Statistics for Business and Economics Chapter 3 Probability

Contents Events, Sample Spaces, and Probability Unions and Intersections Complementary Events The Additive Rule and Mutually Exclusive Events Conditional Probability The Multiplicative Rule and Independent Events Bayes’s Rule As a result of this class, you will be able to ...

Learning Objectives Develop probability as a measure of uncertainty Introduce basic rules for finding probabilities Use probability as a measure of reliability for an inference Provide an advanced rule for finding probabilities As a result of this class, you will be able to ...

DEFINITION OF PROBABILITY PROBABILITY IS THE STUDY OF RANDOM OR NONDETERMINISTIC EXPERIMENTS. IT MEASURES THE NATURE OF UNCERTAINTY.

SOME KEY PROBABILISTIC TERMINOLOGIES RANDOM EXPERIMENT AN EXPERIMENT IN WHICH ALL OUTCOMES (RESULTS) ARE KNOWN BUT SPECIFIC OBSERVATIONS CANNOT BE KNOWN IN ADVANCE. EXAMPLES: TOSS A COIN ROLL A DIE

RANDOM VARIABLE ILLUSTRATION THE OUTCOME OF AN EXPERIMENT IS CALLED A RANDOM VARIABLE. IT CAN ALSO BE DEFINED AS A QUANTITY THAT CAN TAKE ON DIFFERENT VALUES. ILLUSTRATION

Probability As A Long – run Relative Frequency What’s the probability of getting a head on the toss of a single fair coin? Use a scale from 0 (no way) to 1 (sure thing). So toss a coin twice. Do it! Did you get one head & one tail? What’s it all mean?

Many Repetitions!* Total Heads Number of Tosses Number of Tosses 1.00 0.75 0.50 0.25 0.00 25 50 75 100 125 Number of Tosses

Events, Sample Spaces, and Probability 3.1 Events, Sample Spaces, and Probability :1, 1, 3

Experiments & Sample Spaces Process of observation that leads to a single outcome that cannot be predicted with certainty Sample point Most basic outcome of an experiment Sample space (S) Collection of all sample points Sample Space Depends on Experimenter!

Visualizing Sample Space 1. Listing S = {Head, Tail} 2. Venn Diagram H T S

Sample Space Examples Experiment Sample Space Toss a Coin, Note Face {Head, Tail} Toss 2 Coins, Note Faces {HH, HT, TH, TT} Select 1 Card, Note Kind {2♥, 2♠, ..., A♦} (52) Select 1 Card, Note Color {Red, Black} Play a Football Game {Win, Lose, Tie} Inspect a Part, Note Quality {Defective, Good} Observe Gender {Male, Female}

Events Specific collection of sample points Subset of a sample space (S) Simple Event Contains only one sample point Compound Event Contains two or more sample points

S Venn Diagram Experiment: Toss 2 Coins. Note Faces. Sample Space S = {HH, HT, TH, TT} Compound Event: At least one Tail Other compound events could be formed: Tail on the second toss {HT, TT} At least 1 Head {HH, HT, TH} TH HT Outcome HH TT S

Event Examples Experiment: Toss 2 Coins. Note Faces. Sample Space: HH, HT, TH, TT Event Outcomes in Event Typically, the last event (Heads on Both) is called a simple event. 1 Head & 1 Tail HT, TH Head on 1st Coin HH, HT At Least 1 Head HH, HT, TH Heads on Both HH

Probabilities

What is Probability? 1. Numerical measure of the likelihood that event will occur P(Event) P(A) Prob(A) 2. Lies between 0 & 1 3. Sum of sample points is 1 1 Certain .5 Impossible

Probability Rules for Sample Points Let pi = P(A) represent the probability of sample point i, that is, event A 1. All sample point or event probabilities must lie between 0 and 1 (i.e., 0 ≤ pi ≤ 1 or 0 ≤ P(A) ≤ 1 ). 2. The probabilities of all sample points within a sample space must sum to 1 (i.e.,  pi = 1).

Example A COIN IS WEIGHTED SO THAT HEADS IS TWICE AS LIKELY TO APPEAR AS TAILS. FIND P(T) AND P(H). 2. THREE STUDENTS A, B AND C ARE IN A SWIMMING RACE. A AND B HAVE THE SAME PROBABILITY OF WINNING AND EACH IS TWICE AS LIKELY TO WIN AS C. FIND THE PROBABILITY THAT B OR C WINS. Solution

Equally Likely Probability P(Event) = X / T X = Number of outcomes in the event T = Total number of sample points in Sample Space Each of T sample points is equally likely — P(sample point) = 1/T © 1984-1994 T/Maker Co.

Probability of An Event The probability of an event A, denoted by P(A), is obtained by adding the probabilities of the individual outcomes in the event. When all the possible outcomes are equally likely,

Example A PAIR OF FAIR DICE IS TOSSED. FIND THE PROBABILITY THAT THE MAXIMUM OF THE TWO NUMBERS IS GREATER THAN 4. ONE CARD IS SELECTED AT RANDOM FROM 50 CARDS NUMBERED 1 TO 50. FIND THE PROBABILITY THAT THE NUMBER ON THE CARD IS (I) DIVISIBLE BY 5, (II) PRIME, (III) ENDS IN THE DIGIT 2. Solution

Steps for Calculating Probability 1. Define the experiment; describe the process used to make an observation and the type of observation that will be recorded 2. List the sample points 3. Assign probabilities to the sample points 4. Determine the collection of sample points contained in the event of interest 5. Sum the sample points probabilities to get the event probability

Combinations Rule A sample of n elements is to be drawn from a set of N elements. The number of different samples possible is denoted by and is equal to where the factorial symbol (!) means that For example, 0! is defined to be 1.

Example Compute

Unions and Intersections 3.2 Unions and Intersections :1, 1, 3

Unions & Intersections Outcomes in either events A or B or both ‘OR’ statement Denoted by  symbol (i.e., A  B) 2. Intersection Outcomes in both events A and B ‘AND’ statement Denoted by  symbol (i.e., A  B)

ILLUSTRATING THE INTERSECTION AND UNION OF EVENTS THROUGH VENN DIAGRAMS

Example Solution Toss a fair die once. Define the following events: E: event outcomes; O: odd primes; F: event primes; M: multiple of 3; Find E or O; O or F; O and M; F and E; Solution

Event Union: Venn Diagram Experiment: Draw 1 Card. Note Kind, Color & Suit. Ace Black Event Black: 2, 2,..., A Sample Space: 2, 2, 2, ..., A S Event Ace: A, A, A, A Event Ace  Black: A, ..., A, 2, ..., K

Event Union: Two–Way Table Experiment: Draw 1 Card. Note Kind, Color & Suit. Color Simple Event Ace: A, A, A, A Sample Space (S): 2, 2, 2, ..., A Type Red Black Total Ace Ace & Ace & Ace Red Black Non-Ace Non & Non & Non- Red Black Ace Total Red Black S Event Ace  Black: A,..., A, 2, ..., K Simple Event Black: 2, ..., A

Event Intersection: Venn Diagram Experiment: Draw 1 Card. Note Kind, Color & Suit. Ace Black Event Black: 2,...,A Sample Space: 2, 2, 2, ..., A S Event Ace: A, A, A, A Event Ace  Black: A, A

Event Intersection: Two–Way Table Experiment: Draw 1 Card. Note Kind, Color & Suit. Color Simple Event Ace: A, A, A, A Sample Space (S): 2, 2, 2, ..., A Type Red Black Total Ace Ace & Ace & Ace Red Black Non-Ace Non & Non & Non- Red Black Ace Event Ace  Black: A, A Total Red Black S Simple Event Black: 2, ..., A

Compound Event Probability 1. Numerical measure of likelihood that compound event will occur 2. Can often use two–way table Two variables only

Event Probability Using Two–Way Table Total 1 2 A P(A  B ) P(A  B ) P(A ) 1 1 1 1 2 1 A P(A  B ) P(A  B ) P(A ) 2 2 1 2 2 2 Total P(B ) P(B ) 1 1 2 Joint Probability Marginal (Simple) Probability

Two–Way Table: Example Experiment: Draw 1 Card. Note Kind & Color. Color Type Red Black Total Ace 2/52 2/52 4/52 Non-Ace 24/52 24/52 48/52 P(Ace) Total 26/52 26/52 52/52 P(Red) P(Ace  Red)

Thinking Challenge What’s the Probability? P(A) = P(D) = P(C  B) = P(A  D) = P(B  D) = What’s the Probability? Event C D Total A 4 2 6 B 1 3 5 10 Let students solve first. Allow about 20 minutes for this.

Solution* The Probabilities Are: P(A) = 6/10 P(D) = 5/10 P(C  B) = 1/10 P(A  D) = 9/10 P(B  D) = 3/10 Event C D Total A 4 2 6 B 1 3 5 10

3.3 Complementary Events :1, 1, 3

S Complementary Events AC A Complement of Event A The event that A does not occur All events not in A Denote complement of A by AC S AC A

S Rule of Complements AC A The sum of the probabilities of complementary events equals 1: P(A) + P(AC) = 1 S AC A

Complement of Event: Example Experiment: Draw 1 Card. Note Color. Black Sample Space: 2, 2, 2, ..., A S Event Black: 2, 2, ..., A Complement of Event Black, BlackC: 2, 2, ..., A, A

Complement of An Event Example What is the probability of getting at least 1 head in the experiment of tossing a fair coin 3 times? The Sample Space Is HHH HHT HTH HTT THH THT TTH TTT Solution

The Additive Rule and Mutually Exclusive Events 3.4 The Additive Rule and Mutually Exclusive Events :1, 1, 3

Mutually Exclusive Events Events do not occur simultaneously A  B does not contain any sample points 

Mutually Exclusive Events Example Experiment: Draw 1 Card. Note Kind & Suit.  Outcomes in Event Heart: 2, 3, 4 , ..., A Sample Space: 2, 2, 2, ..., A Mutually Exclusive What is the intersection of mutually exclusive events? The null set.  S Event Spade: 2, 3, 4, ..., A Events  and are Mutually Exclusive

Addition And General Addition Rules Used to get compound probabilities for union of events P(A OR B) = P(A  B) = P(A) + P(B) – P(A  B) For mutually exclusive events: P(A OR B) = P(A  B) = P(A) + P(B)

Venn Diagram Illustration of The General Addition Rule

General Addition Rule: Example A class contains 10 men and 20 women of which half the men and half the women have brown eyes. Find the probability that a person chosen at random is a man or has brown eyes. Solution

General Addition Rule Example Experiment: Draw 1 Card. Note Kind & Color. Color Type Red Black Total Ace 2 4 Non-Ace 24 48 26 52 Try other examples using this table. P(Ace  Black) = P(Ace) + P(Black) – P(Ace  Black) 4 26 2 28 = + – = 52 52 52 52

Addition Rule: Example A class contains 5 freshmen, 4 sophomores, 8 juniors and 3 seniors. A student is chosen at random to represent the class. Find the probability that the student chosen is a sophomore or a junior. Solution

Thinking Challenge Using the additive rule, what is the probability? P(A  D) = P(B  C) = Using the additive rule, what is the probability? Event C D Total A 4 2 6 B 1 3 5 10 Let students solve first. Allow about 10 minutes for this.

Solution* Using the additive rule, the probabilities are: 1. 6 5 2 9 P(A  D) = P(A) + P(D) – P(A  D) 6 5 2 9 = + – = 10 10 10 10 2. P(B  C) = P(B) + P(C) – P(B  C) 4 5 1 8 = + – = 10 10 10 10

Conditional Probability 3.5 Conditional Probability :1, 1, 3

Conditional Probability 1. Event probability given that another event occurred 2. Revise original sample space to account for new information Eliminates certain outcomes 3. P(A | B) = P(A and B) = P(A  B) P(B) P(B)

Venn Diagram Illustration

Example In a certain college, 25% of the students failed mathematics, 15% of the students failed chemistry, and 10% of the students failed both mathematics and chemistry. A student is selected at random. (a) If he failed chemistry, what is the probability that he failed mathematics? (b) If he failed mathematics, what is the probability that he failed chemistry? Solution

Example You draw a card at random from a standard deck of 52 cards. Find each of the following probabilities: The card is a heart, given that it is red. The card is red, given that it is a heart. The card is an ace, given that it is red. The card is a queen, given that it is a face card. Solution

Conditional Probability Using Two–Way Table Experiment: Draw 1 Card. Note Kind & Color. Color Type Red Black Total Ace 2 4 Non-Ace 24 48 26 52 Revised Sample Space Try other examples using this table.

Thinking Challenge Using the table then the formula, what’s the probability? P(A|D) = P(C|B) = Event C D Total A 4 2 6 B 1 3 5 10 Let students solve first. Allow about 20 minutes for this.

Solution* Using the formula, the probabilities are:

The Multiplicative Rule and Independent Events 3.6 The Multiplicative Rule and Independent Events :1, 1, 3

Multiplicative Rule 1. Used to get compound probabilities for intersection of events 2. P(A and B) = P(A  B) = P(A)  P(B|A) = P(B)  P(A|B) 3. For Independent Events: P(A and B) = P(A  B) = P(A)  P(B)

Example (Multiplicative rule) (1) A man is dealt 5 cards one after the other from an ordinary deck of 52 cards. What is the probability that they are all spades? (2) The students in a class are selected at random, one after the other, for an examination. Find the probability that the boys and girls in the class alternate if (i) the class consists of 4 boys and 3 girls. (ii) the class consists of 3 boys and 3 girls. Solution

Multiplicative Rule: Example Experiment: Draw 1 Card. Note Kind & Color. Color Type Red Black Total Ace 2 2 4 Try other examples using this table. Non-Ace 24 24 48 Total 26 26 52 P(Ace  Black) = P(Ace)∙P(Black | Ace)

Example(Independence) The baseball World Series ends when a team wins 4 games. Suppose that sports analysts consider one team a bit stronger, with a 55% chance to win any individual game. Find the probability that the underdog will win the series by winning the first 4 games. Solution

Example(Independence)   A slot machine has three wheels that spin independently. Each has 10 equally likely symbols: 4 bars, 3 lemons, 2 cherries, and a bell. If you play, what is the probability that 1. you get three lemons? 2. you get no fruit symbols? 3. you get no bells? 4. you get at least one bar? Solution

Statistical Independence 1. Event occurrence does not affect probability of another event Toss 1 coin twice 2. Causality not implied 3. Tests for independence P(A | B) = P(A) P(B | A) = P(B) P(A  B) = P(A)  P(B)

Thinking Challenge Using the multiplicative rule, what’s the probability? Event C D Total A 4 2 6 B 1 3 5 10 P(C  B) = P(B  D) = P(A  B) = Let students solve first. Allow about 10 minutes for this.

Solution* Using the multiplicative rule, the probabilities are:

Tree Diagram Experiment: Select 2 pens from 20 pens: 14 blue & 6 red. Don’t replace. Dependent! R P(R  R)=(6/20)(5/19) =3/38 5/19 R 6/20 14/19 B P(R  B)=(6/20)(14/19) =21/95 R 6/19 P(B  R)=(14/20)(6/19) =21/95 14/20 B 13/19 B P(B  B)=(14/20)(13/19) =91/190

3.7 Bayes’s Rule :1, 1, 3

Bayes’s Rule Given k mutually exclusive and exhaustive events B1, B1, . . . Bk , such that P(B1) + P(B2) + … + P(Bk) = 1, and an observed event A, then

Bayes Rule Via Tree Diagram (Example)   In a certain college, 4% of the men and 1% of the women are taller than 6 feet. Furthermore, 60% of the students are women. Now if a student is selected at random and is taller than 6 feet, what is the probability that the student is a woman? Solution

Bayes Rule Via Tree Diagram (Example) Suppose a test for TB has the following properties: If the person tested has TB, 99% of time the test will return a positive result; if a person tested does not have TB, 95% of the time the test will return a negative result. Among the test group, 2% actually have TB. What proportion of those tested positive actually have TB? Solution

Bayes’ Rule Example A company manufactures MP3 players at two factories. Factory I produces 60% of the MP3 players and Factory II produces 40%. Two percent of the MP3 players produced at Factory I are defective, while 1% of Factory II’s are defective. An MP3 player is selected at random and found to be defective. What is the probability it came from Factory I?

Bayes’ Rule: Example Defective 0.02 Factory I 0 .6 0.98 Good Defective 0.01 0 .4 Factory II 0.99 Good

Key Ideas Probability Rules for k Sample Points, S1, S2, S3, . . . , Sk 1. 0 ≤ P(Si) ≤ 1 2.

Key Ideas Random Sample All possible such samples have equal probability of being selected.

Key Ideas Combinations Rule Counting number of samples of n elements selected from N elements

Key Ideas Bayes’s Rule