Block 3 Discrete Systems
Lesson 11 –Discrete Probability
Models
The world is an uncertain
place
Random Process


Random – happens by chance, uncertain,
non-deterministic, stochastic
Random Process – a process having an
observable outcome which cannot be
predicted with certainty

one of several outcomes will occur at random
Uses of Probability



Measures uncertainty
Foundation of inferential statistics
Basis for decision models under uncertainty
A manager observing
an uncertainty outcome
Two Approaches to modeling
probability

Sample Space and Random Events


Uses sets and set theory
Random Variables and Probability
Distributions


Discrete case – algebraic
Continuous case - calculus
Sample Spaces and Random
Events



Let S = the set of all possible outcomes
(events) from a random process. Then S is
called the sample space.
Let E = a subset of S. Then E is called a
random event.
Basic problem: Given a random process and
the sample space S, what is the probability of
the event E occurring - P(E).
Example Sample Space &
Random Events





Let S = the set of all outcomes from observing the
number of demands on a given day for a particular
product. S = {0, 1, 2,…n}
Let E1 = the random event, there is one demand.
Then E1 = {1}
Let E2 = the random event of no more than 3
demands. Then E2 = {0, 1, 2, 3}
Let E3 = the random event, there are at least 4
demands. Then E3 = {4, 5, …, n} or E3 = E2c
S x S = the set of outcomes from observing two days
of demands = {(0,0), (1,0), (0,1), …}
What is a probability – P(E)?
Let P(E) =Probability of the event E occurring, then



0 ≤ P(E) ≤ 1
If P(E) = 0, then event will not occur (impossible event)
If P(E) = 1, then the event will occur, i.e. a certain event
So the closer P(E) is to 1,
then the more likely it is
that the event E will
occur?
The Sample Space
The collection of all possible outcomes (events) relative
to a random process is called the sample space, S
where
S = {E1, E2 ... Ek} and P(S) = 1
sampling space
How are probabilities
determined?

Elementary or basic events
1.
2.
3.

Empirical or relative frequency
A priori or equally-likely using counting methods
Subjectively –personal judgment or belief
Compound events formed from unions,
intersections, and complements of basic
events

Laws of probability
Example - Relative Frequency
(empirical)

A coin is tossed 2,000 times and heads appear 1,243
times.


The company’s Web site has been down 5 days out
of the last month (30 days).


P(H) = 1243/2000 = .6215
P(site down) = 5/30 = .16667
2 out of every 20 units coming off the production line
must be sent back for rework.

P(rework) = 2/20 = .1
Will you die this year?
Probability Male Death
0.024
0.019
0.014
0.009
0.004
-0.001
0
10
20
30
40
Age
50
60
70
age
0
5
10
16
17
20
25
30
35
40
45
50
55
60
65
70
75
80
85
90
Male
0.007644
0.000202
0.00011
0.00081
0.000964
0.00129
0.001379
0.001389
0.00177
0.002589
0.003891
0.005643
0.008106
0.012405
0.019102
0.029824
0.046499
0.073269
0.120186
0.192615
Female
0.006275
0.000152
0.000113
0.000375
0.000423
0.000456
0.000499
0.000628
0.000953
0.001514
0.002264
0.003227
0.004884
0.007732
0.012199
0.019312
0.030582
0.050396
0.086443
0.147616
Example - A priori (equally-likely
outcomes)

A pair of fair dice are tossed.
S x S = {(1, 1),(1, 2),(1, 3),(1, 4),(1, 5),(1, 6), (2, 1),(2, 2),(2,
3),(2, 4),(2, 5),(2, 6),(3, 1),(3, 2),(3, 3),(3, 4),(3, 5),(3, 6),(4,
1),(4, 2),(4, 3),(4, 4),(4, 5),(4, 6),(5, 1),(5, 2),(5, 3),(5, 4),(5,
5),(5, 6),(6, 1),(6, 2),(6, 3),(6, 4),(6, 5),(6, 6)}
P(a seven) = 6/36 = .1667

A supply bin contains 144 bolts to be used by the
manufacturing cell in the assembly of an automotive
door panel. The supplier of the bolts has indicated
that the shipment contains 7 defective bolts.
Let E = the event, a defective bolt is selected
P(E) = 7/144 = .04861.
A priori (equally-likely outcomes)
versus Relative Frequency (empirical)
A prior (knowable independently of experience):
n( A)
P( A) 
n( S )
n(A) = the number of ways in which event A can occur
n(S) = total number of outcomes from the random process
Relative frequency:
n( E )
P ( E )  Lim
n 
n
where
n(E) = number of times event E occurs in n trials
Examples – Subjective Probability

8 out of 10 “leading” economists believe that the gross
national product (GNP) will grow by at least 3% this year.



P(GNP  .03) = 8/10 = .8
Bigg Bosse, the CEO for a major corporation, consults his
marketing staff. Together they make the assessment that
there is a 50-50 chance that sales will increase next year.
The House of Congress majority leader, after consulting
with his staff, determines that there is only 25 percent
chance that an important tax bill will be passed.
Computing Probabilities for
Compound Events
Finding the probability of the union,
intersection, and complements of events
Mutually Exclusive Events
A
B
P(Ac ) = 1 - P(A)
P( A  B) = P(A) + P(B) if A and B are mutually
exclusive
P(A  B) = P() = 0
if A and B are mutually exclusive
Note that A’ and B’ are not mutually exclusive.
More Mutually Exclusive
Events
Random process: draw a card at random from an
ordinary deck of 52 playing cards
Let A = the event, an ace is drawn
Let B = the event, a king is drawn
P(A) = 1/13 and P(B) = 1/13
Then P( A  B) = P(A) + P(B) = 1/13 + 1/13 = 2/13 It’s not
P(A  B) = 0
me!
P(A’) = 1 – 1/13 = 12/13; P(B’) = 12/13
P(A  B)’ = P( A’  B’) = ?
the event is not an ace or not king
The Addition Formula
A
B
A B
AB
P(A  B) = P(A) + P(B) - P(A  B)
More of the Addition Formula
Random process: draw a card at random from an ordinary
deck of 52 playing cards
Let A = the event, draw a spade
P(A) = 13/52 =1/4
let B = the event draw an ace
P(B) = 4/52 = 1/13
P(A  B) = 1/52
P(A  B) = P(A) + P(B) - P(A  B)
= 1/4 + 1/13 – 1/52 = 13/52 + 4/52 – 1/52 = 16/52 = .3077
12 1
3
36
The Multiplication rule
Independent Events
Two events, A and B are independent if the
P(A) is not affected by the event B having
occurred (and vice-versa).
If A and B are independent,
then
P(A  B) = P(A) P(B)
Note that A’ and B’ are independent
if A and B are independent.
An Independence
event
Proof by Example
Let E1 = the event, a three or four is rolled on the toss of a
single fair die, P(E1) = 2/6
E2 = the event, a head is tossed from a fair coin, P(E2)
= 1/2
then D x C = {(1,H), (1,T), (2,H), (2,T), (3,H), (3,T), (4,H),
(4,T), (5,H), (5,T), (6,H), (6,T)}; P(E1  E2) = 2/12
Mult. rule: P(E1  E2) = P(E1) P(E2) = (2/6) (1/2) = 2/12
Another Example
Let A = the event, prototype A fails heat stress test
B = the event prototype B fails a vibration test
Given P(A) = .1 ; P(B) = .3
Find P(A  B) = ? and P(A  B) = ?
Assuming independence:
P(A  B) = P(A) P(B) = (.1)(.3) = .03
P(A  B) = P(A) + P(B) - P(A  B) = .1 + .3 - .03 = .37
P(A  B)’ = P(A’  B’) = P(A’)P(B’) = (.9)(.7) = .63
Glee Laundry Detergent
Each box of powered laundry detergent coming off of
the final assembly line is subject to an automatic
weighing to insure that the weight of the contents falls
within specification. Each box is then visually
inspected by a quality assurance technician to insure
that is it properly sealed.
I have been
rejected.
Glee
More Glee



If three percent of the boxes fall outside the weight
specifications and five percent are not properly sealed,
what is the probability that a box will be rejected after final
assembly?
Let A = the event, a box does not meet the weight
specification; P(A) = .03
Let B = the event, a box is not properly sealed; P(B) = .05
P(A  B) = P(A) + P(B) – P(A)P(B)
= .03 + .05 - .0015 = .0785
A Reliability Problem
An assembly is composed of 3 components as shown
below. If A is the event, component A does not fail, B is
the event, component B does not fail, and C is the event,
component C does not fail, find the reliability of the
assembly where P(A) = .8, P(B) = .9, and P(C) = .8.
Assume independence among the components.
A
B
C
P(S) = P[ (A  B)  C] = P(A  B) + P(C) – P(A  B  C)
= P(A) P(B) + P(C) – P(A)P(B)P(C) = (.8)(.9) + .8 – (.8)(.9)(.8)
= .72 + .8 - .576 = .944
Next – random variables and
their probability distributions!
Variables that are random; what
will they think of next?
Discrete Random Variables



A random variable (RV) is a variable which takes on
numerical values in accordance with some probability
distribution.
Random variables may be either continuous (taking on
real numbers) or discrete (usually taking on non-negative
integer values).
The probability distribution which assigns probabilities to
each value of a discrete random variable can be
described in terms of a probability mass function (PMF),
p(x) in the discrete case.
Random Variables - Examples




Y = a discrete random variable, the number of
machines breaking down each shift
X = a discrete random variable, the monthly demand
for a replacement part
Z = a discrete random variable, the number of
hurricanes striking the Gulf Coast each year
Xi = a discrete random variable, the number of
products sold in month i
The PMF
The Probability Mass Function (PMF), p(x), is defined
as
p(x) = Pr(X = x}
and has two properties:
1. p ( x)  0
2.
 p ( x)  1
all x
By convention, capital letters represent the random variable while the
corresponding small letters denote particular values the random
variable may assume.
A Probability Distribution


Let X = a RV, the number of customer
complaints received each day
The Probability Mass Function (PMF) is
.1
.3

p( x)  
.5
.1
if x  0
if x  1
if x  2
if x  3
4
 p ( x)  1
x 1
The CDF
The cumulative distribution function (CDF), F(x) is
x
defined where
F ( x)  Pr( X  x)   p(i )
i 0
Example:
.1
.3

p( x)  
.5
.1
if x  0
if x  1
if x  2
if x  3
.1 if x  0
.4 if x  1

F ( x)  
.9 if x  2
1.0 if x  3
Rolling the dice
X = the outcome from rolling a pair of dice
number of ways
2
(1,1)
1
3
(1,2) , (2,1)
2
4
(1,3) , (2,2) , (3,1)
3
5
(1,4) , (2,3) , (3,2) , (4,1)
4
6
(1,5) , (2,4) , (3,3) , (4,2) , (5,1)
5
7
(1,6) , (2,5) , (3,4) , (4,3) , (5,2) , (6,1)
6
8
(2,6) , (3,5) , (4,4) , (5,3) , (6,2)
5
9
(3,6) , (4,5) , (5,4) , (6,3)
4
10
(4,6) , (5,5) , (6,4)
3
11
(5,6) , (6,5)
2
12
(6,6)
1
Total
36
An Example
Let X = a random variable, the sum resulting from
the toss of two fair dice;
X = 2, 3, …, 12
(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)
(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)
(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)
(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)
(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)
(6,1) (6,2) (6,3) (6,4) (6,5) (6,6)
x
p(x)
F(x)
2
1/36
1/36
3
2/36
3/36
4
3/36
6/36
=S
5
6
7
8
9
10
11
12
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
10/36 15/36 21/36 26/36 30/36 33/36 35/36 36/36
Probability Histogram for the
Random Variable X
Expected Value or Mean
The expected value of a random variable (or
equivalently the mean of the probability distribution)
is defined as

E[ X ]     xp( x)
x 0
Don’t you get it? The expected value
is just a weighted average of the
values that the random variable
takes on where the probabilities are
the weights.
Example – Expected Value
I expected this to
have a little value!
.1
.3

p( x)  
.5
.1
if x  0
if x  1
if x  2
if x  3
E[X] = 0 (.1) + 1 (.3) + 2 (.5) + 3 (.1)
= 1.6
Example –Expected Value
12
E[ X ]   xp ( x) 
Dice example:
x 2
= 2(1/36) + 3(2/36) + 4(3/36) + 5(4/36) + 6(5/36) + 7(6/36)
+ 8(5/36) + 9(4/36) + 10(3/36) + 11(2/36) + 12(1/36)
= (1/36) (2 + 6 + 12 + 20 + 30 + 42 + 40 + 36 + 30 + 22 + 12)
= (252/36) = 7
x
p(x)
F(x)
2
1/36
1/36
3
2/36
3/36
4
3/36
6/36
5
6
7
8
9
10
11
12
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
10/36 15/36 21/36 26/36 30/36 33/36 35/36 36/36
Yet Another Discrete
Distribution
Let X = a RV, the number of customers per day
x
p ( x) 
; x  1, 2,..., 20
210
x  1 
 1   x( x  1)
F ( x)   2 
  ( x  1) 
 
2   210 
420
 210  
F(20) = (20)(21) / 420 = 1
n
recall the
arithmetic series:
n
Sn  a   ( j  1)d  [2a  (n  1)d ]
2
j 1
More of yet another discrete
distribution
Let X = a RV, the number of customers per day
x
; x  1, 2,..., 20
210
x
x( x  1)
F ( x) 
 2  ( x  1)(1) 
2  210 
420
p( x) 
n  n  1 2n  1
i 

6
i 1
n
2

20  21 41
 2870;
6
2870 / 210  13.67
Pr{X = 15} = p(15) = 15/210 = .0714
Pr{X  15} = F(15) = (15)(16)/420 = .5714
Pr{10 < X  15} = F(15) – F(10)
= .5714 – (10)(11)/420 = .5714 - .2619 = .3095
20
20
x
1 20 2
2
E[ X ]   xp ( x)   x

x  13

210 210 x 1
3
x 1
x 1
A most enjoyable
lesson.
So ends our discussion on
Discrete Probabilities
Coming soon to a classroom near
you – discrete optimization
models