ch. 4a narrated powerpoints

CHAPTER FOUR: 

CHAPTER FOUR INTRODUCTION TO PROBABILITY

Chapter 4 Introduction to Probability: 

Chapter 4 Introduction to Probability 4.1 Experiments, Counting Rules, and Assigning Probabilities 4.2 Events and Their Probability 4.3 Some Basic Relationships of Probability 4.4 Conditional Probability 4.5 Bayes’ Theorem

Slide 3: 

Uncertainties Managers often base their decisions on an analysis of uncertainties such as the following: What are the chances that sales will decrease if we increase prices? What is the likelihood a new assembly method method will increase productivity? What are the odds that a new investment will be profitable?

Probability as a Numerical Measure of the Likelihood of Occurrence: 

Probability as a Numerical Measure of the Likelihood of Occurrence 0 1 .5 Increasing Likelihood of Occurrence Probability: The event is very unlikely to occur. The occurrence of the event is just as likely as it is unlikely. The event is almost certain to occur. Probability values are always on a scale of 0 to 1

Statistics v. Probabilities: 

Statistics v. Probabilities Statistics: Decisions are made about populations based on sample information. Based on the sample, we make inferences, forecasts, and projections about the population. Probabilities: Decisions are made about samples based on population information. We assume that a population is known and calculate the chances of obtaining sample results from the population.

4.1 An Experiment and Its Sample Space: 

4.1 An Experiment and Its Sample Space An experiment is any process of observation that generates well-defined outcomes. Examples: Tossing a coin: heads or tails Rolling one die: 1, 2, 3, 4, 5, or 6 Playing a game: win, lose, or tie (in some sports) Inspect a part: defective, not defective Making an investment: gain, loss, break even The probability space or sample space for an experiment is the set of all experimental outcomes. An individual experimental outcome is also called a sample point .

Outcomes: One-step experiments: 

Outcomes: One-step experiments We determine the possible outcomes of any particular experiment by identifying all possible outcomes. Outcomes Tossing a coin: heads or tails 2 Rolling one die: 1, 2, 3, 4, 5, or 6 6 Playing a game: win, lose, or tie 3 Inspect a part: defective, not defective 2 I nvestment: gain, loss, break even 3 We have not assigned a probability to any of these outcomes yet; for now, we are just identifying the number of possible outcomes to any experiment.

Outcome Counting Rules: 

Outcome Counting Rules We have three “counting rules” for determining the number of possible outcomes of certain experiments. They are: Combinations Permutations Multiple-step experiments Remember: we still are not determining probabilities yet; we are still merely determining the possible number of outcomes from an experiment.

Outcome Counting Rules: Combinations: 

The combination counting rule enables us to determine the number of possible outcomes when n objects are to be selected from a set of N objects (no replacement). Outcome Counting Rules: Combinations Number of Combinations of N Objects Taken n at a Time where: N ! = N ( N - 1)( N - 2) . . . (2)(1) n ! = ( n) ( n - 1)( n - 2) . . . (2)(1) 0! = 1 The notation ! means factorial: the number x itself times (itself – 1) until and including 1. So, 5! = (5)(4)(3)(2)(1) = 120

Outcome Counting Rules: B. Combinations: 

Outcome Counting Rules: B. Combinations Doing the math: Of six students, I am going to select 2 to go to show their work on the blackboard. How many different combination outcomes of students are there from my selection process? ____6!___ = (6)(5)(4)(3)(2)(1) = 720 = 15 2! (6 – 2)! (2)(1)[(4)(3)(2)(1)] 48 Prove it (show the sample set): AB AC AD AE AF BA BC BD BE BF CA CB CD CE CF DA DB DC DE DF EA EB EC ED EF FA FB FC FD FE Note: the same combination is excluded ; we want different combination outcomes. AB is the same as BA; AC the same as CA, etc.

Combinations: 

Combinations Another example from the textbook: a lottery selects six winning numbers from a set of 53 possible numbers. How many different outcomes are possible? . 53! . = (53)(52)(51)(50)(49)(48)(47!) 6! (53 – 6)! (6)(5)(4)(3)(2)(1) x 47! = 22,957,480 possible outcomes

Slide 12: 

Number of Permutations of N Objects Taken n at a Time where: N ! = N ( N - 1)( N - 2) . . . (2)(1) n ! = n ( n - 1)( n - 2) . . . (2)(1) 0! = 1 Outcome Counting Rules: Permutations Another useful counting rule enables us to count the number of possible outcomes when n objects are to be selected from a set of N objects, where the order of selection is important.

Outcome Counting Rules: Permutations: 

Outcome Counting Rules: Permutations Doing the math: Of six students, I am going to select 2 to go to show their work on the blackboard. One will start the problem, go halfway, and the second will finish the problem. Therefore, how many different permutations (order matters) of students are there from my selection process? ____6!___ = (6)(5)(4)(3)(2)(1) = 720 = 30 (6 – 2)! (4)(3)(2)(1) 24 Prove it: AB AC AD AE AF BA BC BD BE BF CA CB CD CE CF DA DB DC DE DF EA EB EC ED EF FA FB FC FD FE Note: unlike combinations, the same combination is included. AB is the not the same as BA; AC is not the same as CA, etc.

Permutations: 

Permutations What if in our lottery example, you had to select the exact six winning numbers from a set of 53 possible numbers in an exact order? How many different outcomes are possible? . 53! . = (53)(52)(51)(50)(49)(48)(47!) (53 – 6)! 47! = 16,529,385,600 possible outcomes

Outcome Counting Rules: Multiple-step experiments: 

Outcome Counting Rules: Multiple-step experiments What if we have multiple step experiments? We can still determine the number of possible outcomes if we know how many outcomes are possible for each step of the experiment. If an experiment consists of k samples in which there are n 1 possible results for the first sample, n 2 possible results for the second sample, and so on, then the total number of experimental outcomes is given by: ( n 1 )(n 2 ) . . . ( n k ).

Outcome Counting Rules: Multiple-step Experiments: 

Outcome Counting Rules: Multiple-step Experiments Doing the math: First, determine the number of possible outcomes (the sample space) for each step of the experiment. n 1 = number of possible outcomes for sample #1. For tossing a coin, there are 2 possible outcomes. Next, determine the number of possible outcomes for the next experimental sample. n 2 = number of possible outcomes for sample #2. For tossing a coin, n 1 and n 2 happen to be equal because they are identical events. The counting rule for multiple event experiments tells us that the number of possible, different outcomes is ( n 1 )( n 2 ) = (2)(2) = 4. Tossing three coins has 8 different outcomes; four has 16; six has 64.

Outcome Counting Rules: Multiple –step experiments: 

Outcome Counting Rules: Multiple – step experiments For example, tossing a coin three times: k = 3; n 1 = 2; n 2 = 2; n 3 = 2 … so (2)(2)(2) = 8 there are eight possible outcomes: (H , H, H) (H, H, T) (H, T, H) (H,T,T ), ( T, H, H) (T, H, T) (T, T, H) (T, T, T) This is our sample space of eight (8) possible experimental events or outcomes (sample points). KP&L example in the book: Possible outcomes for design stage = 3 Possible outcomes for the construction stage = 3 Total possible outcomes = (3)(3) = 9

Outcome Counting Rules: Multiple-step Experiments: 

Outcome Counting Rules: Multiple-step Experiments REM: total # of experimental outcomes = ( n 1 )( n 2 ) ...( n k ). # of possible outcomes for flipping a coin and then rolling a die? (2)(6) = 12 # of possible outcomes for flipping a coin and then drawing a card? (2)(52) = 104 # of possible outcomes for drawing a card, returning it to the deck, and drawing again? (52)(52) = 2,704 # of possible outcomes for drawing a card, not returning it, and drawing again? (52)(51) = 2,622 REM: these are outcomes that tell us how many sample points are in the sample set, not probabilities A helpful graphical representation of a multiple-sample experiment is a tree diagram .

Tree Diagram: 

Tree Diagram H, H, H H, H, T H H H H H H H T T T T T T T H, T, H H, T, T T, H, H T, H, T T, T, H T, T, T

Assigning Probabilities to Experimental Outcomes: 

Assigning Probabilities to Experimental Outcomes Classical Method Relative Frequency Method Subjective Method Assigning probabilities based on the assumption of equally likely outcomes Assigning probabilities based on experimentation or historical data Assigning probabilities based on judgment 1 2 3 Whichever we use, 0 < P(E i ) < 1 ; and P(E n ) = 1.

1. Classical Method: 

1. Classical Method If an experiment has n possible outcomes, and each outcome must be equally likely, this method would assign a probability of 1/ n to each outcome. Experiment: Rolling a die Sample Space: S = {1, 2, 3, 4, 5, 6} Probabilities: Each sample point has a 1/6 chance of occurring Example Other classical method experiments: flipping a coin; drawing a card when the card selected is returned to the deck. Few events in business avail themselves to the classical method of probability assignment .

1. Classical Method: Multiple Steps: 

1. Classical Method: Multiple Steps We can look at the probability of an outcome for each step … Possible outcomes: 1, 1 1,2 1, 3 1,4 1, 5 1,6 2, 1 2,2 2, 3 2,4 2, 5 2,6 3, 1 3,2 3, 3 3,4 3, 5 3,6 4, 1 4,2 4, 3 4,4 4, 5 4,6 5, 1 5,2 5, 3 5,4 5, 5 5,6 6 , 1 6 ,2 6 , 3 6 ,4 6 , 5 6 ,6 Prob ( 6 on the first die) = 6/36 = 1/6 = .167 Prob ( odd # on the 2 nd ) = 18/36 = .50 Prob (1,2,3, or 4 on 1 st ) = 24/36 = .6667

1. Classical Method: Multiple Steps: 

1. Classical Method: Multiple Steps … or the probability of an outcome comprising of both steps. It all depends on what we are interested in. Possible outcomes: 1,1 1,2 1,3 1,4 1,5 1,6 2,1 2,2 2,3 2,4 2,5 2,6 3,1 3,2 3,3 3,4 3,5 3,6 4,1 4,2 4,3 4,4 4,5 4,6 5,1 5,2 5,3 5,4 5,5 5,6 6,1 6,2 6,3 6,4 6,5 6,6 Prob (7) = 6/36 = 1/6 = .167 Prob (“Snake eyes”) = 1/36 = .028 Prob (Not 7) = 30/36 = .833

2. Relative Frequency Method: 

2. Relative Frequency Method Number of Cars Rented Number of Days 0 1 2 3 4 4 6 18 10 2 Lucas Car Rental would like to assign probabilities to the number of cars it rents each day. Office records show the following frequencies of daily rentals for the last 40 days. Experience data or observations over time are available to provide probabilities.

2. Relative Frequency Method (cont.): 

Each probability assignment is given by dividing the frequency (number of days) by the total frequency (total number of days). 2. Relative Frequency Method (cont.) 4/40 Probability Number of Cars Rented Number of Days 0 1 2 3 4 4 6 18 10 2 40 .10 .15 .45 .25 .05 1.00

2. Relative Frequency Method: Multiple Steps: 

2. Relative Frequency Method: Multiple Steps We can also assign probabilities by the relative frequency method for events containing multiple steps. See the “KP&L” example in your book (Tables 4.2 and 4.3): Stage 1 Stage 2 Sample Point Past Freq. Probability 2 6 (2,6) = 8 mos. 6 (6/40) .15 2 7 (2,7) = 9 6 .15 2 8 (2,8) = 10 2 .05 3 6 (3,6) = 9 4 .10 3 7 (3,7) = 10 8 .20 3 8 (3,8) = 11 2 .05 4 6 (4,6) = 10 2 .05 4 7 (4,7) = 11 4 .10 4 8 (4,8) = 12 6 .15

3. Subjective Method: 

3. Subjective Method When economic conditions and a company’s circumstances change rapidly it might be inappropriate to assign probabilities based solely on historical data. We can use any data available as well as our experience and intuition, but ultimately a probability value should express our degree of belief that the experimental outcome will occur. The best probability estimates often are obtained by combining the estimates from the classical or relative frequency approach with the subjective estimate. It’s our best, educated guess!

Slide 28: 

An event is a collection of sample points. The probability of any event is equal to the sum of the probabilities of each of the sample points in the event. If we can identify all the sample points of an experiment and assign a probability to each (either through classical, relative frequency, or subjectively), we can compute the probability of a event made up of all of those sample points occurring. 4.2 Events and Their Probabilities A single event outcome satisfies only one criterion.

Events: 

Events An “event” is a collection of sample points. An event can be almost anything depending on how we define the event: A red ace drawn from a deck A voter who voted for Obama A consumer who purchased a Guinness The number of successes / failures The project is completed late / on time The investment makes money/losses money/breaks even

Events: 

Events A fair die is thrown Sample Space: {1, 2, 3, 4, 5, 6} Now define the events as if the number is even, you win. If the number is odd, I win. Event you win: {2, 4, 6} Event you lose: {1, 3, 5} Note: an event can contain one or more sample points. An event is a specific collection of sample points. 2 4 6 1 3 5

Events and their probabilities: 

Events and their probabilities

Events and their probabilities: 

Events and their probabilities Example : Probability of a nonusable manufactured product e 1 = Broken .05 e 2 = Defective .02 e 3 = Good .93 If we define E nonusable = { e 1 + e 2 } then P( E nonusable ) = P( e 1 ) + P( e 2 ) Example: Probability of bad weather at a football game e 1 = No precipitation .65 e 2 = Rain .25 e 3 = Snow .10 If we define E bad weather = { e 2 + e 3 } then P( E bad ) = P( e 2 ) + P( e 3 ) What’s the probability of not needing an umbrella?

Events and Their Probabilities: 

Events and Their Probabilities Using the “KP&L” example in your book, what is the probability of Stage 1 being completed in 3 months or less? Stage 1 Stage 2 Sample Point Past Freq. Probability 2 6 (2,6) 6 .15 2 7 (2,7) 6 .15 2 8 (2,8) 2 .05 3 6 (3,6) 4 .10 3 7 (3,7) 8 .20 3 8 (3,8) 2 .05 4 6 (4,6) 2 .05 4 7 (4,7) 4 .10 4 8 (4,8) 6 .15

Events and Their Probabilities: 

Events and Their Probabilities Using the “KP&L” example in your book, what is the probability of Stage 2 being completed in 8 months? Stage 1 Stage 2 Sample Point Past Freq. Probability 2 6 (2,6) 6 .15 2 7 (2,7) 6 .15 2 8 (2,8) 2 .05 3 6 (3,6) 4 .10 3 7 (3,7) 8 .20 3 8 (3,8) 2 .05 4 6 (4,6) 2 .05 4 7 (4,7) 4 .10 4 8 (4,8) 6 .15

Events and Their Probabilities: 

Events and Their Probabilities Using the “KP&L” example in your book, if we define “on time” as 10 months or less, then P(on time) = .70 Stage 1 Stage 2 Sample Point Past Freq. Probability 2 6 (2,6) 6 .15 2 7 (2,7) 6 .15 2 8 (2,8) 2 .05 3 6 (3,6) 4 .10 3 7 (3,7) 8 .20 3 8 (3,8) 2 .05 4 6 (4,6) 2 .05 4 7 (4,7) 4 .10 4 8 (4,8) 6 .15

4.3 Some Basic Relationships of Probability: 

4.3 Some Basic Relationships of Probability There are two basic probability relationships that can be used to compute the probability of an event without knowledge of all the sample point probabilities . Complement of an Event Addition Law Union of Two Events Intersection of Two Events Mutually Exclusive Events

Slide 37: 

The complement of A is denoted by A c . The complement of event A is defined to be the event consisting of all sample points that are not in A. Complement of an Event Event A A c Sample Space S Venn Diagram If we know the probability of A c , and we know A c cant occur in A, then we can figure the probability of Event A = 1 – P ( A c )

Joint Events: Unions & Intersections: 

Joint Events: Unions & Intersections An event can often be viewed as a composition of two or more events. Such events are called joint or compound events , and involve two concepts: Union of two events: the union of two events, A and B, is the event that occurs if either A or B or both occur on a single performance of each experiment. Intersection of two events: the intersection of two events, A and B, is the event that occurs if both A and B occur on a single performance of each experiment. If the events share an outcome, we have an intersection.

Joint Events: 

Joint Events A joint event is a collection of sample points that satisfies two or more criteria. The joint events can be almost anything depending on how we define the events: A red card or a face card drawn from a deck A voter who voted for Obama & is over the age of 65 A consumer who purchased a Guinness & has income over $35,000 The Cardinals & the Cubs both winning on the same day The project is completed on time & on budget

Slide 40: 

The union of events A and B is denoted by A  B  The union of events A and B is the event containing all sample points that are in A or B or both. Union of Two Events Sample Space S Event A Event B We want to know the probability of the grey area within the “figure 8.” Some sample points are contained in both Event A & Event B. Venn Diagram

Slide 41: 

The addition law provides a way to compute or verify the probability of joint events A or B, or both A and B occurring by math if we only have the probabilities. Addition Law The law is written as: P ( A  B ) = P ( A ) + P ( B ) - P ( A  B  Where P ( A  B ) = probability of either A or B happening P ( A ) = probability of A happening P ( B ) = probability of B happening P ( A  B ) = probability of A and B happening

Joint Events: Unions & Intersections: 

Joint Events: Unions & Intersections Rolling a die. Event A: Toss an even number Event B: Toss a number less than or equal to 3. Event A: {2, 4, 6}; Event B: {1, 2, 3} A  B: {1, 2, 3, 4, 6} A  B: {2} Probability of A  B: 5/6 ths Probability of A  B: 1/6 th 2 1 3 4 6 5 A B

Addition Law: Solving for the Union: 

Addition Law: Solving for the Union A deck of cards; what’s the probability of drawing red or a face card? Sample set: A♣ 2♣ 3♣ 4♣ 5♣ 6♣ 7♣ 8♣ 9♣ 10♣ J♣ Q♣ K♣ A♠ 2♠ 3♠ 4♠ 5♠ 6♠ 7♠ 8♠ 9♠ 10♠ J♠ Q♠ K♠ A♥ 2♥ 3♥ 4♥ 5♥ 6♥ 7♥ 8♥ 9♥ 10♥ J♥ Q♥ K♥ A♦ 2♦ 3♦ 4♦ 5♦ 6♦ 7♦ 8♦ 9♦ 10♦ J♦ Q♦ K♦ By math, we find P ( A  B ) = P ( A ) + P ( B ) - P ( A  B ) = (26/52) + (12/52) – P (6/52) = .5 + .23 - .115 = .615

Slide 44: 

The intersection of events A and B is denoted by A    The intersection of events A and B is the set of all sample points that are in both A and B . Sample Space S Event A Event B Intersection of Two Events Intersection of A and B ; we are looking only at the blue area of overlap When two separate events happen at the same time

Use the Addition Law: 

Use the Addition Law Addition Law: P ( A  B ) = P ( A ) + P ( B ) - P ( A  B ) Rolling a die. Event A: Toss an even number Event B: Toss a number less than or equal to 3. Event A: {2, 4, 6}; Event B: {1, 2, 3} A  B: {1, 2, 3, 4, 6} A  B: {2} P(A  B) = .5 + .5 - .167 = .833 How do we solve for P ( A  B ) ? For now, we can only solve for the intersection if we know P ( A  B ), P ( A ), and P ( B )

Slide 46: 

Mutually Exclusive Events Two events are said to be mutually exclusive if the events have no sample points in common. Two events are mutually exclusive if, when one event occurs, the other cannot occur. Sample Space S Event A Event B

Slide 47: 

Mutually Exclusive Events If events A and B are mutually exclusive, P ( A  B  = 0. The addition law for mutually exclusive events is: P ( A  B ) = P ( A ) + P ( B ) there’s no need to include “ - P ( A  B  ”

Mutually Exclusive Events: 

Mutually Exclusive Events The Cardinals win and the Cubs win when they are playing each other; absolute mutual exclusivity. On days the two teams are not playing each other, these events are not mutually exclusive St. Louis wins Cubs win

Summary of the Addition Law: 

Summary of the Addition Law Addition Law: P ( A  B ) = P ( A ) + P ( B ) - P ( A  B ) If we know three of the four probabilities of the addition law, we can solve for the unknown probability. Memory suggestion: Union =  nion =  Intersection =  O r A nd

What kind of event is it? It all depends on how you define the event(s): 

What kind of event is it? It all depends on how you define the event(s) Complementary Event Joint Event Mutually Exclusive Roll a die, get a 6 Roll a die, get a 6 or an even number (union); roll a 6 (intersection) Roll a die, get a 6 or a 2 Cardinals win (either win or lose) Cards or Cubs win (union); Cards and Cubs win (intersection); when not playing each other Cardinals or Cubs win when playing each other (absolute mutual exclusivity) Product not defective (either good or bad) Product not defective and a special order (intersection) Make a profit Make a profit & project done on time Make a profit, loss, or break even P(A C )= 1 – P(A) P(A  B)= P(A)+P(B)- P(A ∩ B) P(A  B)= P(A)+P(B)

Multiple Sample Experiments Example: Bradley Investments: 

Multiple Sample Experiments Example: Bradley Investments Bradley has invested in two stocks, Markley Oil and Collins Mining. Bradley has determined that there are only certain possible outcomes for each of these investments three months from now, and they are: Investment Gain or Loss in 3 Months (in $000) Markley Oil Collins Mining 10 5 0 - 20 8 - 2

Counting Rules: A. Multiple-Sample Experiments: 

Bradley Investments can be viewed as a two-sample experiment. It involves two stocks, each with a set of independent experimental outcomes. Markley Oil: n 1 = 4 Collins Mining: n 2 = 2 Total Number of Experimental Outcomes: n 1 n 2 = (4)(2) = 8 Counting Rules: A. Multiple-Sample Experiments REM: someone determined that there are four possible outcomes for the oil stock, and two possible outcomes for the mining stock; therefore, using the counting rule in multiple-sample (two samples here), there are 8 possible outcomes. What are those possible outcomes?

Event Tree Diagram illustrating the possible outcomes (the sample space): 

Event Tree Diagram illustrating the possible outcomes (the sample space) Gain 5 Gain 8 Gain 8 Gain 10 Gain 8 Gain 8 Lose 20 Lose 2 Lose 2 Lose 2 Lose 2 Even Markley Oil* (Stage 1) Collins Mining # (Stage 2) Experimental Outcomes (10 + 8) = Gain $18,000 (10 - 2) = Gain $8,000 (5 + 8) = Gain $13,000 (5 - 2) = Gain $3,000 (0 + 8) = Gain $8,000 (0 - 2) = Lose $2,000 (-20 + 8) = Lose $12,000 (-20 - 2) = Lose $22,000 * Four possible outcomes # Two possible outcomes Notice we sum along each branch of the tree to get the end result outcomes; event “gain” and event “lose.”

Probability Assigned: Subjective Method (Bradley Investments Example): 

Probability Assigned: Subjective Method (Bradley Investments Example) Applying the subjective method, an analyst made the following probability assignments. Exper . Outcome Net Gain or Loss Subjective Probability (10, 8) (10, - 2) (5, 8) (5, - 2) (0, 8) (0, - 2) ( - 20, 8) ( - 20, - 2) $18,000 Gain $8,000 Gain $13,000 Gain $3,000 Gain $8,000 Gain $2,000 Loss $12,000 Loss $22,000 Loss .20 .08 .16 .26 .10 .12 .02 .06

Events and Their Probabilities: Probability of Markley Oil being profitable?: 

Events and Their Probabilities: Probability of Markley Oil being profitable? Event M = Markley Oil Profitable Alone M = {(10, 8), (10, - 2), (5, 8), (5, - 2)} P ( M ) = P (10, 8) + P (10, - 2) + P (5, 8) + P (5, - 2) = .20 + .08 + .16 + .26 = .70 Outcomes Result Subj. Prob. (10 , 8 ) + $18,000 .20 (10 , -2 ) + $8,000 .08 (5 , 8 ) + $13,000 .16 (5 , -2 ) + $3,000 .26 We only use data where 1 st variable, Markley, is > $0 (profit)

Slide 56: 

Events and Their Probabilities: Probability of Collins Mining being profitable? Event C = Collins Mining Profitable Alone* C = {( 10 , 8), ( 5 , 8), ( 0 , 8), ( - 20 , 8)} P ( C ) = P (10, 8) + P (5, 8) + P (0, 8) + P ( - 20, 8) = .20 + .16 + .10 + .02 = .48 REM: The probability of any event is equal to the sum of the probabilities of each of the applicable sample points in the event. * We are actually only looking at the second sample point and four events.

Union of Two Events: 

Union of Two Events L et’s look at the probability of the union of two events when we know all the sample points. Later we will use the Addition Law to calculate for probabilities without knowledge of all the sample point, but knowing the probabilities. In our Bradley Investment example, we know all the sample points and their individual subjective probabilities. Which events show the union between either company making a profit? $18,000 Gain $8,000 Gain $13,000 Gain $3,000 Gain $8,000 Gain $2,000 Loss $12,000 Loss $22,000 Loss (10, 8) (10, - 2) (5, 8) (5, - 2) (0, 8) (0, - 2) ( - 20, 8) ( - 20, - 2) .20 .08 .16 .26 .10 .12 .02 .06

Sample Points in Bradley Investments: 

Sample Points in Bradley Investments 10, 8 10, -2 5, 8 5, -2 0, 8 0, -2 -20, 8 -20, -2 Profit for Markley Profit for Collins Zero or loss Venn diagram for all possibilities for Markley and Collins:

Slide 59: 

Union of Two Events by sample points Event M = Markley Oil Profitable Event C = Collins Mining Profitable M  C = Markley Oil Profitable or Collins Mining Profitable M  C = {(10, 8), (10, - 2), (5, 8), (5, - 2), (0, 8), ( - 20, 8)} P ( M  C) = P (10, 8) + P (10, - 2) + P (5, 8) + P (5, - 2) + P (0, 8) + P ( - 20, 8) = .20 + .08 + .16 + .26 + .10 + .02 = .82 The probability of either M or C being profitable It’s the sum of all probabilities except (0, -2) [.12] and (-20, -2) [.06]

Slide 60: 

Intersection of Two Events: Inspecting sample points to verify Addition Law Event M = Markley Oil Profitable Event C = Collins Mining Profitable M  C = Markley Oil Profitable and Collins Mining Profitable M  C = {(10, 8), (5, 8)} P ( M  C) = P (10, 8) + P (5, 8) = .20 + .16 = .36 What’s the probability that both companies will not be profitable at the same time ? = 1 - .36 = .64 or 64%

Use the Addition Law: 

Use the Addition Law Addition Law: P ( A  B ) = P ( A ) + P ( B ) - P ( A  B ) What’s the probability of M and C both being profitable? (We’re solving for the intersection) In our investment example, we now know P ( M  C) = .82; P ( M ) = .70; P ( C ) = .48 Use the Addition Law to solve for P ( A  B ): .82 = .70 + .48 - P ( M  C ) P ( M  C ) = 1.18 - .82 = .36 The probability of both M and C being profitable is 36%. We’ve just solved by math for the probability of the intersection of the two events occurring simultaneously by using the addition law.

Slide 62: 

Event M = Markley Oil Profitable Event C = Collins Mining Profitable M  C = Markley Oil Profitable or Collins Mining Profitable We know: P ( M ) = .70, P ( C ) = .48, P ( M  C ) = .36 Thus: P ( M  C) = P ( M ) + P( C ) - P ( M  C ) .82 = .70 + .48 - .36 . 82 = .82 Does the Addition Law Check? (This formula verifies, by math, what we were able to see by inspecting the sample points in our sample space.) What’s the probability of either M or C profitable?