Probability (Part 2) Chapter55 Contingency Tables Tree Diagrams Bayes’ Theorem Counting Rules McGraw-Hill/Irwin Copyright © 2009 by The McGraw-Hill Companies, Inc.

5B-2 A contingency table is a cross- tabulation of frequencies into rows and columns.A contingency table is a cross- tabulation of frequencies into rows and columns. Row 1 Row 2 Row 3 Row 4 Variable 2 Variable 1 Col 1 Col 2 Col 3 Cell A contingency table is like a frequency distribution for two variables. Contingency Tables What is a Contingency Table? What is a Contingency Table?

5B-3 Consider the following cross-tabulation table for n = 67 top-tier MBA programs: (Table 5.4) Contingency Tables Example: Salary Gains and MBA Tuition Example: Salary Gains and MBA Tuition

5B-4 Are large salary gains more likely to accrue to graduates of high-tuition MBA programs?Are large salary gains more likely to accrue to graduates of high-tuition MBA programs? The frequencies indicate that MBA graduates of high-tuition schools do tend to have large salary gains.The frequencies indicate that MBA graduates of high-tuition schools do tend to have large salary gains. Also, most of the top-tier schools charge high tuition.Also, most of the top-tier schools charge high tuition. More precise interpretations of this data can be made using the concepts of probability.More precise interpretations of this data can be made using the concepts of probability. Contingency Tables Example: Salary Gains and MBA Tuition Example: Salary Gains and MBA Tuition

5B-5 The marginal probability of a single event is found by dividing a row or column total by the total sample size.The marginal probability of a single event is found by dividing a row or column total by the total sample size. For example, find the marginal probability of a medium salary gain (P(S 2 )).For example, find the marginal probability of a medium salary gain (P(S 2 )). P(S 2 ) =33/67 =.4925 Conclude that about 49% of salary gains at the top-tier schools were between $50,000 and $100,000 (medium gain).Conclude that about 49% of salary gains at the top-tier schools were between $50,000 and $100,000 (medium gain). Contingency Tables Marginal Probabilities Marginal Probabilities

5B-6 Find the marginal probability of a low tuition P(T 1 ).Find the marginal probability of a low tuition P(T 1 ). P(T 1 ) = 16/67 =.2388 There is a 24% chance that a top-tier school’s MBA tuition is under $ There is a 24% chance that a top-tier school’s MBA tuition is under $ Contingency Tables Marginal Probabilities Marginal Probabilities

5B-7 A joint probability represents the intersection of two events in a cross-tabulation table.A joint probability represents the intersection of two events in a cross-tabulation table. Consider the joint event that the school has low tuition and large salary gains (denoted as P(T 1  S 3 )).Consider the joint event that the school has low tuition and large salary gains (denoted as P(T 1  S 3 )). Contingency Tables Joint Probabilities Joint Probabilities

5B-8 So, using the cross-tabulation table,So, using the cross-tabulation table, P(T 1  S 3 ) = 1/67 =.0149 There is less than a 2% chance that a top-tier school has both low tuition and large salary gains.There is less than a 2% chance that a top-tier school has both low tuition and large salary gains. Contingency Tables Joint Probabilities Joint Probabilities

5B-9 Found by restricting ourselves to a single row or column (the condition).Found by restricting ourselves to a single row or column (the condition). For example, knowing that a school’s MBA tuition is high (T 3 ), we would restrict ourselves to the third row of the table.For example, knowing that a school’s MBA tuition is high (T 3 ), we would restrict ourselves to the third row of the table. Contingency Tables Conditional Probabilities Conditional Probabilities

5B-10 Find the probability that the salary gains are small (S 1 ) given that the MBA tuition is large (T 3 ).Find the probability that the salary gains are small (S 1 ) given that the MBA tuition is large (T 3 ). P(S 1 | T 3 ) = 5/32 =.1563 What does this mean?What does this mean? Contingency Tables Conditional Probabilities Conditional Probabilities

5B-11 To check for independent events in a contingency table, compare the conditional to the marginal probabilities.To check for independent events in a contingency table, compare the conditional to the marginal probabilities. For example, if large salary gains (S 3 ) were independent of low tuition (T 1 ), then P(S 3 | T 1 ) = P(S 3 ).For example, if large salary gains (S 3 ) were independent of low tuition (T 1 ), then P(S 3 | T 1 ) = P(S 3 ). ConditionalMarginal P(S 3 | T 1 )= 1/16 =.0625P(S 3 ) = 17/67 =.2537 What do you conclude about events S 3 and T 1 ?What do you conclude about events S 3 and T 1 ? Contingency Tables Independence Independence

5B-12 Calculate the relative frequencies below for each cell of the cross-tabulation table to facilitate probability calculations.Calculate the relative frequencies below for each cell of the cross-tabulation table to facilitate probability calculations. Contingency Tables Relative Frequencies Relative Frequencies Symbolic notation for relative frequencies:Symbolic notation for relative frequencies:

Here are the resulting probabilities (relative frequencies). For example,Here are the resulting probabilities (relative frequencies). For example, Contingency Tables Relative Frequencies Relative Frequencies P(T 1 and S 1 ) = 5/67 P(T 2 and S 2 ) = 11/67 P(T 3 and S 3 ) = 15/67 P(S 1 ) = 17/67 P(T 2 ) = 19/67 5B-13

Contingency Tables Relative Frequencies Relative Frequencies The nine joint probabilities sum to since these are all the possible intersections.The nine joint probabilities sum to since these are all the possible intersections. Summing the across a row or down a column gives marginal probabilities for the respective row or column.Summing the across a row or down a column gives marginal probabilities for the respective row or column. 5B-14

5B-15 A small grocery store would like to know if the number of items purchased by a customer is independent of the type of payment method the customer chooses to use.A small grocery store would like to know if the number of items purchased by a customer is independent of the type of payment method the customer chooses to use. Why would this information be useful to the store manager?Why would this information be useful to the store manager? The manager collected a random sample of 368 customer transactions.The manager collected a random sample of 368 customer transactions. Contingency Tables Example: Payment Method and Purchase Quantity Example: Payment Method and Purchase Quantity

5B-16 Here is the contingency table of frequencies: Contingency Tables Example: Payment Method and Purchase Quantity Example: Payment Method and Purchase Quantity

5B-17 Calculate the marginal probability that a customer will use cash to make the payment.Calculate the marginal probability that a customer will use cash to make the payment. Let C be the event cash.Let C be the event cash. P(C) =126/368 =.3424 Contingency Tables Example: Payment Method and Purchase Quantity Example: Payment Method and Purchase Quantity Now, is this probability the same if we condition on number of items purchased?Now, is this probability the same if we condition on number of items purchased?

5B-18 P(C | 1-5)= 30/88=.3409 P(C | 6-10)= 46/135=.3407 P(C | 10-20)= 31/89=.3483 P(C | 20+)= 19/56=.3393 P(C) =.3424, so what do you conclude about independence?P(C) =.3424, so what do you conclude about independence? Based on this, the manager might decide to offer a cash-only lane that is not restricted to the number of items purchased.Based on this, the manager might decide to offer a cash-only lane that is not restricted to the number of items purchased. Contingency Tables Example: Payment Method and Purchase Quantity Example: Payment Method and Purchase Quantity

5B-19 Contingency tables require careful organization and are created from raw data.Contingency tables require careful organization and are created from raw data. Consider the data of salary gain and tuition for n = 67 top-tier MBA schools. Contingency Tables How Do We Get a Contingency Table? How Do We Get a Contingency Table?

5B-20 The data should be coded so that the values can be placed into the contingency table.The data should be coded so that the values can be placed into the contingency table. Contingency Tables How Do We Get a Contingency Table? How Do We Get a Contingency Table? Once coded, tabulate the frequency in each cell of the contingency table using MINITAB’s : Stat | Tables | Cross Tabulation

5B-21 A tree diagram or decision tree helps you visualize all possible outcomes.A tree diagram or decision tree helps you visualize all possible outcomes. Start with a contingency table.Start with a contingency table. Tree Diagrams What is a Tree? What is a Tree? For example, this table gives expense ratios by fund type for 21 bond funds and 23 stock funds.For example, this table gives expense ratios by fund type for 21 bond funds and 23 stock funds.

To label the tree, first calculate conditional probabilities by dividing each cell frequency by its column total.To label the tree, first calculate conditional probabilities by dividing each cell frequency by its column total. For example,For example, P(L | B) = 11/21 =.5238 Here is the table of conditional probabilitiesHere is the table of conditional probabilities Tree Diagrams What is a Tree? What is a Tree? 5B-22

5B-23 To calculate joint probabilities, useTo calculate joint probabilities, use P(A  B) = P(A | B)P(B) = P(B | A)P(A) The joint probability of each terminal event on the tree can be obtained by multiplying the probabilities along its branch.The joint probability of each terminal event on the tree can be obtained by multiplying the probabilities along its branch. The tree diagram shows all events along with their marginal, conditional and joint probabilities.The tree diagram shows all events along with their marginal, conditional and joint probabilities. For example,For example, P(B  L)P(B  L) = P(L | B)P(B) = (.5238)(.4773) =.2500 Tree Diagrams What is a Tree? What is a Tree?

5B-24 Tree Diagrams Tree Diagram for Fund Type and Expense Ratios Tree Diagram for Fund Type and Expense Ratios Figure 5.11

5B-25 Thomas Bayes ( ) provided a method (called Bayes’s Theorem) of revising probabilities to reflect new probabilities.Thomas Bayes ( ) provided a method (called Bayes’s Theorem) of revising probabilities to reflect new probabilities. The prior (marginal) probability of an event B is revised after event A has been considered to yield a posterior (conditional) probability.The prior (marginal) probability of an event B is revised after event A has been considered to yield a posterior (conditional) probability. Bayes’s formula is:Bayes’s formula is: Bayes’ Theorem

5B-26 Bayes’ formula begins as:Bayes’ formula begins as: In some situations P(A) is not given. Therefore, the most useful and common form of Bayes’ Theorem is:In some situations P(A) is not given. Therefore, the most useful and common form of Bayes’ Theorem is: Bayes’ Theorem

If a woman is not pregnant, what is the test’s “track record”?If a woman is not pregnant, what is the test’s “track record”? Consider an over-the-counter pregnancy testing kit and it’s “track record” of determining pregnancies.Consider an over-the-counter pregnancy testing kit and it’s “track record” of determining pregnancies. Bayes’ Theorem How Bayes’ Theorem Works How Bayes’ Theorem Works 96% of time 1% of time 4% of time 99% of time False Negative False Positive If a woman is actually pregnant, what is the test’s “track record”?If a woman is actually pregnant, what is the test’s “track record”? Table B-27

5B-28 Intuitively, if 1,000 women use this test, the results should look like this.Intuitively, if 1,000 women use this test, the results should look like this. Bayes’ Theorem How Bayes’ Theorem Works How Bayes’ Theorem Works Suppose that 60% of the women who purchase the kit are actually pregnant.Suppose that 60% of the women who purchase the kit are actually pregnant.

5B-29 Of the 580 women who test positive, 576 will actually be pregnant. So, the desired probability is: P(Pregnant│Positive Test) = 576/580 =.9931 Bayes’ Theorem How Bayes’ Theorem Works How Bayes’ Theorem Works

5B-30 Now use Bayes’s Theorem to formally derive the result P(Pregnant | Positive) =.9931:Now use Bayes’s Theorem to formally derive the result P(Pregnant | Positive) =.9931: First define A = positive test B = pregnant A' = negative test B' = not pregnantFirst define A = positive test B = pregnant A' = negative test B' = not pregnant P(A | B) =.96 P(A | B') =.01 P(B) =.60 And the compliment of each event is:And the compliment of each event is: P(A' | B) =.04 P(A' | B') =.99 P(B') =.40 From the contingency table, we know that:From the contingency table, we know that: Bayes’ Theorem How Bayes’ Theorem Works How Bayes’ Theorem Works

5B-31 P(B | A) = P(A | B)P(B) P(A | B)P(B) + P(A | B')P(B') =(.96)(.60) (.96)(.60) + (.01)(.40) = = =.9931 So, there is a 99.31% chance that a woman is pregnant, given that the test is positive.So, there is a 99.31% chance that a woman is pregnant, given that the test is positive. Bayes’ Theorem How Bayes’ Theorem Works How Bayes’ Theorem Works

5B-32 Bayes’s Theorem shows us how to revise our prior probability of pregnancy to get the posterior probability after the results of the pregnancy test are known.Bayes’s Theorem shows us how to revise our prior probability of pregnancy to get the posterior probability after the results of the pregnancy test are known. Prior Before the test Posterior After positive test result P(B) =.60  P(B | A) =.9931 Bayes’s Theorem is useful when a direct calculation of a conditional probability is not permitted due to lack of information.Bayes’s Theorem is useful when a direct calculation of a conditional probability is not permitted due to lack of information. Bayes’ Theorem How Bayes’ Theorem Works How Bayes’ Theorem Works

5B-33 A tree diagram helps visualize the situation.A tree diagram helps visualize the situation. Bayes’ Theorem How Bayes’ Theorem Works How Bayes’ Theorem Works

5B-34 The 2 branches showing a positive test (A) comprise a reduced sample space B  A and B'  A, Bayes’ Theorem How Bayes’ Theorem Works How Bayes’ Theorem Works so add their probabilities to obtain the denominator of the fraction whose numerator is P(B  A).

5B-35 A generalization of Bayes’s Theorem allows event B to be polytomous (B 1, B 2, … B n ) rather than dichotomous (B and B').A generalization of Bayes’s Theorem allows event B to be polytomous (B 1, B 2, … B n ) rather than dichotomous (B and B'). Bayes’ Theorem General Form of Bayes’ Theorem General Form of Bayes’ Theorem

5B-36 Based on historical data, the percent of cases at 3 hospital trauma centers and the probability of a case resulting in a malpractice suit are as follows:Based on historical data, the percent of cases at 3 hospital trauma centers and the probability of a case resulting in a malpractice suit are as follows: let event A = a malpractice suit is filed B i = patient was treated at trauma center ilet event A = a malpractice suit is filed B i = patient was treated at trauma center i Bayes’ Theorem Example: Hospital Trauma Centers Example: Hospital Trauma Centers (Table 5.18)

5B-37 Applying the general form of Bayes’ Theorem, find P(B 1 | A).Applying the general form of Bayes’ Theorem, find P(B 1 | A). Bayes’s Theorem Example: Hospital Trauma Centers Example: Hospital Trauma Centers 0.

5B-38 Conclude that the probability that the malpractice suit was filed in hospital 1 is.1389 or 13.89%.Conclude that the probability that the malpractice suit was filed in hospital 1 is.1389 or 13.89%. All the posterior probabilities for each hospital can be calculated and then compared:All the posterior probabilities for each hospital can be calculated and then compared: Bayes’ Theorem Example: Hospital Trauma Centers Example: Hospital Trauma Centers (Table 5.19)

5B-39 = 1, = 3, = 5, = 2,000 x.008 = 3,000 x.005 = 5,000 x.001 = 10,000x.2 = 10,000x.3 = 10,000x.5 Intuitively, imagine there were 10,000 patients and calculate the frequencies:Intuitively, imagine there were 10,000 patients and calculate the frequencies: Hospital Malpractice Suit Filed No Malpractice Suit Filed Total 154,9955, ,9853, ,9842,000 Total369,96410,000 Bayes’ Theorem Example: Hospital Trauma Centers Example: Hospital Trauma Centers

5B-40 Now, use these frequencies to find the probabilities needed for Bayes’ Theorem.Now, use these frequencies to find the probabilities needed for Bayes’ Theorem. Hospital Malpractice Suit Filed No Malpractice Suit Filed Total 1P(B 1 |A)=5/36=.1389P(B 1 |A')=.5012P(B 1 )=.5 2P(B 2 |A)=15/36=.4167P(B 2 |A')=.2996P(B 2 )=.3 3P(B 3 |A)=16/36=4444P(B 3 |A')=.1991P(B 3 )=.2 TotalP(A)=36/10000=.0036P(A')= For example,For example, Bayes’ Theorem Example: Hospital Trauma Centers Example: Hospital Trauma Centers

5B-41 Consider the following visual description of the problem:Consider the following visual description of the problem: Bayes’ Theorem Example: Hospital Trauma Centers Example: Hospital Trauma Centers

5B-42 The initial sample space consists of 3 mutually exclusive and collectively exhaustive events (hospitals B 1, B 2, B 3 ).The initial sample space consists of 3 mutually exclusive and collectively exhaustive events (hospitals B 1, B 2, B 3 ). Bayes’ Theorem Example: Hospital Trauma Centers Example: Hospital Trauma Centers

5B-43 As indicated by their relative areas, B 1 is 50% of the sample space, B 2 is 30% and B 3 is 20%.As indicated by their relative areas, B 1 is 50% of the sample space, B 2 is 30% and B 3 is 20%. Bayes’ Theorem Example: Hospital Trauma Centers Example: Hospital Trauma Centers 50% 30% 20%

5B-44 But, given that a malpractice case has been filed (event A), then the relevant sample space is reduced to the yellow area of event A.But, given that a malpractice case has been filed (event A), then the relevant sample space is reduced to the yellow area of event A. The revised probabilities are the relative areas within event A.The revised probabilities are the relative areas within event A. Bayes’ Theorem Example: Hospital Trauma Centers Example: Hospital Trauma Centers P(B 1 | A) P(B 2 | A) P(B 3 | A)

5B-45 If event A can occur in n 1 ways and event B can occur in n 2 ways, then events A and B can occur in n 1 x n 2 ways.If event A can occur in n 1 ways and event B can occur in n 2 ways, then events A and B can occur in n 1 x n 2 ways. In general, m events can occur n 1 x n 2 x … x n m ways.In general, m events can occur n 1 x n 2 x … x n m ways. Counting Rules Fundamental Rule of Counting Fundamental Rule of Counting

5B-46 How many unique stock-keeping unit (SKU) labels can a hardware store create by using 2 letters (ranging from AA to ZZ) followed by four numbers (0 through 9)?How many unique stock-keeping unit (SKU) labels can a hardware store create by using 2 letters (ranging from AA to ZZ) followed by four numbers (0 through 9)? For example, AF1078: hex-head 6 cm bolts – box of 12 RT4855: Lime-A-Way cleaner – 16 ounce LL3319: Rust-Oleum primer – gray 15 ounceFor example, AF1078: hex-head 6 cm bolts – box of 12 RT4855: Lime-A-Way cleaner – 16 ounce LL3319: Rust-Oleum primer – gray 15 ounce Counting Rules Example: Stock-Keeping Labels Example: Stock-Keeping Labels

5B-47 View the problem as filling six empty boxes:View the problem as filling six empty boxes: There are 26 ways to fill either the 1st or 2nd box and 10 ways to fill the 3rd through 6 th. Therefore, there are 26 x 26 x 10 x 10 x 10 x 10 = 6,760,000 unique inventory labels. Counting Rules Example: Stock-Keeping Labels Example: Stock-Keeping Labels

5B-48 L.L. Bean men’s cotton chambray shirt comes in 6 colors (blue, stone, rust, green, plum, indigo), 5 sizes (S, M, L, XL, XXL) and two styles (short and long sleeves).L.L. Bean men’s cotton chambray shirt comes in 6 colors (blue, stone, rust, green, plum, indigo), 5 sizes (S, M, L, XL, XXL) and two styles (short and long sleeves). Their stock might include 6 x 5 x 2 = 60 possible shirts.Their stock might include 6 x 5 x 2 = 60 possible shirts. However, the number of each type of shirt to be stocked depends on prior demand.However, the number of each type of shirt to be stocked depends on prior demand. Counting Rules Example: Shirt Inventory Example: Shirt Inventory

5B-49 The number of ways that n items can be arranged in a particular order is n factorial.The number of ways that n items can be arranged in a particular order is n factorial. n factorial is the product of all integers from 1 to n.n factorial is the product of all integers from 1 to n. Factorials are useful for counting the possible arrangements of any n items.Factorials are useful for counting the possible arrangements of any n items. n! = n(n–1)(n–2)...1 There are n ways to choose the first, n-1 ways to choose the second, and so on.There are n ways to choose the first, n-1 ways to choose the second, and so on. Counting Rules Factorials Factorials

5B-50 As illustrated below, there are n ways to choose the first item, n-1 ways to choose the second, n-2 ways to choose the third and so on.As illustrated below, there are n ways to choose the first item, n-1 ways to choose the second, n-2 ways to choose the third and so on. Counting Rules Factorials Factorials

5B-51 A home appliance service truck must make 3 stops (A, B, C).A home appliance service truck must make 3 stops (A, B, C). In how many ways could the three stops be arranged?In how many ways could the three stops be arranged? 3! = 3 x 2 x 1 = 6 List all the possible arrangements:List all the possible arrangements: {ABC, ACB, BAC, BCA, CAB, CBA} How many ways can you arrange 9 baseball players in batting order rotation?How many ways can you arrange 9 baseball players in batting order rotation? 9! = 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 362,880 Counting Rules Factorials Factorials

5B-52 A permutation is an arrangement in a particular order of r randomly sampled items from a group of n items and is denoted by n P rA permutation is an arrangement in a particular order of r randomly sampled items from a group of n items and is denoted by n P r In other words, how many ways can the r items be arranged, treating each arrangement as different (i.e., XYZ is different from ZYX)?In other words, how many ways can the r items be arranged, treating each arrangement as different (i.e., XYZ is different from ZYX)? Counting Rules Permutations Permutations

5B-53 n = 5 home appliance customers (A, B, C, D, E) need service calls, but the field technician can service only r = 3 of them before noon.n = 5 home appliance customers (A, B, C, D, E) need service calls, but the field technician can service only r = 3 of them before noon. The order is important so each possible arrangement of the three service calls is different.The order is important so each possible arrangement of the three service calls is different. The number of possible permutations is:The number of possible permutations is: Counting Rules Example: Appliance Service Cans Example: Appliance Service Cans

5B-54 The 60 permutations with r = 3 out of the n = 5 calls can be enumerated.The 60 permutations with r = 3 out of the n = 5 calls can be enumerated. ABC ABD ABE ACD ACE ADE BCD BCE BDE CDE There are 10 distinct groups of 3 customers:There are 10 distinct groups of 3 customers: Each of these can be arranged in 6 distinct ways: ABC, ACB, BAC, BCA, CAB, CBA Since there are 10 groups of 3 customers and 6 arrange- ments per group, there are 10 x 6 = 60 permutations. Counting Rules Example: Appliance Service Cans Example: Appliance Service Cans

5B-55 A combination is an arrangement of r items chosen at random from n items where the order of the selected items is not important (i.e., XYZ is the same as ZYX).A combination is an arrangement of r items chosen at random from n items where the order of the selected items is not important (i.e., XYZ is the same as ZYX). A combination is denoted n C rA combination is denoted n C r Counting Rules Combinations Combinations

5B-56 n = 5 home appliance customers (A, B, C, D, E) need service calls, but the field technician can service only r = 3 of them before noon.n = 5 home appliance customers (A, B, C, D, E) need service calls, but the field technician can service only r = 3 of them before noon. This time order is not important.This time order is not important. Thus, ABC, ACB, BAC, BCA, CAB, CBA would all be considered the same event because they contain the same 3 customers.Thus, ABC, ACB, BAC, BCA, CAB, CBA would all be considered the same event because they contain the same 3 customers. The number of possible combinations is:The number of possible combinations is: Counting Rules Example: Appliance Service Calls Revisited Example: Appliance Service Calls Revisited

5B combinations is much smaller than the 60 permutations in the previous example.10 combinations is much smaller than the 60 permutations in the previous example. The combinations are easily enumerated:The combinations are easily enumerated: ABC, ABD, ABE, ACD, ACE, ADE, BCD, BCE, BDE, CDE Counting Rules Example: Appliance Service Calls Revisited Example: Appliance Service Calls Revisited

Applied Statistics in Business & Economics End of Chapter 5B 5B-58