Download Examples for Chapter 4

Chapter 4 Introduction to Probability
I. Basic Definitions
p.150
II. Identify Sample Space with Counting Rules p.151
III. Probability of Outcome
p.155
IV. Relationship and Probabilities of Events
V. Conditional Probability
VI. Bayes’ Theorem
p.178
p.171
p.164
I. Basic Definitions
p.150
• Experiment: a process that generates well-defined
outcomes (how many, what, mutually exclusive). (p.150)
Example: Toss a coin, roll a die, select a part for
inspection.
• Sample Space (U for Universe): all possible outcomes
for an experiment.
(p.150)
Example: Toss a coin: {Head, Tail}
• Sample Point: a particular outcome in the sample
space.
(p.150)
Example: Toss a coin: two sample points.
II. Identify Sample Space with Counting Rules
1. Counting Rule for Multiple-Step Experiment p.151
It is a sequence of k independent sub-experiments
(steps). Given the number of outcomes for each step
(n1, n2, …, nk), the number of outcomes for overall
experiment = ?
(n1)( n2) …( nk)
Example: Toss coin twice.
Example: Three sales calls (Yes, No).
II. Identify Sample Space with Counting Rules
2. Combination
p.154
The experiment is to select n objects from a set of N
different objects. Each combination is an outcome.
The number of outcomes = ?
N!
C 
n!( N  n)!
N
n
Factorial: N!=N(N-1)…(2)(1)
0! = 1
n!=n(n-1)…(2)(1)
Example: Elect 2 committee members from 3
professors. How many election results can be?
II. Identify Sample Space with Counting Rules
3. Permutation
p.154
The experiment is to select n objects from a set of N
different objects where the order of selection is
important. Each permutation is an outcome. The
number of outcomes = ?
N!
P 
( N  n)!
N
n
Example: Elect one committee chair and one member
from 3 professors. How many election results can be?
Homework: p.158 #1, #2, #3
III. Probability
Definition: A measure of “chance”, 0  P  1, P(U) = 1.
1. Probability of an outcome
p.155
• Classical Method: outcomes are equally likely to
occur.
P(A) = 1/n
n: the # of all possible outcomes
A: Outcome A.
• Relative Frequency Method: empirical probability is
a frequency obtained from a sample or historical
data.
P(A) = x/n
n: sample size, the # of elements.
x: Outcome A occurred x times.
• Subjective Method
III. Probability
1. Probability of an outcome (Examples)
(1) “Equally likely"
Example: Toss a coin.
Example: Roll a die.
P(Head) = ?
P(1) = ?
(2) Relative frequency
Example: A sample of 40 students and 5 students got
“A”.
(1) P(A) = ? (What is the probability that a
student will get an A in this class?)
(2) What is the probability that a student will
not have an A?
Homework: p.160 #13
III. Probability
2. Probability of an event
p.160
Event: a collection of sample points (outcomes). (p.160)
As previous definition, all outcomes are mutually
exclusive. For Event A,
(p.161)
(1) Generally, P(A) =  Pi
(Sample point i in event A)
i
(2) If all sample points (outcomes) in the sample space
are equally likely to occur, then
P(A) = x/n
n: the # of all sample points (outcomes) in sample
space.
x: the number of sample points for the event A.
III. Probability (continued)
2. Probability of an event (Examples)
Example: Roll a die. What is the probability that the
experiment ends with 3 points or less?
Example: p.164 #20
Homework: p.162 #14, p.163 #18.
IV. Relationship of Events and Their Probabilities
1. Contingency table for a sample with two variables
2. Relationship of events - Given probabilities, find
other probabilities
(1) Relationships between events
• Complement
• Intersection
• Union
(2) Rules
• Probability of complement
• Addition law for “AND”
• Conditional probability and multiplication law for
“OR”
Relationships between events
• Complement of event A: Ac consists all sample
points that are not in event A.
(p.164)
• Intersection of events A AND B: AB consists of all
sample points belonging to both A and B. (p.166)
• Union of events A and B: AB consists of all sample
points belonging to A OR B OR both.
(p.165)
Example: p.169 #23
(for revised questions)
Find events Bc, AB, and AB.
1. Contingency Table Approach
Example: p.171 Table 4.4
Given: A sample of 1200 police officers.
Two characteristics: Gender, Promotion.
(1) Use characteristics to define events.
(2) Find probabilities for events: P(M), P(Mc),
P(WA), P(AAc), P(WA), P(MW).
Promoted
(A)
Not Promoted (Ac)
Totals
Men (M)
288
672
960
Women (W)
36
204
240
Totals
324
876
1200
Summary: contingency table for empirical probability
(1) Use characteristics to define events in contingency
table:
Simple events: events defined by one characteristic.
Based on one characteristic, all elements are
assigned to mutually exclusive events.
Joint events: events defined by two characteristics.
(2) Find probabilities for events: P(M), P(Mc),
P(WA), P(AAc), P(WA), P(MW).
Simple probability (Marginal Probability) and Joint
Probability
Homework: p.176 #33 a, b
2. Relationship of probabilities
• Probability of complement (p.165)
P(Ac) = 1 - P(A)
• Probability of intersection
Multiplication Law (coming soon)
• Probability of union and the Addition Law
Addition Law
- General format: (p.166)
P(AB) = P(A) + P(B) - P(AB)
- If A and B are mutually exclusive,
(p.168)
P(AB) = P(A) + P(B)
Mutually exclusive  P(AB) = 0
Example: p.169 #22
Example: p.170 #26
Homework: p.169 #23, p.170 #28
V. Conditional Probability
p.171
1. Conditional Probability
• Condition provides more information - “Given”, “If”,
“of”, “among”.
• Condition may lead to a restricted sample space.
• Condition versus Intersection: restricted sample
space and one event; two events.
Example: Take one student from my class at random.
(1) If the student is a junior, what is the probability for
an A-student?
(2) What is the probability of a student being a junior
and an A-student?
2. Multiplication Law for Intersection (AND)
V. Conditional Probability
(Formulas)
1. Conditional Probability
P( A  B)
• P( A | B) 
P( B)
(p.173)
P(B|A) = ?
• Independent events if P(A|B) = P(A)
(p.174)
- A card is King and a card is King given Club?
2. Multiplication Law for Intersection (AND)
(p.174)
• P(AB) = P(A)P(B|A)
P(BA) = ?
• If events A and B are independent,
(p.175)
P(AB) = P(A)P(B)
• Jordan is A and Jerry is A? Jordan is the first and
Jerry is the second (no tie)?
Two Approaches:
(1) Contingency Table Approach
Example: p.171 Table 4.4
Given: a sample in contingency table.
Find:
(1) Probability that an officer is a man. (.8)
(2) Probability that an officer is a man who got
promotion. (.24)
(3) Probability that an officer is promoted given that
the officer is a man. (.3)
(4) Probability that an officer is a man given that
the officer is not promoted.
(.7671)
(5) Are events “man” and “promotion” independent?
Homework: p.176 #33
(2) Rules: given probabilities, find other probabilities.
Example: p.177 #36
What probabilities are given?
Make the first shot: A
P(A) = .89
Make the second shot: B P(B) = .89
Assume events A and B are independent.
Find:
a. Assume events A and B are independent, P(AB)=?
b. P(AB) = ?
c. P((AB)c) = 1 - P(AB)
Answer:
a. Assume events A and B are independent,
P(AB)=P(A)P(B)=(.89)(.89)=.7921
b. P(AB) = P(A)+P(B)-P(AB)=.89+.89-.7921=.9879
c. P((AB)c) = 1 - P(AB)=1-.9879 = .0121
Homework: p.175 #30, P.176 #31
VI. Bayes’ Theorem (Two-Event Case) p.181
• When to use:
Prior Probability
for events Ai
P(A1)
P(A2)
New info
for event B
P(B|A1)
P(B|A2)
Update (Posterior)
for events Ai
P(A1|B)
P(A2|B)
Two events A1 and A2 that are mutually exclusive
(P(A1A2)=0), and A1 and A2 are collectively exhaustive
(P(A1)+P(A2)=1).
We often have two variables to define events on one sample
space. One variable defines two events A1 and A2. Another
variable defined event B.
VI. Bayes’ Theorem (Two-Event Case Formulas) p.181
• Bayes’ Theorem:
If A1 and A2 are mutually exclusive (P(A1A2)=0), and A1
and A2 are collectively exhaustive (P(A1)+P(A2)=1), then
P( A1) P( B | A1)
P( A1 | B) 
P( A1) P( B | A1)  P( A2) P( B | A2)
P( A2) P( B | A2)
P( A2 | B) 
P( A1) P( B | A1)  P( A2) P( B | A2)
• A useful formula: P(B) = ? Why?
P( B)  P( A1) P( B | A1)  P( A2) P( B | A2)
VI. Bayes’ Theorem (Examples)
Example: p.183 #39
Think: Why Bayes’s theorem can work for this problem?
Prior Probability
for events Ai
P(A1)=.4
P(A2)=.6
New info
for event B
P(B|A1)=.2
P(B|A2)=.05
Update (Posterior)
for events Ai
P(A1|B)=?
P(A2|B)=?
P(B) = ?
Two events A1 and A2 that are mutually exclusive
(P(A1A2)=0), and A1 and A2 are collectively exhaustive
(P(A1)+P(A2)=1).
Answer: c. & d.
P(B) = (.4)(.2)+(.6)(.05) = .11
P(A1|B)= (.4)(.2)/.11 = .7273 P(A2|B)= (.6)(.05)/.11 = .2727
VI. Bayes’ Theorem (Examples)
Example: p.183 #42
Think: Why Bayes’s theorem can work for this problem?
Prior Probability
New info
Update (Posterior)
for events Ai
for event B
for events Ai
P(A1)=.05 (default)
P(B|A1)=1
P(A1|B)=?
P(A2)=? (not default) P(B|A2)=.20 P(A2|B)=?
B: miss payments.
P(B) =?
We have two variables to define events on one sample
space. One variable defines two events A1 and A2 that are
mutually exclusive (P(A1A2)=0), and A1 and A2 are
collectively exhaustive (P(A1)+P(A2)=1). Another variable
defined event B.
Answer: a. P(?) = .2083.
Follow-up: Probability that a
cardholder will not miss any payment? P(Bc)
VI. Bayes’ Theorem (General Case Formulas) p.181
• Bayes’ Theorem:
If A1, A2, …, An are mutually exclusive, and A1, A2, …, An
are collectively exhaustive, then
P( A1) P( B | A1)
P( A1 | B) 
P( A1) P( B | A1)  ...  P( An) P( B | An)
P( A2) P( B | A2)
P( A2 | B) 
P( A1) P( B…...
| A1)  ...  P( An) P( B | An)
•A useful formula: P(B) = ? Why?
P( B)  P( A1) P( B | A1)  ...  P( An) P( B | An)
Homework: p.183 #41, p.184 #43