Probability
Chapter 2
Towards a theory of probability…
Will the apple fall?
It is a random
event.
Chapter 2B
Today’s Menu
2-2 Axioms – one more time
2-0 Complementary Rule
2-2.2 Axioms of Probability
Mutually Exclusive Events
A Generalization


A collection of events is mutually exclusive if for all
pairs i and k, Ei Ek = 
For a collection of mutually exclusive events,
P(E1 E2 . . .

Ek) = P(E1) + P(E2) + . . . + P(Ek)
We often rely on mutually exclusive events in a
sample space


They partition the space
They make many computations feasible and simple
The Complimentary Rule
Either an event will occur or it will not occur.
That is P(E) + P(Ec) = 1
Or
P(E) = 1 – P(Ec)
It follows from
E  Ec = S and E, Ec are mutually exclusive
P(E) + P(Ec) = P(S) = 1
I knew that.
The Complimentary Rule
Four members of the ENM faculty agree to go
to one of 6 Los Vegas gambling casinos. What
is the probability that at least two will end up in
the same casino?
Let E = the event, at least two meet, then
Ec = the event, none meet
P(E) = 1 - P(Ec)
P(at least two meet) = 1 – P(none meet)
Picking your Casino
Sampling with replacement
I have 6 choices
I have 6 choices
I have 6 choices
I have 6 choices
6 x 6 x 6 x6 = 64
Picking your Casino
Sampling without replacement
I have 6 choices
Picking your Casino
Sampling without replacement
I choose
Now I only have 5 choices
Picking your Casino
Sampling without replacement
I choose
I choose
and I now have 4 choices
Picking your Casino
Sampling without replacement
I choose
I choose
I choose
and I now have 3 choices
6 x 5 x 4 x 3 = P(6,4)
The Complimentary Rule
P(E) = 1 - P(Ec)
P(at least two meet) = 1 – P(none meet)
P(6,4)
6!
P(none meet) =
=
= .2778
4
4
6
2!6
P(at least 2 meet) =1-.2778 =.7222
The Birthday Problem



The Enlightened Company gives each of its employees the
day off on their birthday. What is the probability that at least
two workers in the same office will be off on their birthdays the
same day.
Assume every day of the year is equally likely.
Let n = the number of workers in an office and A = the event
at least two people have a common birthday; then Ac is the
event that none do.
P(365, n)
P(365, n)
P( A ) 
; P( A)  1 
n
365
365n
c
n
4
P(A) 0.02
P(n)
16
0.28
23
0.51
32
0.75
40
0.89
56
0.99
More Birthday Problems
How many people are needed in order to have a
better than 50% chance that two people have a
birthday within k days of each other?
within k days
# people required
0
23
1
14
2
11
3
9
4
8
5
7
7
6
2-3 The Addition Rule
In general,
P(A  B) = P(A) + P(B) – P(A  B)
S
A
Basically, we avoid
double counting the
overlap area
B
If A & B do not intersect, i.e., P(A  B) = 0
P(A  B) = P(A) + P(B)
 A & B are mutually exclusive
The Addition Rule – Example 1
Random process: draw a card at random from a deck
of 52 playing cards
Let A = the event, draw a spade:
P(A) = 13/52 =1/4
P(B) = 4/52 = 1/13
let B = the event draw an ace:
P(A  B) = 1/52
P(A  B) = P(A) + P(B) - P(A  B)
= 13/52 + 4/52 – 1/52 = 16/52 = .3077
The Addition Rule – Example 2

In studying the cause of power failures, these data
have been gathered:




20% are due to transformer damage (event A)
70% are due to line damage (event B)
2% involve both problems
What is the probability that a given power failure
involves transformer or line damage?
P(A  B) = P(A) + P(B) – P(A  B) = .20 + .70 -.02 = .88
The three-event case
S
A
B
C
A three-event example

The Pandemonium Company (TPC) produces powdered laundry
detergent (SOPE). During final inspection a box of sope may be
found to be underweight (Event A), not correctly labeled (Event B)
or not properly sealed (Event C). Quality Control has provided the
following data:
Event
Probability
A
5%
B
7%
C
12%
A and B
3%
A and C
4%
B and C
5%
A and B and C 1%
1. (easy) Underweight or
incorrectly labeled boxes are
rejected.* What fraction falls In
this category?
2. (harder) What fraction of the
boxes have no problems?
*unsealed boxes are resealed and
shipped to the customers.
A three-event example solved
Event
Probability
A
5%
B
7%
C
12%
A and B
3%
A and C
4%
B and C
5%
A and B and C 1%
1. (easy) Underweight or
incorrectly labeled boxes are
rejected. What fraction falls In
this category?
2. (harder) What fraction of the
boxes have no problems?
1. P(A  B) = P(A) + P(B) - P(A  B) = .05 + .07 - .03 = .09
2. P(Ac  Bc  Cc) = 1 - P(Ac  Bc  Cc)c =1 – P(A  B  C)
= 1 – [ P(A) + P(B) + P(C) – P(A  B) – P(A  C) – P(B  C)
+ P(A  B  C)] = 1 – [.05 + .07 + .12 - .03 - .04 - .05 + .1]
= 1 - .13 = .87
2-4 Conditional Probability


Conditional Probabilities are based upon a
reduced sample space
P(B|A) is the probability B occurs given that
A has occurred.
S
A
B
More Conditional Probability
Let Ni = the number of equally likely outcomes in event I
NT = total number of outcomes in the sample space
A
NA
B
NT
NB
NAB
P(A) = NA/NT
P(A
B) = NAB/NT
P(B|A) = P(A
B) / P(A)
= (NAB/NT)/(NA/NT)
= NAB/ NA
A Conditional Example


In a study of waters near power plants and other industrial
plants,
 5% showed signs of both chemical and thermal pollution.
 40% showed signs of chemical pollution (event A)
 30% showed signs of thermal pollution. (event B)
What is the probability that a stream showing chemical
pollution will show signs of thermal pollution?
 Desired: P(B|A)
 Given: P(A) = .4, P(B) = .3, P(A  B) = P(B  A) = .05
P(B|A) = P(B  A)/ P(A) = .05 / .40 = 1/8 = .125
Complementary
(A  B )
Law
c
S
A
B
(A  B)
P( B | A)  P( B | A)  1
c
c
P( A  B)
P
(
A

B
)
c
P( B | A) 
and P( B | A) 
P( A)
P( A)
c
P
(
A

B
)

P
(
A

B
)
c
P( B | A)  P( B | A) 
P( A)

P ( A  B)  ( A  B c ) 
P( A)

P  A  ( B  B c ) 
P( A)
P  A  S  P( A)


1
P ( A)
P ( A)
More of A Conditional Example


In a study of waters near power plants and other industrial
plants,
 5% showed signs of chemical and thermal pollution.
 40% showed signs of chemical pollution (event A)
 30% showed signs of thermal pollution. (event B)
What is the probability that a stream showing chemical
pollution will not show signs of thermal pollution?
 Desired: P(Bc |A)
P(BC|A) = 1 - P(B|A) = 1-.125 = .875
A another conditional probability
A company has 150 employees that are
categorized according to their education and work
assignment. Education
Production Office Staff Sales
totals
High School
2-yr tech school
4-yr college degree
Masters degree
totals
25
14
45
0
84
2
20
18
12
52
0
8
2
4
14
An employee is selected at random:
1.
What is the probability that the employee has a
Master’s degree?
2.
What is the probability that the employee works in
sales given that the employee has a Master’s
degree?
3.
Given that the employee works in the office, what
is probability the employee has a 2-yr technical
degree?
27
42
65
16
150
2-5 The Multiplication Rule
P(A  B) = P(A|B) P(B)
Since P(A|B) = P(B  A) / P(B)
And P(B  A) = P(A  B)
Multiplication Rule Example 1
A parts bin contains 5 defective items from among 20 items in
the bin. If 2 are selected at random, what is the probability
that both are defective?
Let Ai = the event, the ith item selected is defective. Desired
P(A1 A2)
I
Since P(A1  A2 ) = P(A2  A1 ) = P(A2|A1) P(A1)
selected
P(A1) = 5/20 and P(A2|A1) = 4/19
poorly.
Therefore
P(A1  A2 ) = P(A2|A1) P(A1) = (5/20) (4/19) = .0526
Is there an alternate approach to this problem?
An Alternate Approach
A parts bin contains 5 defective items from among 20 items in
the bin. If 2 are selected at random, what is the probability
that both are defective?
Let Ai = the event, the ith item selected is defective. Desired
P(A1 A2)
P(A1  A2 ) = P(A2|A1) P(A1) = (5/20) (4/19) = .0526
 5 15 
   10 1
2  0    

P(both defective) 

 .0526
 20 
190 
 
2 
Multiplication Rule Example 2





Following aircraft accidents, a detailed investigation is
conducted. The probability that an accident due to structural
failure is correctly identified is .9.
If 25% of all accidents are due to structural failures, find the
probability that an aircraft accident is due to structural failure
and is diagnosed as due to structural failure.
Define: A = the event, structural failure and B = the event,
diagnosed correctly
Given: P(B|A) = .9 and P(A) = .25
Required: P(A  B )= P(B|A) P(A) = (.9) (.25) = .225
Multiplication Rule Example 3
In a machine shop, there are four automatic screw machines.
Past inspection reports yields the following data:
Machine
1
2
3
4
Percent
Production
15
30
20
35
Percent
Defectives
Produced
4
3
5
2
I’m a nondefective machine
3 screw.
If a screw is randomly picked from inventory, what is the probability
that It will be from machine 3 and be defective?
Let Ei = the event screw came from machine i
D = the event, screw is defective
Given: P(E3) = .20 and P(D | E3) = .05
Required: P(E3  D) = (.05) (.20) = .01
A harder question
In a machine shop, there are four automatic screw machines. Past
inspection reports yields the following data:
Machine
1
2
3
4
Percent
Production
15
30
20
35
Percent
Defectives
Produced
4
3
5
2
I’m a nondefective machine
3 screw.
If a screw is randomly picked and found to be defective, what is the
probability that it was produced by machine 3?
Let Ei = the event screw came from machine i
D = the event, screw is defective
Given: P(Ei) and P(D | Ei)
need P(D)
Required: P(E3 |D) = P(E3  D) / P(D)
Continuing with a harder question
Let D = (E1  D)  (E2  D)  (E3 D)  (E4  D)
Then
P(D) = P(E1  D) + P(E2  D) + P(E3 D) + P(E4  D)
= P(D|E1) P(E1) + P(D|E2) P(E2) + P(D|E3) P(E3) + P(D|E4)
P(E4)
= (.04)(.15) + (.03)(.30) + (.05)(.20) + (.02)(.35) = .032
Machine
1
2
3
4
Percent
Production
15
30
20
35
Percent
Defectives
Produced
4
3
5
2
Required:
P(E3 |D) = P(E3  D) / P(D)
= .01/.032 = .3125
2-5 Multiplication and Total
Probability Rules
This is pretty
powerful stuff!
We must use this
power wisely!
Total Probability Rule (two
events)
Total Probability Rule
(multiple events)
The events Ei , i = 1,…,k, form a partition
Example



Blood type distribution in the United States is Type A 41%,
Type B 9%, Type AB 4%, and Type O 46%.
4% of the Army recruits with Type O are classified as Type A,
4% with Type B are classified as Type A, 10% with Type AB
are classified as Type A, and 88% with Type A are classified
as Type A
Define the relevant events, express the above as probabilities
of these events, and find the probability a recruit is classified
as type A.
Let A = event, blood type A
B = event, blood type B
AB = event, blood typeA B
O = event, blood type O
T = event, classified as type A
P(A) = .41, P(B) = .09, P(AB) = .04,
P(O) = .46
P(T|A) = .88, P(T|O) = .04
P(T|AB) = .10, P(T|B) = .04
P(T) = (.88)(.41) + (.04)(.46)
+ (.04)(.09) + (.10) (.04) = .3868
The Multiplication Rule
applied to the Addition Rule
P(A  B) = P(A) + P(B) – P(A  B)
= P(A) + P(B) – P(A|B) P(B)
= P(A) + P(B) – P(B|A) P(A)
Gosh, both rules
at once.
But if A and B are mutually
exclusive…
P(A  B) = P(A) + P(B) – P(A  B)
Where P(A  B) =  and
P(B|A) = P(A  B) / P(A) = 
A
B
Note that Ac and Bc are not mutually exclusive.
P(A  B) = P(A) + P(B) – P(B|A) P(A)
17th
A Two Rule Example




The 17th National Bank & Trust Company has observed that customers
who have sufficient funds in their checking accounts postdate their
checks, by mistake, once every 200 times.
Customers who write checks on insufficient funds postdate them 95%
of the time.
Seven percent of the customers have insufficient funds.
What is the probability of the bank receiving a postdated check or a
check from an account having insufficient funds?
Ye Olde Solution Box
Let A = the event, check is postdated
B = the event, customer has insufficient funds
Given: P(A|Bc) = .005, P(A|B) = .95 P(B) = .07
Required: P(A  B) = P(A) + P(B) – P(A|B) P(B)
P(A) = P(A|Bc) P(Bc) + P(A|B)P(B) = (.005)(.93) + (.95)(.07) = .07115
P(A  B) = .07115 + .07 – (.95)(.07) = .07465
One Last Example




Twenty percent of a shipment of preformatted
CD’s contain defects.
Twenty-five percent of the defective CD’s also
contain a virus.
Thirty-three percent of the non-defective disks
also contain a virus.
What is the probability of selecting at random a
non-defective disk without a virus?
One Last Example

Twenty percent of a shipment of formatted CD’s contain
defects. Twenty-five percent of the defective CD’s also
contain a virus. Thirty-three percent of the non-defective
disks also contain a virus. What is the probability of
selecting at random a non-defective disk without a virus?
Ye Olde Solution Box:
Let A = the event, CD has a defect
B = the event, CD contains a virus
Given: P(A) = .2, P(B|A) = .25, P(B|Ac) = .33
Required: P(Ac  Bc) = 1 – P(A  B) = 1 – [P(A) + P(B) – P(B|A)P(A)]
P(B) = P(B|A)P(A) + P(B|Ac )P(Ac) = (.25)(.2) + (.33)(.8) = .314
P(Ac  Bc) = 1 – [.2 + .314 – (.25)(.2)] = 1 - .464 = .536
Towards a Theory of
Probability…
This has been great!
What do we have to
look forward to next
time?
And many more examples!