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Probability cont’d…
1
Objectives





To construct and understand a contingency/crosstabulation table
To understand that compound events involve the
occurrence of more than one event
To describe and compute a conditional probability
To understand and be able to utilize the complement
rule
To compute the probabilities of compound events
using the addition and the multiplication rules
2
Bivariate Data


The values of two different variables
that are obtained from the same
population element
Each of the two variables may be either
qualitative or quantitative
3
Cross-tabulation or Contigency table

When bivariate data result from two
qualitative (attribute or categorical)
variables, the data are often arranged
on a cross-tabulation or contingency
table.
4
ExampleK


Thirty students from RTU-CEIT were randomly
identified and classified according to two
variables: gender (M/F) and major (CE, ECE,
EE) as shown in the table (next slide). These 30
bivariate data can be summarized on a 2x3
cross-tabulation table, where the two rows
represent the two genders, male and female,
and the three columns represent the three
major categories
of CE, ECE and EE.
5
ExampleK cont’d…
Genders and Majors of 30 college students
Name
Gender
Major
Name
Gender
Major
Name
Gender
Major
Adrian M
CE
Fer
M
EE
Mac
M
ECE
Abeil
ECE
Flan
M
CE
Myr
F
ECE
Bernie M
CE
Heidi
F
CE
Orly
M
EE
Bheng F
CE
Hune
M
EE
Pam
F
CE
Ben
M
EE
Jane
F
EE
Pul
M
EE
Brian
M
ECE
Kim
M
ECE
Ray
M
ECE
Chen
F
CE
Kler
M
CE
Sher
F
CE
Carl
M
EE
Lois
M
ECE
Son
F
EE
Cres
F
ECE
Lint
F
CE
Tate
M
ECE
Ellis
F
ECE
Lope
M
EE
Yate
M
CE
F
6
Cross-tabulation table of Gender
and Major (frequencies)
Major
Gender
CE
ECE
EE
Row Total
M
5
6
7
18
F
6
4
2
12
Column
Total
11
10
9
30
7

The resulting 2x3 cross-tabulation
(contingency) table shows the
frequency for each cross-category of
the two variables along with the row
and column totals, called marginal
totals or marginals. The total of the
marginal totals is the grand total and is
equal to n, the sample size.
8
Cross-tabulation table
of Gender and
Major (relative frequencies; % of grand total)
Major
Gender
CE
ECE
EE
Row Total
M
17%
20%
23%
60%
F
20%
13%
7%
40%
Column
Total
37%
33%
30%
100%
9
Another example of a
Age/Rank
Contingency table
Full
Professor
R1
Associate
Professor
R2
Assistant
Professor
R3
Instructor
R4
Total
A1
20-29
0
3
57
6
66
A2
30-39
52
170
163
17
402
A3
40-49
156
125
61
6
348
A4
50-59
145
68
36
4
253
A5
60-69
75
15
3
0
93
Total
428
381
320
33
1162
10
Rules of Probability


Often, one wants to know the
probability of a compound event and
the only data available are the
probabilities of the related simple
events.
Compound events

Combinations of more than one simple
event
11
Finding the

Complementary events



Probability of “Not A”
Complement of an event A,
A:
The set of all sample points (simple
events) in the sample space that do not
belong to the event A
A read as “A complement”
12
Examples of complementary events
Event A
Complement of event A, A
success
failure
selected voter is a Filipino
selected voter is not a Filipino
no heads on 10 tosses
at least one head
13
Complement Rule




Probability of A complement
= one – probability of A
Or
P(A ) = 1 – P(A)
14
Example


Two dice are rolled. What is the
probability that the sum is at least 3
(that is, 3, 4, 5, …, 12)?
P(sum is at least 3) = P( A ) = 1 – P(A)


P(sum of 2) = P(A) = 1/36
Therefore (“less than 3” and “at least 3” are complementary events)

P(sum is at least 3) = P(A )

= 1 – 1/36 = 35/36
15
Finding the

Probability of “A or B”
An hourly wage earner wants to
estimate the chances of “receiving a
promotion or getting a pay raise.” The
worker would be happy with either
outcome.
16
Addition RuleNMEE






Let A and B be two events in a sample
space, S
Probability of A or B
= probability of A + probability of B
- probability of A and B
Or
P(A or B) = P(A) + P(B) – P(A and B)
17
Addition RuleMEE



Let A and B be two events in a sample
space, S
Probability of A or B
= probability of A + probability of B


Or
P(A or B) = P(A) + P(B)

because P(A and B) = 0
18
Example
From the data (see next slide), find the
probability that a faculty selected is an
associate professor or in his 40’s.
 Solution:
 Let R2 = event that the faculty selected
is an associate professor

A3 = event that the faculty selected
is in his 40’s
19
Joint probability distribution for the cross
classification of faculty by age and rank

Age/Rank
Full
Professor
R1
Associate
Professor
R2
Assistant
Professor
R3
Instructor
R4
Total
A1
20-29
0
3
57
6
66
A2
30-39
52
170
163
17
402
A3
40-49
156
125
61
6
348
A4
50-59
145
68
36
4
253
A5
60-69
75
15
3
0
93
Total
428
381
320
33
1162
20
Solution

cont’d…
R2 and A3 = events that the faculty
selected is an associate
professor in his 40’s
381 348 125
P( R2 or A3 ) 


1162 1162 1162


P(R2 or A3) = 0.329 + 0.299 – 0.108
= 0.520
21
Another solution



Refer to the contigency table, page 13
R2 = 381
A3 = 348
f
P( R2 or A 3 ) 
n
381  348  125

1,162
 0.52
22
ExampleM

The table (next slide) is a contingency
table that shows 500 TV sets classified
by screen size and whether each set is
colored or not. What is the probability
that a set selected at random will be
either a colored set or have a screen
size of less than 21 inches.
23
TV sets classified by screen size
and whether colored or not
Kind
Screen Size
21 inches or
over
Under 21 inches
Total
Colored
175
125
300
Black & white
50
150
200
Total
225
275
500
24
Solution

Events of interest:






A: a colored TV is selected
B: a set with screen size of less that 21
inches is selected
P(A) = 300/500 = 0.60
P(B) = 275/500 = 0.55
P(A&B) = 125/500 = 0.25
P(A or B) = 0.60 + 0.55 – 0.25 = 0.90
25
Finding the


Probability of “A and B”
Suppose a criminal justice professor wants his
class to determine the likeliness of the event
“a driver is ticketed for a speeding violation
and the driver had previously attended a
defensive driving class.”
The students are confident they can find the
probabilities of “a driver being ticketed for
speeding” and “a driver has attended a
defensive driving class separately.
26
Conditional Probability of Events


the relative frequency with which an
event can be expected to occur under
the condition that additional preexisting
information is known about some other
event.
P(A│B) symbolyzes

The probability of event A occurring under the
condition that event B is known to already exist
27
Ways to say or express the
conditional probability, P(A│B)



The “probability of A, given B.”
The “probability of A, knowing B.”
The “probability of A happening,
knowing B has already occurred.”
28
ExampleK

The data from a National Election exit
poll of 13,660 voters in 250 precincts
across the country on May, 2004, are
shown in the table (see next slide).
29
Example
Gender
cont’d…
Percentage
of Voters
Percent for Percent for Percent for
candidate A candidate B Others
Men
46
55
44
1
Women
54
48
51
1
18-29
17
45
54
1
30-44
29
53
46
1
45-59
30
51
48
1
60 and older
24
54
46
0
Age
30
ExampleK

cont’d…
One person is to be selected at random
from the sample of 13,600 voters.
Using the table, find the answers to the
following probability questions (see next slide).
31
ExampleK




cont’d…
1. What is the probability the person selected
is a man?
2. What is the probability the person selected
is from age 18 to 29?
3. Knowing the voter selected was a woman,
what is the probability she voted for
candidate B?
4. What is the probability the person selected
voted for candidate A if the voter was 60 or
older?
32
Solution






1. P(voter selected is a man) = 0.46
2. P(voter selected is of age 18 to 29)
= 0.17
3. P(candidate B│woman) = 0.51
4. P(candidate A│60 and older) = 0.54
Note:
The first two are simple probabilities, whereas the last two
are conditional probabilities.
33
ExampleK

The data from a nationwide exit poll of
1000 voters in 25 precincts across the
country on May, 2004, are shown in the
table (see next slide).
34
ExampleK cont’d…
Education
No. for
candidate A
No. for
candidate B
No. for
Others
No. of
Voters
No high school
19
20
1
40
High school
graduate
114
103
3
220
Some college
172
147
1
320
College graduate
135
119
6
260
Postgraduate
70
88
2
160
Total
510
477
13
1000
35
ExampleK cont’d…


One person is to be selected at random
from the preceding sample of 1000
voters.
Using the table, find the answer to the
following probability questions (next
slide):
36
ExampleK cont’d…


1. Knowing the voter selected was a
high school graduate, what is the
probability the person voted for
candidate B?
2. Knowing the voter selected had some
college education, what is the
probability the person voted for
candidate A?
37
ExampleK cont’d…


3. Knowing the selected person voted
for candidate B, what is the probability
the voter has a postgraduate
education?
4. Knowing the selected person voted
for candidate A, what is the probability
the voter does not have a high school
education?
38
Solution




1. P(candidate B│high school graduate)
= 103/220 = 0.46818 = 0.47
2. P(candidate A│some college)
= 172/320 = 0.5375 = 0.54
3. P(post graduate│candidate B)
= 88/477 = 0.1844 = 0.18
4. P(no high school│candidate A)
= 19/510 = 0.0372 = 0.04
39
Note

When finding a conditional probability, some
outcomes from the list of possible outcomes
will be eliminated as possibilities as soon as
the condition is known. Consider question 4.
As soon as the conditional stated “knowing
the selected person voted for candidate A,”
the 477 who voted for candidate B and the
13 voting for Others were eliminated, leaving
the 510 possible outcomes.
40
Independent Events


Two events are independent if the occurrence
(or nonoccurrence) of one gives us no
information about the likeliness of occurrence
for the other.
Or


The probability of A remains unchanged after we
know that B has happened (or has not happened)
P(A) = P(A│B) = P(A│not B)
41
Dependent Events

Events that are not independent. That
is, the occurrence of one event does
have an effect on the probability for
occurrence of the other event.
42
OR

Two events A and B are said to be
independent if and only if either
P(A│B) = P(A)
P(B│A) = P(B)

Otherwise



the events are said to be dependent
43
Multiplication RuleIE




Let A and B be two events defined in
sample space, S
Probability of A and B
= Probability of A x probability of B
When two events are involved, either event can be identified as A, with
the other identified as B.

P(A and B) = P(A).P(B)
44
Multiplication RuleDE




Let A and B be two events defined in
sample space, S
Probability of A and B
= Probability of A x probability of B,
knowing A
When two events are involved, either event can be identified as A, with
the other identified as B.
P(A and B) = P(A).P(B│A)
Or P(B and A) = P(B).P(A│B)

45
Conditional probability of A,
given that B has occurred
P(A & B)
P( A B) 
if P(B)  0
P(B)
46
Conditional probability of B,
given that A has occurred
P( A & B)
P( B A) 
if P(A)  0
P( A)
47
ExampleM


Using the data and events A and B as
defined in the table on page 18, compute
P(A│B) and P(B│A).
Solution:
P( A & B) 0.25
P( A B) 

 0.45
P( B)
0.55
P( A & B) 0.25
P( B A) 

 0.42
P( A)
0.60
48
ExampleM

There are 10 balls in a container, 3 of which
are red and dotted, 1 is red and striped, 2 are
gray and dotted and 4 are gray and striped.
Calculate the following probabilities: (a)
probability of red given dotted, (b) probability
of red given striped, © probability of gray
given dotted, (d) probability of gray given
striped, and (e) probability of red and dotted
(joint probability)
49
Solution

a. Probability of red given dotted
P(red & dotted )
P( R D)  P (red dotted ) 
P(dotted )
3
3
10

  0.60
5 5
10
50
Solution cont’d…

b. Probability of red given striped
P(red & striped )
P( R S )  P(red striped ) 
P ( striped )
1
1
10

  0.20
5 5
10
51
Solution cont’d…

c. Probability of gray given dotted
P( gray & dotted )
P(G D)  P( gray dotted ) 
P(dotted )
2
2
10

  0.40
5 5
10
52
Solution cont’d…

d. Probability of gray given striped
P ( gray & striped )
P(G S )  P ( gray striped ) 
P( striped )
4
4
10

  0.80
5 5
10
53
Solution cont’d…



e. Probability of red and dotted (joint
probability)
P(R and D) = P(red and dotted)
= P(R│D) P(D) = 0.6 (5/10) = 0.30
54
ExampleB


Two cards are drawn from a deck of 52
cards. Calculate the probability that the
draw includes an ace and a ten.
Solution:

Events of interest
A: Draw an ace and a ten
B: Draw the ace on the first draw and the ten on the
second
C: Draw the ten on the first and the ace on the
second
55
Solution







B = B1 and B2
C = C1 and C2
where
B1: Draw an ace on the first draw
B2: Draw a ten on the second draw
C1: Draw a ten on the first draw
C2: Draw an ace on the second draw
56
Solution cont’d…

Applying the Multiplicative Rule:
P ( B1 and B2 )  P ( B1 ) P ( B2 B1 )
4 4
  
52  51 
4 4
P (C1 and C 2 )   
52  51 
57
Solution cont’d…


Applying the Additive Rule:
P(A) = P(B) + P©
 4  4   4  4 
P( A)        
 52  51   52  51 
8

663
58
ExampleM

Two cards are drawn from a deck of 52
cards. Calculate the probability that the
draw includes a Jack and a Queen.
59
SolutionM

Consider the events of interest (see
table below):
Event
First draw
Second draw
B
Jack: P(J) = 4/52
Queen: P(Q) = 4/51
C
Queen: P(Q) = 4/52
Jack: P(J) = 4/51
60
Solution



a. Draw a Jack and a Queen. Find
P(J&Q) and P(Q&J).
b. Draw the Jack on the first draw and
a Queen on the second draw. Find P(J)
then P(Q).
c. Draw a Queen on the first draw and
a Jack on the second draw. Find P(Q)
then P(J).
61
Solution cont’d…

In event B:



P(B) = P(J&Q) = P(J) P(Q)
Since J & Q are independent
P(B) =P(J&Q) = (4/52)(4/51)
= 4/663
62
Solution cont’d…

In event C:



P© = P(Q&J) = P(Q) P(J)
P© = (4/52)(4/51) = 4/663
P(A) = P(B) + P©
= 4/664 + 4/663
= 8/663
= 0.012
63