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Chapter 8
Counting , Probability Distributions,
and Further Probability Topics
1.
2.
3.
8.5 Probability Distributions
& Expected Value
8.1, 8.2 Multiplication
Principle, Permutations,
& Combinations
8.4 Binomial Probability
8.5 Probability Distributions & Expected
.......Value
Random Variables




A random variable is a numerical description of the
outcome of an experiment.
A random variable can be classified as being either
discrete or continuous depending on the numerical
values it assumes.
A discrete random variable may assume either a
finite number of values or an infinite sequence of
values.
A continuous random variable may assume any
numerical value in an interval or collection of
intervals.
Discrete Random Variables

Discrete random variable with a finite number of
values
Let x = number of passengers on a flight to New York
where x can take on the values 0, 1, 2,…, 450.

Discrete random variable with an infinite sequence of
values
Let x = number of customers visiting Disney World
in one day where x can take on the values
0, 1, 2, . . .
We can count the customers arriving, but there is no
finite upper limit on the number that might arrive.
Discrete Probability Distributions




The probability distribution for a random variable
describes how probabilities are distributed over the
values of the random variable.
The probability distribution is defined by a
probability function, denoted by f(x), which provides
the probability for each value of the random variable.
The required conditions for a discrete probability
function are:
0 ≤ f(xi) ≤ 1
f(x1) + f(x2) + f(x3) + ∙ ∙ ∙ + f(xn) = 1
We can describe a discrete probability distribution
with a table, graph, or equation.
Example: JSL Appliances

Using past data on TV sales (below left), a tabular
representation of the probability distribution for TV
sales (below right) was developed.
Number
Units Sold (x) of Days
0
80
1
50
2
40
3
10
4
20
200
x
0
1
2
3
4
P(x)
.40
.25
.20
.05
.10
1.00
Example: JSL Appliances
Graphical Representation of the
Probability Distribution (Histogram)
Probability, f(x)

.50
.40
.30
.20
.10
0
1
2
3
4
Values of Random Variable x (TV sales)
Example: JSL Appliances
Graphical Representation of the
Probability Distribution (Histogram)
Probability, f(x)

P  x  1  P 1  P  2  P  3  P  4
 .25  .20  .05  .10  .60
.50
.40
.30
.20
.10
0
1
2
3
4
Values of Random Variable x (TV sales)
Discrete Uniform Probability
Distribution


The discrete uniform probability distribution is the
simplest example of a discrete probability
distribution given by an equation.
The discrete uniform probability function is
P(x) = 1/n
where:

n = the number of values the random
variable may assume
Note that the values of the random variable are
equally likely.
Expected Value

The expected value E(x), or average, of a random
variable is a measure of its central location.
 If a random variable x can take on n values (x1, x2,
x3, …, xn)
 with corresponding probabilities P(x1), P(x2),
P(x3),…, P(xn),
 then the expected value of the random variable is
E  x   x1P  x1   x2 P  x2   x3P  x3     xn P  xn 
or
E  x   xi P  xi 
Example: JSL Appliances

Expected Value of a Discrete Random Variable
x
0
1
2
3
4
P(x)
xP(x)
.40
.00
.25
.25
.20
.40
.05
.15
.10
.40
xP(x) = 1.20 = E(x)
The expected value (or average) of the distribution is
1.2 TVs.
8.1, 8.2 The Multiplication Principle,
Permutations, Combinations

Multiplication Principle: For experiments
involving multiple steps.

Combinations: For experiments in which r
elements are to be selected from a larger set of n
objects.
Permutations: For experiments in which r
elements are to be selected from a larger set of n
objects, in a specific order of selection.

A Counting Rule for
Multiple-Step Experiments
THE MULTIPLICATION PRINCIPLE

If an experiment consists of a sequence of k steps
in which there are n1 possible results for the first
step, n2 possible results for the second step, and
so on, then the total number of experimental
outcomes is given by (n1)(n2)...(nk).
Example: Roll a 6-sided die, and toss a coin. How many
experimental outcomes are possible?
Example: Multiplication Principle

2-Step Experiment
n1 = 6
1
2
3
4
5
n2 = 2
H
H
H
H
T
(1,H)
(1,T)
(2,H)
(2,T)
T
(3,H)
(3,T)
T
T
H
T
6
H
T
(4,H)
(4,T)
(5,H)
(5,T)
(6,H)
(6,T)
6  2 = 12
Counting Rule for Combinations
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
Another useful counting rule enables us to count the
number of experimental outcomes when r elements are
to be selected from a larger set of n objects.
The number of combinations of n elements taken r at a
time is
n
n!
n Cr  C    
 r  r !(n  r )!
n
r
FACTORIAL NOTATION
For any natural number n
n! = n(n - 1)(n - 2) . . . (3)(2)(1)
Also,
0! = 1
Combination - Example



In the Florida Lottery contestants choose 6
numbers from 1 to 53. How many possible
combinations of 6 numbers are there?
n = 53
r=6
 53 
53!
 22,957, 480
53 C6  

 6  6! 53  6 !
Using the TI-30XA
Factorial
10!  10 2nd
 x!
 3,628,800
Using the TI-30XA
Combination
53
C6  53 2nd
 nCr 
6 
Counting Rule for Permutations


A third useful counting rule enables us to count the
number of experimental outcomes when r elements are
to be selected from a larger set of n objects, in a specific
order of selection.
The number of permutations of n objects taken r at a
time is
n!
n Pr  P  n, r  
(n  r )!
Example - Permutation

If the Florida Lottery required contestants to
pick 6 numbers out of 53 in the correct order,
how many experimental outcomes would
there be?
53!
P  53,6  
 16,529,385,600
 53  6 !
Using the TI-30XA
Permutation
P  53 2nd
53 6
 nPr 
6 
 1.65293856 10
10
 1.65293856 10
 16,529,385, 600
Now You Try
1.
2.
3.
An automobile manufacturer produces 8 models, each
available in 7 different exterior colors, with 4 different
upholstery fabrics, and 5 interior colors. How many
varieties of automobiles are available?
Hamburger Hut sells hamburgers with a selection of
cheese, relish, lettuce, tomato, mustard, or ketchup.
How many different hamburgers can be ordered with
exactly 3 extras?
A horse race consists of 12 horses. What is the
probability of selecting the 1st, 2nd, and 3rd place horses
in the correct order?
8.4 Binomial Probability Distribution

Properties of a Binomial Experiment

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The experiment consists of a sequence of n identical
trials.
Two outcomes, success and failure, are possible on
each trial.
The probability of a success, denoted by p, does not
change from trial to trial.
The trials are independent.
Notation for Binomial Probability
Distributions
P(S) = p
P(F) = 1 – p = q
n
x
p
p = probability of a success
q = probability of a failure
denotes the fixed number of trials
denotes a specific number of successes
in n trials, so x can be any whole
number between 0 and n, inclusive.
denotes the probability of success in one
of the n trials.
q
denotes the probability of failure in one
of the n trials.
P(x)
denotes the probability of getting
exactly x successes among the n trials.
INT 221 – B. Potter
23
Example: Evans Electronics
Binomial Probability Distribution
 Evans is concerned about a low retention rate for
employees. On the basis of past experience,
management has seen a turnover of 10% of the
hourly employees annually. Thus, for any hourly
employee chosen at random, management estimates
a probability of 0.1 that the person will not be with
the company next year.
Choosing 3 hourly employees at random, what is
the probability that 1 of them will leave the company
this year?
Let:
p = .10, n = 3, x = 1
Example: Evans Electronics
Success = leave within 1 year; Fail = not leave
1.
Determine the number of experimental outcomes
involving 1 success in 3 trials (x = 1 success in n = 3
trials).
2.
Determine the probability of each of the above
outcomes.
Example: Evans Electronics
1.
Determine the number of experimental outcomes
involving 1 success in 3 trials (x = 1 success in n = 3
trials).
n
n!
3!

3
 
 x  x ! n  x  ! 1! 3  1!
Example: Evans Electronics
2.
Determine the probability of each of the above outcomes.
P  S   p, P  F   1  p
P  SFF  p 1  p 1  p 
  0.1 0.9  0.9    0.1 0.9   0.081
2
Probability of a particular sequence of trial outcomes =
p 1  p 
x
n x
With x successes in n trials
Binomial Probability Distribution

Binomial Probability Function
n x
(n x )
P ( x)    p (1  p )
 x
where:
f(x) = the probability of x successes in n trials (x = 1)
n = the number of trials (3)
p = the probability of success on any one trial (.10)
Binomial Probability Distribution

Binomial Probability Function
n x
(n x )
P ( x)    p (1  p )
 x
Step 1
Step 2
Example: Evans Electronics

Using the Binomial Probability Function
n x
P( x)    p (1  p)( n x )
 x
3!
P(1) 
(0.1)1 (0.9)2
1!(3  1)!
= (3)(0.081)
= .243
Example: Evans Electronics

Using a Tree Diagram
First
Worker
Second
Worker
Leaves (.1)
Leaves (.1)
Third
Worker
3
.0010
S (.9)
2
.0090
2
.0090
1
.0810
2
.0090
1
.0810
L (.1)
1
.0810
S (.9)
0
.7290
L (.1)
S (.9)
L (.1)
.081 3S(.9).243
Leaves (.1)
Stays (.9)
Probab.
L (.1)
Stays (.9)
Stays (.9)
Value
of x
Binomial Probability Distribution

Expected Value

E(x) =  = np
Evan’s Electronics:
E(x) =  = 3(.1) = .3
Binomial Probability Distribution

Seventy percent of the students applying to a university
are accepted.
1.
What is the probability that among the next 18
applicants exactly 10 will be accepted?
n  18, x  10, p  .70
Solution #1
n!
 n x
x
P  x 
p 1  p 
x ! n  x  !
n  18, x  10, p  .7
18!
1810
10
P 10  
.7 1  .7 
10!18  10 !
  43, 758.028247525.00006561  .0811
Binomial Probability Distribution

Seventy percent of the students applying to a university
are accepted.
1.
What is the probability that among the next 18
applicants exactly 10 will be accepted?
2.
Determine the expected number of acceptances if n =
18.
Solution #2
E  x     np  18 .70  12.6
Binomial Probability Distribution

Seventy percent of the students applying to a university
are accepted.
1.
What is the probability that among the next 18
applicants exactly 10 will be accepted?
2.
Determine the expected number of acceptances if n =
18.
3.
What is the probability that among the next 5 applicants
no more than 3 are accepted?
P  0  x  3  P  3  P  2  P 1  P  0
Or , 1   P  4   P  5  
Solution #3
5!
 53
3
P  3 
 0.7 1  0.7 
 .3087
3! 5  3!
P  2   .1323
P 1  .0284
P  0   .0024
P  0  x  3  P  3  P  2   P 1  P  0 
 .3087  .1323  .0284  .0024  .4718
Now You Try

1.
2.
In 2003, the percentage of children under 18 years of age
who lived with both parents was approximately 70%.
Find the probabilities that the following number of
persons selected at random from 10 children under 18
years of age in 2003 lived with both parents.
Exactly 6 (x = 6)
At most 4 (0  x  4)
End of Chapter 8