1. Contingency Tables Tree Diagrams Bayes’ Theorem Counting Rules5
Probability (Part 2)
Chapter
Contingency Tables
Tree Diagrams
Bayes’ Theorem
Counting Rules
McGraw-Hill/Irwin Copyright © 2009 by The McGraw-Hill Companies, Inc. All rights reserved.

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

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

4. Contingency Tables Example: Salary Gains and MBA Tuition
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.
Also, most of the top-tier schools charge high tuition.
More precise interpretations of this data can be made using the concepts of probability.

5. Contingency Tables Marginal Probabilities
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(S2)).
P(S2) =
33/67 =
.4925
Conclude that about 49% of salary gains at the top-tier schools were between $50,000 and $100,000 (medium gain).

6. Contingency Tables Marginal Probabilities
Find the marginal probability of a low tuition P(T1).
P(T1) =
16/67 =
.2388
There is a 24% chance that a top-tier school’s MBA tuition is under $

7. Contingency Tables Joint Probabilities
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(T1  S3)).

8. Contingency Tables Joint Probabilities
So, using the cross-tabulation table,
P(T1  S3) =
1/67 =
.0149
There is less than a 2% chance that a top-tier school has both low tuition and large salary gains.

9. Contingency Tables Conditional Probabilities
Found by restricting ourselves to a single row or column (the condition).
For example, knowing that a school’s MBA tuition is high (T3), we would restrict ourselves to the third row of the table.

10. Contingency Tables Conditional Probabilities
Find the probability that the salary gains are small (S1) given that the MBA tuition is large (T3).
P(T1 | S3) =
5/32 =
.1563
What does this mean?

11. Contingency Tables Independence
To check for independent events in a contingency table, compare the conditional to the marginal probabilities.
For example, if large salary gains (S3) were independent of low tuition (T1), then P(S3 | T1) = P(S3).
Conditional
Marginal
P(S3 | T1)= 1/16 = .0625
P(S3) = 17/67 = .2537
What do you conclude about events S3 and T1?

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

13. Contingency Tables Relative Frequencies
Here are the resulting probabilities (relative frequencies). For example,
P(T1 and S1) = 5/67
P(T2 and S2) = 11/67
P(T3 and S3) = 15/67
P(S1) = 17/67
P(T2) = 19/67
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14. Contingency Tables Relative Frequencies
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.
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15. Contingency Tables Example: Payment Method and Purchase Quantity
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?
The manager collected a random sample of 368 customer transactions.

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

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

18. Contingency Tables Example: Payment Method and Purchase Quantity
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?
Based on this, the manager might decide to offer a cash-only lane that is not restricted to the number of items purchased.

19. Contingency Tables How Do We Get a Contingency Table?
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.

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

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

22. Tree Diagrams What is a Tree?
To label the tree, first calculate conditional probabilities by dividing each cell frequency by its column total.
For example,
P(L | B)
= 11/21
= .5238
Here is the table of conditional probabilities
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23. Tree Diagrams What is a Tree? = .2500
The tree diagram shows all events along with their marginal, conditional and joint probabilities.
To 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.
For example,
P(B  L)
= P(L | B)P(B)
= (.5238)(.4773)
= .2500

24. Tree Diagrams Tree Diagram for Fund Type and Expense Ratios
Figure 5.11

25. Bayes’ Theorem
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.
Bayes’s formula is:

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

27. Bayes’ Theorem How Bayes’ Theorem Works
Consider an over-the-counter pregnancy testing kit and it’s “track record” of determining pregnancies.
If a woman is actually pregnant, what is the test’s “track record”?
If a woman is not pregnant, what is the test’s “track record”?
False Positive
False Negative
Table 5.17
96% of time
1% of time
4% of time
99% of time
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28. Bayes’ Theorem How Bayes’ Theorem Works
Suppose that 60% of the women who purchase the kit are actually pregnant.
Intuitively, if 1,000 women use this test, the results should look like this.

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

30. Bayes’ Theorem How Bayes’ Theorem Works
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 pregnant
From the contingency table, we know that:
And the compliment of each event is:
P(A' | B) = .04 P(A' | B') = .99 P(B') = .40
P(A | B) = .96 P(A | B') = .01 P(B) = .60

31. Bayes’ Theorem How Bayes’ Theorem Works P(A | B)P(B) P(B | A) =
P(A | B)P(B) + P(A | B')P(B') =
(.96)(.60)
(.96)(.60) + (.01)(.40) =
.576 =
.576
.580 =
.9931
So, there is a 99.31% chance that a woman is pregnant, given that the test is positive.

32. After positive test result
Bayes’ Theorem
How Bayes’ Theorem Works
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.

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

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

35. Bayes’ Theorem General Form of Bayes’ Theorem
A generalization of Bayes’s Theorem allows event B to be polytomous (B1, B2, … Bn) rather than dichotomous (B and B').

36. Bayes’ Theorem Example: Hospital Trauma Centers
(Table 5.18)
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 Bi = patient was treated at trauma center i

37. Bayes’s Theorem Example: Hospital Trauma Centers
Applying the general form of Bayes’ Theorem, find P(B1 | A). 0.

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

39. Malpractice Suit Filed No Malpractice Suit Filed
Bayes’ Theorem
Example: Hospital Trauma Centers
Intuitively, imagine there were 10,000 patients and calculate the frequencies:
Hospital
Malpractice Suit Filed
No Malpractice Suit Filed
Total 1 5
4,995
5,000 2 15
2,985
3,000 3 16
1,984
2,000 36
9,964
10,000
= 5,000 x .001
= 5,
= 10,000x.5
= 3,000 x .005
= 3,
= 10,000x.3
= 2,000 x .008
= 1,
= 10,000x.2

40. Malpractice Suit Filed No Malpractice Suit Filed
Bayes’ Theorem
Example: Hospital Trauma Centers
Now, use these frequencies to find the probabilities needed for Bayes’ Theorem.
For example,
Hospital
Malpractice Suit Filed
No Malpractice Suit Filed
Total 1
P(B1|A)=5/36=.1389
P(B1|A')=.5012
P(B1)=.5 2
P(B2|A)=15/36=.4167
P(B2|A')=.2996
P(B2)=.3 3
P(B3|A)=16/36=4444
P(B3|A')=.1991
P(B3)=.2
P(A)=36/10000=.0036
P(A')=.9964
1.0000

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

42. Bayes’ Theorem Example: Hospital Trauma Centers
The initial sample space consists of 3 mutually exclusive and collectively exhaustive events (hospitals B1, B2, B3).

43. Bayes’ Theorem Example: Hospital Trauma Centers
As indicated by their relative areas, B1 is 50% of the sample space, B2 is 30% and B3 is 20%.
30%
50%
20%

44. Bayes’ Theorem Example: Hospital Trauma Centers
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.
P(B2 | A)
P(B1 | A)
P(B3 | A)

45. Counting Rules Fundamental Rule of Counting
If event A can occur in n1 ways and event B can occur in n2 ways, then events A and B can occur in n1 x n2 ways.
In general, m events can occur n1 x n2 x … x nm ways.

46. Counting Rules Example: Stock-Keeping Labels
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 ounce

47. Counting Rules Example: Stock-Keeping Labels
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 6th.
Therefore, there are 26 x 26 x 10 x 10 x 10 x 10 = 6,760,000 unique inventory labels.

48. Counting Rules Example: Shirt Inventory
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.
However, the number of each type of shirt to be stocked depends on prior demand.

49. Counting Rules Factorials n! = n(n–1)(n–2)...1
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! = n(n–1)(n–2)...1
Factorials are useful for counting the possible arrangements of any n items.
There are n ways to choose the first, n-1 ways to choose the second, and so on.

50. Counting Rules Factorials
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.

51. Counting Rules Factorials
A home appliance service truck must make 3 stops (A, B, C).
In how many ways could the three stops be arranged?
3! = 3 x 2 x 1 = 6
List all the possible arrangements:
{ABC, ACB, BAC, BCA, CAB, CBA}
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

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

53. Counting Rules Example: Appliance Service Cans
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 number of possible permutations is:

54. Counting Rules Example: Appliance Service Cans
The 60 permutations with r = 3 out of the n = 5 calls can be enumerated.
There are 10 distinct groups of 3 customers:
Each of these can be arranged in 6 distinct ways:
ABC ABD ABE ACD
ACE
ADE BCD BCE BDE
CDE
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.

55. Counting Rules Combinations
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 nCr

56. Counting Rules Example: Appliance Service Calls Revisited
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.
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:

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

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