
will win $80. Find the expectation if the number 123 is
played for $1 and boxed. (When a number is “boxed”, it
can win when the digits occur in any order.)
There are six different possibilities when a
number with all different digits is boxed,
(3 • 2 • 1 = 6); hence:
$80.00 
6
– $1.00 = $0.48 – $1.00 = – $0.52
1000
Alternate solution :
 6   994  = – $0.52
E ( X ) = 79 
 – 1

1000
1000

 

5-3 The Binomial Distribution

Many types of probability problems have
only two possible outcomes or they can
be reduced to two outcomes.

Examples include: when a coin is tossed
it can land on heads or tails, when a baby
is born it is either a boy or girl, etc.
Bluman, Chapter 5
37
The Binomial Distribution
The binomial experiment is a probability
experiment that satisfies these requirements:
1. Each trial can have only two possible
outcomes—success or failure.
2. There must be a fixed number of trials.
3. The outcomes of each trial must be
independent of each other.
4. The probability of success must remain the
same for each trial.
Bluman, Chapter 5
38
Notation for the Binomial Distribution
P(S)
The symbol for the probability of success
P(F)
The symbol for the probability of failure
p
The numerical probability of success
q
The numerical probability of failure
P(S) = p and P(F) = 1 – p = q
n
The number of trials
X
The number of successes
Note that X = 0, 1, 2, 3, ... , n
Bluman, Chapter 5
39
The Binomial Distribution
In a binomial experiment, the probability of
exactly X successes in n trials is
n!
X
n X
P X  
 p q
 n  X ! X !
or
P X  
n
Cx
number of possible
desired outcomes
Bluman, Chapter 5
 p q
X
n X
probability of a
desired outcome
40
Chapter 5
Discrete Probability Distributions
Section 5-3
Example 5-16
Page #280
Bluman, Chapter 5
41
Example 5-16: Survey on Doctor Visits
A survey found that one out of five Americans say
he or she has visited a doctor in any given month.
If 10 people are selected at random, find the
probability that exactly 3 will have visited a doctor
last month.
n!
P X  
 p X  q n X
 n  X ! X !
n  10,"one out of five"  p  15 , X  3
10!  1 
P  3 
 
7!3!  5 
3
7
4
    0.201
5
Bluman, Chapter 5
42
Chapter 5
Discrete Probability Distributions
Section 5-3
Example 5-17
Page #281
Bluman, Chapter 5
43
Example 5-17: Survey on Employment
A survey from Teenage Research Unlimited
(Northbrook, Illinois) found that 30% of teenage
consumers receive their spending money from
part-time jobs. If 5 teenagers are selected at
random, find the probability that at least 3 of them
will have part-time jobs.
n  5, p  0.30,"at least 3"  X  3, 4,5
5!
3
2
P  3 
  0.30    0.70   0.132 P  X  3  0.132
2!3!
0.028
5!
4
1
P  4 
  0.30    0.70   0.028
0.002
1!4!
5!
5
0
 0.162
P  5 
  0.30    0.70   0.002
0!5!
Bluman, Chapter 5
44
Exercise #11
R. H. Bruskin Associates Market Research found that 40%
of Americans do not think that having a college education
is important to succeed in the business world. If a random
sample of five Americans is selected, find
these possibilities.
a. Exactly two people will agree.
b. At most three people will agree.
c. At least two people will agree.
d. Fewer than three people will agree.
a. Exactly two people will agree
n = 5, p = 0.40
5!
 (0.4) 2 (1– 0.4)5–2
P(2 people) =
(5–2)!2!
5!
 (0.4) 2 (0.6)3
=
3!2!
= 0.346
b. At most three people will agree.
P(X ) = 0.078 + 0.259 + 0.346 + 0.230
= 0.913
c. At least two people will agree.
X = 2, 3, 4, or 5 people
P(X ) = 0.346 + 0.230 + 0.077 + 0.01
= 0.663
d. Fewer than three people will agree.
X = 0, 1, or 2 people
P(X ) = 0.078 + 0.259 + 0.346
= 0.683
Exercise #19
A survey found that 21% of Americans watch fireworks
on television on July 4. Find the mean, variance, and
standard deviation of the number of individuals who
watch fireworks on television on July 4 if
a random sample of 1000 Americans
is selected.
n = 1000, p = 0.21
 = n • p = 1000(0.21) = 210
 2 = n • p • q = 1000(0.21)(0.79)
= 165.9
 = 165.9 = 12.9
Exercise #25
In the past year, 13% of businesses have eliminated jobs.
If five businesses are selected at random, find the
probability that at least three have eliminated jobs during
the last year. n = 5, p = 0.13, X = 3,4,5
3
2
P(X) = 5! (0.13) (0.87) +
2!3!
5! (0.13) 4 (0.87)1 +
1!4!
5! (0.13) 5 (0.87) 0
0!5!
= 0.018
Chapter 5
Discrete Probability Distributions
Section 5-3
Example 5-18
Page #281
Bluman, Chapter 5
50
Example 5-18: Tossing Coins
A coin is tossed 3 times. Find the probability of
getting exactly two heads, using Table B.
n  3, p  12  0.5, X  2  P  2   0.375
Bluman, Chapter 5
51
The Binomial Distribution
The mean, variance, and standard deviation
of a variable that has the binomial distribution
can be found by using the following formulas.
Mean:   np
Variance:   npq
2
Standard Deviation:   npq
Bluman, Chapter 5
52
Chapter 5
Discrete Probability Distributions
Section 5-3
Example 5-23
Page #284
Bluman, Chapter 5
53
Example 5-23: Likelihood of Twins
The Statistical Bulletin published by Metropolitan
Life Insurance Co. reported that 2% of all American
births result in twins. If a random sample of 8000
births is taken, find the mean, variance, and
standard deviation of the number of births that
would result in twins.
  np  8000  0.02   160
 2  npq  8000  0.02  0.98  156.8  157
  npq  8000  0.02  0.98  12.5  13
Bluman, Chapter 5
54