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 p.164

V. Conditional Probability p.171

VI. Bayes’ Theorem p.178

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 = ?

Factorial: N!=N(N-1)…(2)(1) n!=n(n-1)…(2)(1)0! = 1

Example: Elect 2 committee members from 3 professors. How many election results can be?

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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 = ?

Example: Elect one committee chair and one member from 3 professors. How many election results can be?

Homework: p.158 #1, #2, #3

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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/nn: the # of all possible outcomesA: Outcome A.

• Relative Frequency Method: empirical probability is a frequency obtained from a sample or historical

data. P(A) = x/nn: 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. P(Head) = ?Example: Roll a die. 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) = (Sample point i in event A)

(2) If all sample points (outcomes) in the sample space are equally likely to occur, then

P(A) = x/nn: the # of all sample points (outcomes) in sample space.x: the number of sample points for the event A.

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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: AB consists of all sample points belonging to both A and B. (p.166)

• Union of events A and B: AB consists of all sample points belonging to A OR B OR both. (p.165)

Example: p.169 #23 (for revised questions)

Find events Bc, AB, and AB.

1. Contingency Table Approach

Example: p.171 Table 4.4Given: 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(WA), P(AAc), P(WA), P(MW).

Men (M) Women (W) TotalsPromoted (A) 288 36 324Not Promoted (Ac) 672 204 876Totals 960 240 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(WA), P(AAc), P(WA), P(MW).

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 intersectionMultiplication Law (coming soon)

• Probability of union and the Addition LawAddition Law- General format: (p.166)

P(AB) = P(A) + P(B) - P(AB)- If A and B are mutually exclusive, (p.168)

P(AB) = P(A) + P(B)

Mutually exclusive P(AB) = 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.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(AB) = P(A)P(B|A) P(BA) = ?• If events A and B are independent, (p.175)

P(AB) = P(A)P(B)• Jordan is A and Jerry is A? Jordan is the first and

Jerry is the second (no tie)?

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Two Approaches:

(1) Contingency Table ApproachExample: 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 #36What probabilities are given? Make the first shot: A P(A) = .89Make the second shot: B P(B) = .89Assume events A and B are independent.Find:a. Assume events A and B are independent, P(AB)=?b. P(AB) = ?c. P((AB)c) = 1 - P(AB)

Answer:a. Assume events A and B are independent,

P(AB)=P(A)P(B)=(.89)(.89)=.7921b. P(AB) = P(A)+P(B)-P(AB)=.89+.89-.7921=.9879c. P((AB)c) = 1 - P(AB)=1-.9879 = .0121

Homework: p.175 #30, P.176 #31

VI. Bayes’ Theorem (Two-Event Case) p.181

• When to use:

Prior Probability New info Update (Posterior)for events Ai for event B for events Ai P(A1) P(B|A1) P(A1|B) P(A2) P(B|A2) P(A2|B)

Two events A1 and A2 that are mutually exclusive (P(A1A2)=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(A1A2)=0), and A1 and A2 are collectively exhaustive (P(A1)+P(A2)=1), then

• A useful formula: P(B) = ? Why?

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VI. Bayes’ Theorem (Examples)Example: p.183 #39Think: 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)=.4 P(B|A1)=.2 P(A1|B)=? P(A2)=.6 P(B|A2)=.05 P(A2|B)=?

P(B) = ?Two events A1 and A2 that are mutually exclusive (P(A1A2)=0), and A1 and A2 are collectively exhaustive (P(A1)+P(A2)=1).

Answer: c. & d. P(B) = (.4)(.2)+(.6)(.05) = .11P(A1|B)= (.4)(.2)/.11 = .7273 P(A2|B)= (.6)(.05)/.11 = .2727

VI. Bayes’ Theorem (Examples)Example: p.183 #42Think: 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(A1A2)=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

…... •A useful formula: P(B) = ? Why?

Homework: p.183 #41, p.184 #43

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