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CHAPTER 5
DISCRETE
PROBABILITY
DISTRIBUTIONS
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5.3 THE BINOMIAL PROBABILITY
DISTRIBUTION
Suppose we want to find the probability that head will turn up 6 times
when a coin is tossed 10 times. The binomial probability distribution is
best suited for this type of question. To use the binomial probability
distribution, the experiment must meet the following conditions:
1. There are n identical trials. In other words, the experiment is
repeated n times and each repetition is performed under identical
conditions. Note that each repetition of the experiment is called a trial
or a Bernoulli trial.
2. Each trial has only two possible outcomes. That is, the outcome is
either a “success” or a “failure.”
3. The probability of each outcome remains constant. The probability of
success is denoted by p and probability of failure is denoted by q.
Therefore, p + q = 1.
4. The trials are independent. In other words, the occurrence of one
trial does not affect the probability of the other.
Note:
a. Success refers to the outcome to which the question is stated. Then
failure refers to the outcome that the question does not stated.
b. Success does not mean that the outcome is favorable. Likewise,
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failure does not mean that the outcome is unfavorable.
The Binomial Probability Distribution
Example #14
Which is the following are binomial experiment? Explain why.
a. Rolling a die many times and observing the number of spots.
b. Rolling a die many times and observing whether the number
obtained is even or odd?
c. Selecting a few voters from a very large population of voters and
observing whether or not each of them favors a certain proposition in
an election when 54% of all voters are known to be in favor of this
proposition.
Solution
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The Binomial Probability Distribution
Solution
b.
Rolling a die many times and observing whether the number obtained is even
or odd
c.
Selecting a few voters from a very large population of voters and observing
whether or not each of them favors a certain proposition in an election when
54% of all voters are known to be in favor of this proposition
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The Binomial Probability Distribution and
Binomial Formula
So, knowing that an experiment is a binomial experiment, then we can
apply the Binomial probability distribution to find the probability of
exactly x successes in n trials by using the binomial formula
x nx
P ( x )  n Cx p q
where
n = total number of trials
p = probability of success
q = 1 – p = probability of failure
x = number of successes in n trials
n - x = number of failures in n trials
n
n!
n Cx 
x !(n  x )!
Cx  number of ways to obtain x success from n trials
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The Binomial Probability Distribution and
Binomial Formula
Example #15
Solution
According to a Harris Interactive
poll, 52% of American college
graduates have Facebook
accounts. Suppose that this
result is true for the current
population of American college
graduates.
a.
b.
Let x be a binomial random
variable that denotes the number
of American college graduates in
a random sample of 15 who have
Facebook accounts. What are the
possible values that x can
assume?
Find the probability that exactly 9
American college graduates in a
sample of 15 have Facebook
accounts.
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The Binomial Probability Distribution and
Binomial Formula
Example #16
According to an October 27, article in Newsweek, 65% of Americans
said they take expired medicines. Suppose this result is true of the
current population of Americans. Find the probability that the number
of Americans in a random sample of 22 who take expired medicines is
a.
exactly 17
b. none
c. exactly 9
Solution
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Using the Table of Binomial Probabilities
So, rather than calculate the probability of x success in n trials by using
the Binomial formula, we can use Table I-A of Appendix A – Tables of
Binomial Probabilities. Lets do this by solving the following problem.
Example #17
According to a March 25, 2007 Pittsburgh Post-Gazette article, 30% to
40% of U.S. taxpayers cheat on their returns. Suppose that 30% of all
current U.S. taxpayers cheat on their returns. Use the binomial
probabilities table (Table I of Appendix C) or technology to find the
probability that the number of U.S. taxpayers in a random sample of
14 who cheat on their taxes is
a.
at least 8
b. at most 3
c. 3 to 7
Solution
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Probability of Success and the Shape of the
Binomial Distribution
The shape of a binomial distribution
depends on the value of the probability of
success.
If p = 0.50, then the binomial probability
distribution is symmetric.
If p < 0.50, then binomial probability
distribution is right skewed.
If p > 0.50, then binomial probability
distribution is left skewed.
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5.4 Mean and Standard Deviation of the
Binomial Distribution
For a binomial distribution, we can still use the formulas in Sections
5.3 and 5.4 to compute its mean and standard deviation.
µ = E(x) = Σ x P(x)
   x 2P ( x )   2
However a simpler methods are:
  np
where,
  npq
n = total number of trials
p = probability of success
q = probability of failure.
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Mean and Standard Deviation of the Binomial
Distribution
Example #18
A fast food chain store conducted a taste survey before marketing a
new hamburger. The results of the survey showed that 70% of the
people who tried hamburger liked it. Encouraged by this result, the
company decided to market the new hamburger. Assume that 70% of
all people like this hamburger. On a certain day, eight customers
bought it for the first time.
a.
Let x denote the number of customers in this sample of eight who will like
the hamburger. Using the binomial probabilities table, obtain the probability
distribution of x and draw a graph of the probability distribution. Determine
the mean and standard deviation of x.
a.
Using the probability distribution of part a, find the probability that exactly
three of the eight customers will like this hamburger.
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Mean and Standard Deviation of the Binomial
Distribution
Solution
Bar Graph for the probability distribution
a. Using Table I of Appendix
0.35
A, the probability
0.3
distribution of x is,
0.25
0.2
P(x)
0
0.0001
1
0.0012
0.1
2
0.0100
0.05
3
0.0467
0
4
0.1361
5
0.2541
6
0.2965
7
0.1977
8
0.0576
P(x)
x
0.15
0
1
2
3
4
5
6
7
8
x
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FACTORIALS, COMBINATIONS, AND
PERMUTATIONS
Factorial is denoted by the symbol “!”. The factorial of a number
is calculated by multiplying all integers from the number to 1.
Formal Definition
The symbol n!, is define as the product of all the integers from n
to 1. In other words,
n! = n(n - 1)(n – 2)(n – 3) · · · 3 · 2 · 1
Also note that by definition,
0! = 1
Example #9
3!  3  2  1  6
(9  3)!  6!  6  5  4  3  2  1  720
9!  9  8  7  6  5  4  3  2  1  362,880
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Combinations
Suppose we want to select two books from 6 books, and the
following conditions applies:
1. This type of selection is without replacement.
2. The order in which selection is made is not important. That is,
it is the same selection whether we select book 1 first (1,2) or
book 2 first (2,1).
Then, the question is how many ways can we make this selection? Of
course, we can do it the old fashion way, write the selection as,
(1,2) (1,3) (1,4) (1,5) (1,6)
(2,3) (2,4) (2,5) (2,6)
(3,4) (3,5) (3,6)
(4,5) (4,6)
(5,6)
Therefore, we can select 2 books from 6 books 15 different ways.
Note:
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Each of the possible selection is called a combination.
Combinations
Definition
Combinations give the number of ways x element can be selected
from n distinct elements. The total number of combinations is given
by,
C
n
x
and is read as “the number of combinations of n elements selected x
at a time.”
The formula for the number of combinations for selecting x from n
distinct elements is,
n!
n Cx 
x !(n  x )!
Note:
n!
n!
n!

 1
n Cn 
n !(n  n)! n ! 0! n !
n C0 
n!
n!
n!


1
0!(n  0)! 0! n ! n ! 17
Combinations
Example #10
2
5!
5!
5  4  3  2 1


 10
5 C3 
3!(5  3)! 3! 2! 3  2  1  2  1
7!
7!
7  6  5  4  3  2 1


 35
7 C4 
4!(7  4)! 4! 3! 4  3  2  1  3  2  1
4
C0  1
3
C3  1
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Permutations
Permutation is similar to combination except that the order of selection
is important. For example, how many ways can we select two letters
from “A”, “B”, and “C” without replacement and if the order does not
matter?
This selection is a combination and the total number of combinations is
3!
 3 ways.
3 C2 
2!1!
These 3 ways are “AB”, “AC, and “BC”. Now suppose the order matters,
then the two letter-combination can be selected or arranged in 6 ways:
“AB”, “BA”, “AC”, “CA”, “BC” and “CB”
These 6 selections are called 6 permutations or arrangements.
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Permutations
Definition
Permutations give the total selection of x elements from n different
elements such that the order of selection matter. The notation and
formula is
n!
n
Px 
(n  x )!
Example #11
6! 6  5  4 !

 6  5  30
6 P2 
4!
4!
6! 8  7  6  5  4 !

 8  7  6  5  1,680
8 P4 
4!
4!
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Permutations
Example #12
A ski patrol unit has nine members available for duty, and two of them
are to be sent to rescue an injured skier. In how many ways can two of
these nine members be selected? Now suppose the order of selection is
important. How many arrangements are possible in this case?
Solution
n  9, x  2
4
a.
9 C2 
9! 9  8  7!

 9  4  36
2!7!
2  7!
b.
9 P2 
9! 9  8  7!

 9  8  72
7!
7!
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Permutations
Example #13
An environmental agency will randomly select 4 houses from a block
containing 25 houses for a radon check. How many total selections are
possible? How many permutations are possible?
Solution
n  25, x  4
25! 25  24  23  22  21!

 25  23  22  12,650
4!21!
4  3  2  21!
a.
25 C 4 
b.
25! 25  24  23  22  21!

 303,600
25 P4 
21!
21!
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