1. Chapter 13
Probability Rules!

2. Objectives: The student will be able to:
Apply the General Addition and General Multiplication rules.
Compute conditional probabilities and use the rule to test for independence.
Use a tree diagram to understand conditional probabilities and to reverse conditioning.

3. 13.1
The General Addition Rule

4. Or But Not Disjoint Your Wallet S = {$1, $2, $5, $10, $20, $50, $100}
A = {odd numbered value} = {$1, $5}
B = {bill with a building} = {$5, $10, $20, $50, $100}
Why is P(A or B) ≠ P(A) + P(B)?
Answer: A and B are not disjoint.
The intersection A and B = {$5} is double counted.
To find P(A or B), subtract P(A and B).

5. The General Addition Rule
P(A or B) = P(A) + P(B) – P(A and B)
The General Addition Rule in words: Add the probabilities of the two events and then subtract the probability of their intersection.
P(odd amount or bill with a building) = P(A) + P(B) – P(A and B}
= P({$1, $5}) + P({$5, $10, $20, $50, $100}) – P({$5})

6. General Addition Rule Example
Survey
Are you currently in a relationship?
Are you involved in sports?
Results
33% are in a relationship.
25% are involved in sports.
11% answered “yes” to both.
Problem
Find the probability that a randomly selected student is in a relationship or is involved in sports.

7. 33% Relationship, 25% Sport, 11% Both
Events
R = {in a relationship}
S = {involved in sports}
Calculations
P(R or S) = P(R) + P(S) – P(R and S) = – = 0.47
Conclusion
There is a 47% chance that a randomly selected student is in a relationship or is involved sports.

8. Using Venn Diagrams P(not in relationship and no sports) P(RC and SC)
This is the part outside of both circles:
P(in a relationship but no sports)
P(R and SC)
This is the part in the circle R that is outside S:
P(in a relationship or involved in sports but not both)
P((R and SC) or (RC and S))
This is the combination of the circles minus the intersection: = 0.36

9. E-mail or Text? 73% use e-mail, 62% text, 49% do both
Find the probability that a randomly selected person:
Either texts or s?
Either texts or s, but not both?
Doesn’t use and doesn’t text?
Think →
Plan:
A = {uses }
B = {texts}

10. 73% Use E-mail, 62% Text, 49% Both
Think →
Plot:
P(A) = 0.73
P(B) = 0.62
P(A and B) = 0.49
P(A and BC) = 0.73 – 0.49 = 0.24
P(AC and B) = 0.62 – 0.49 = 0.13
P(AC and BC) = 1 – ( ) = 0.14

11. 73% Use E-mail, 62% Text, 49% Both
1. P(text or )
Show →
Mechanics
P(A or B) = P(A) + P(B) – P(A and B) = – = 0.86
Tell →
Conclusion: 86% send text messages or .

12. 73% Use E-mail, 62% Text, 49% Both
2. P(text or but not both)
Show →
Mechanics
P(A or B) – P(A and B) = 0.86 – 0.49 = 0.37
P((A and BC) or (AC and B)) = = 0.37
Tell →
Conclusion: 37% send text messages or , but not both.

13. 73% Use E-mail, 62% Text, 49% Both
3. P(neither text nor )
Show →
Mechanics
1 – P(A or B) = 1 – 0.86 = 0.14
P(AC and BC) = 0.14
Tell →
Conclusion: 14% neither text nor .

14. Example
Example: 50% of the students in statistics have taken another college math class, 60% have taken an English class, and 30% have taken both.
What is the probability that a student chosen at random has taken a math or an English class?
Hint: use a Venn Diagram to help!

15. Practice
College dorm rooms on a large college campus had the following: 38% refrigerators, 52% tvs, and 21% had both. What is the probability that a randomly selected dorm room had:
A tv but no refrigerator?
A TV or a refrigerator but not both?
Neither a tv nor a refrigerator?
Employment data at a large company revealed that 72% of workers are married, 44% are college graduates, and half of the college graduates are married. What’s the probability that a randomly selected worker
Is neither married nor a college graduate?
Is married but not a college graduate?
Is married or a college graduate?

16. 13.2
Conditional Probability and the General Multiplication Rule

17. Contingency Table
A table that displays the results of two categorical questions is called a contingency table.
P(girl) = 251/478 = 0.525
P(girl and popular) = 91/478 = 0.190
P(sports) = 90/478 = 0.188

18. Conditional Probability
What if we knew the chosen person was a girl? Would that change the probability that the girl’s goal was sports?
Yes! We write P(sports | girl)
Only look at Girl row: P(sports | girl) = 30/251 = 0.120
Find the probability of selecting a boy given the goal is grades.
P(boy | grades) = 117/247 = 0.474

19. Conditional Probability Formula
Probability of B Given A:
Example: P(girl | popular) =

20. What is the probability that a man has both conditions?
The probabilities that an adult man has high blood pressure and/or high cholesterol are shown in the table:
What is the probability that a man has both conditions?
What is the probability that he has high blood pressure?
What is the probability that a man with high blood pressure has high cholesterol
What’s the probability that a man has high blood pressure if it’s known that he has high cholesterol?
High BP
OK BP
High Chol.
0.11
0.21
OK. Chol.
0.16
0.52

21. College Students: Relationships and Sports
33% in relationships, 25% in sports, 11% in both
If you see a student athlete, what is the probability that this athlete is in a relationship?
P(R) = 0.33, P(S) = 0.25, P(R and S) = 0.11
There is a 0.44 probability the student involved in sports is in a relationship. Notice this is higher than the probability for the general student (0.33).

22. The General Multiplication Rule
For A and B independent, we had: P(A and B) = P(A) × P(B)
Rearranging the conditional probability equation, we get the General Multiplication Rule: P(A and B) = P(A) × P(B | A)
Equivalently, P(A and B) = P(B) × P(A | B)

23. Examples: General Multiplication Rule: P(A and B) = P(A) × P(B | A)
Suppose the probability of a child having a cold is 0.1 and the probability of a child having a fever if s/he has a cold is What is the probability that child has both a cold and a fever?
Suppose there are 5 marbles in a jar, 2 green, 2 yellow, and 1 red. If I draw 2 without replacing the marbles, what is the probability that both are green?

24. More practice
Given a standard deck of 52 cards: (13 cards of each suit: hearts, spades, clubs, and diamonds)
If I draw 3 cards without replacement…
What is the probability of drawing 3 red cards?
What is the probability that the first heart I draw is on my third draw?

25. 13.3
Independence

26. Definition of Independence
Events A and B are independent if knowing A happened does not change the probability of B. In symbols: A and B are independent ↔ P(B | A) = P(B)
Equivalent formulas for independence:
P(A | B) = P(A)
P(A and B) = P(A) × P(B)

27. Grades and Girl Independent?
Determine if the goal of good grades and sex are independent.
P(grades | girl) = 130/251 ≈ 0.52
P(grades) = 247/478 ≈ 0.52
To two decimal places, they are independent.
Are the goal of sports and sex independent?
P(sports | boy) = 60/227 ≈ 0.26
P(sports) = 90/478 ≈ 0.19
No, the goal of sports and sex are dependent.

28. Relationships, Sports, and Independence
33% in a relationship, 25% involved in sports, 11% both
Are being in a relationship and being involved in sports independent?
P(relationship) = 0.33
P(sports) = 0.25
P(relationship and sports) = 0.11
0.33 × = ≠ 0.11
No, they are dependent.
Are they disjoint?
P(relationship and sports) = 0.11 ≠ 0
No, they are not disjoint.

29. Independent ≠ Disjoint
Disjoint events cannot be independent.
Consider the events:
Course grade A
Course grade B
Disjoint: You can’t get both.
Not independent: P(A | B) = 0 ≠ P(A)
A and B are disjoint (also called mutually exclusive) but not independent.

30. Examples – Testing for Independence
If you draw a card at random from a well-shuffled deck is the probability of getting an ace independent of the suit?
E.g. Does P(ace) = P(ace | suit is hearts)?
Approximately 58.2% of US adults have both a landline an a cell phone. 2.8% have only cell phone, 1.6% have no phone service at all
What proportion of US households have a landline?
Are having a cell phone and having a landline independent?
Does P(cell phone)=P(cell phone | have a landline)?
Or P(landline)=P(landline | cell phone)?

31. 13.4
Picturing Probability

32. Tree Diagrams 44% binge drink, 37% drink moderately, 19% don’t drink
Binge drinkers: 17% in an alcohol related accident
Non-bingers: 9% in an alcohol-related accident
Find the probability of being a binge drinker and has had an alcohol-related accident.
Venn diagrams and tables are not great for conditional probabilities.
Use a tree diagram.

33. Tree Diagrams The three branches are shown.
This tree diagram gives the complete information.
Notice the sums:
= 1
= 1
= 1
= 1
Conditional Probabilities:
P(none | binge) = 0.83.

34. Probabilities From Trees
P(moderate and accident) = × =
P(abstain and accident) = × = 0
P(none) = (0.44 × 0.83) (0.37 × 0.91) + (0.19 × 1.0) =

35. Tree Diagram Facts
The sum of the probabilities emanating from any branch is 1.
The final outcomes are disjoint.
To find a conditional probability, multiply across.

36. 13.5
Reversing the Conditioning and Bayes’ Rule

37. Going Backwards Find P(binge | accident)
Work backwards through the tree

38. Bayes’ Rule for Two Branches
Used for reversing the condition
Expressed as an algebraic formula instead of a tree
The tree is easier to use.
A longer version works for more than 2 branches.

39. Tree Diagrams Practice – text #42 using a tree diagram
An airline offers discounted “advance-purchase” fares to customers who buy tickets more than 30 days before travel and charges “regular” fares for tickets purchased during those last 30 days. The company has noticed that 60%of its customers buy advance purchase tickets. The “no-show” rate among people who purchase regular tickets is 30% but only 5% of customers with advance tickets are no-shows.
What percent of all ticket holders are no-shows?
What’s the probability that an customer who didn’t show had an advance-purchase ticket?
Is being a no-show independent of the type of ticket a passenger holds?

40. TB Tests Think → Plan: TB →has tuberculosis, + → tests positive
False Positive: P(+ | TBC) = 0.01
False Negative: P(− | TB) = 0.001
Has TB: P(TB) =
Probability of having TB given a positive test = ?
P(TB | +) = ?

41. Show → Plot Use the Complement Rule. Use the Multiplication Rule.
× = , etc.
Mechanics
P(+) = P(+ | TB) + P(+ | TBC) = =

42. P(TB | +) = ? Show → Mechanics P(+) = 0.01004945 P(TB | +) =
Tell → Conclusion
The probability of having TB after a positive test is less than 0.5%.

43. Reversing the Conditioning (cont.)
Practice: #46
Suppose a polygraph can detect 65% of lies, but incorrectly identifies 15% of true statements as lies
A certain company believes that 95%of its job applicants are trustworthy. The company gives everyone a polygraph. What is the probability that a job applicant rejected due to failing the polygraph was actually trustworthy?
Want: P(trustworthy | failed polygraph)

44. Wear a Seatbelt! In 77% of accidents, driver was wearing a seatbelt.
92% of those drivers escaped serious injury.
63% of non-belted drivers escaped serious injury.
Question: Find the probability a driver who was seriously injured wasn’t wearing a seatbelt?
B = Wore a seatbelt
NB = Did not wear a seatbelt
I = Serious injury
OK = No Serious injury
P(NB | I) = ?

45. Wear a Seatbelt!
Answer
We have: P(B) = 0.77, P(NB) = 0.23, P(OK | B) = 0.92, P(I | B) = P(OK | NB) = 0.63, P(I | NB) = 0.37

46. Wear a Seatbelt!
Conclusion: Even though only 23% don’t wear seatbelts, they account for 58% of all the deaths and serious injuries.

47. What Can Go Wrong?
Don’t use a simple probability rule where a general rule is appropriate.
Don’t assume independence or disjoint until after you have done the verification.
Don’t reverse conditioning naïvely.
P(A | B) may not be the same as P(B | A).
Don’t confuse “disjoint” with “independent.”
Disjoint events cannot happen at the same time.
Independent events must be able to happen at the same time.