1. Chapter 4
Probability

2. Probability 4.1 The Concept of Probability
4.2 Sample Spaces and Events
4.3 Some Elementary Probability Rules
4.4 Conditional Probability and Independence

3. The Concept of Probability
An experiment is any process of observation with an uncertain outcome
The possible outcomes for an experiment are called the experimental outcomes
Probability is a measure of the chance that an experimental outcome will occur when an experiment is carried out

4. Probability
If E is an experimental outcome, then P(E) denotes the probability that E will occur and:
Conditions
0  P(E)  1 such that:
If E can never occur, then P(E) = 0
If E is certain to occur, then P(E) = 1
The probabilities of all the experimental outcomes must sum to 1

5. Assigning Probabilities to Experimental Outcomes
Classical Method
For equally likely outcomes
Long-run relative frequency
In the long run
Subjective
Assessment based on experience, expertise or intuition

6. Classical Method
All the experimental outcomes are equally likely to occur
Example: tossing a “fair” coin
Two outcomes: head (H) and tail (T)
If the coin is fair, then H and T are equally likely to occur any time the coin is tossed
So P(H) = 0.5, P(T) = 0.5
0 < P(H) < 1, 0 < P(T) < 1
P(H) + P(T) = 1

7. Long-Run Relative Frequency Method
Let E be an outcome of an experiment
If it is performed many times, P(E) is the relative frequency of E
P(E) is the percentage of times E occurs in many repetitions of the experiment
Use sampled or historical data
Example: Of 1,000 randomly selected consumers, 140 preferred brand X
The probability of randomly picking a person who prefers brand X is 140/1000 = 0.14 or 14%

8. Subjective Probability
Using experience, intuitive judgment, or expertise to assess a probability
May or may not have relative frequency interpretation

9. Sample Spaces and Events
A sample space of an experiment is the set of all possible experimental outcomes
Heads and tails when flipping a coin
The experimental outcomes in the sample space are often called sample space outcomes

10. Example 4.2: Genders of Two Children
Let: B be the outcome that child is boy G be the outcome that child is girl
Sample space S = {BB, BG, GB, GG}
If B and G are equally likely , then P(B) = P(G) = ½ and
P(BB) = P(BG) = P(GB) = P(GG) = ¼

11. A Tree Diagram of the Genders of Two Children

12. Events An event is a set (or collection) of sample space outcomes
The probability of an event is the sum of the probabilities of the sample space outcomes that correspond to the event

13. Example 4.4: Gender of Two Children
Experimental Outcomes: BB, BG, GB, GG
All outcomes equally likely: P(BB) = … = P(GG) = ¼
P(one boy and one girl) = P(BG) + P(GB) = ¼ + ¼ = ½
P(at least one girl) = P(BG) + P(GB) + P(GG) = ¼+¼+¼ = ¾

14. Probabilities: Equally Likely Outcomes
If the sample space outcomes (or experimental outcomes) are all equally likely, then the probability that an event will occur is equal to the ratio:
The number of ways the event can occur
Over the total number of outcomes

15. Example 4.7: AccuRatings Case
Of 5528 residents sampled, 445 prefer KPWR
Estimated Share P(KPWR) = 445 / 5528 =
So the probability that any resident chosen at random prefers KPWR is
Assuming 8,300,000 Los Angeles residents aged 12 or older:
# Listeners = Population x Share so 8,300,000 x = 668,144

16. Example: AccuRatings Case Continued

17. Some Elementary Probability Rules
Complement
Union
Intersection
Addition
Conditional probability
Multiplication

18. Complement
The complement (Ā) of an event A is the set of all sample space outcomes not in A
P(Ā) = 1 – P(A)

19. Union and Intersection
The union of A and B are elementary events that belong to either A or B or both
Written as A  B
The intersection of A and B are elementary events that belong to both A and B
Written as A ∩ B

20. Some Elementary Probability Rules

21. Mutually Exclusive
A and B are mutually exclusive if they have no sample space outcomes in common
In other words: P(A∩B) = 0

22. The Addition Rule
If A and B are mutually exclusive, then the probability that A or B (the union of A and B) will occur is P(AB) = P(A) + P(B)
If A and B are not mutually exclusive: P(AB) = P(A) + P(B) – P(A∩B) where P(A∩B) is the joint probability of A and B both occurring together

23. Example: Newspaper Subscribers #1
Define events:
A = event that a randomly selected household subscribes to the Atlantic Journal
B = event that a randomly selected household subscribes to the Beacon News
Given:
total number in city, N = 1,000,000
number subscribing to A, N(A) = 650,000
number subscribing to B, N(B) = 500,000
number subscribing to both, N(A∩B) = 250,000

24. Example: Newspaper Subscribers #2
Use the relative frequency method to assign probabilities

25. Example: Newspaper Subscribers #3
Refer to the contingency table in Table 4.3 for all data
For example, the chance that a household does not subscribe to either newspaper
Want , so from middle row and middle column of Table 4.3

26. Example: Newspaper Subscribers #4
The chance that a household subscribes to either newspaper:
Note that if the joint probability was not subtracted the would have gotten 1.15, which is greater than 1, which is absurd
The subtraction avoids double counting the joint probability

27. Conditional Probability and Independence
The probability of an event A, given that the event B has occurred, is called the conditional probability of A given B
Denoted as P(A|B)
Further, P(A|B) = P(A∩B) / P(B)
P(B) ≠ 0

28. Interpretation Restrict sample space to just event B
The conditional probability P(A|B) is the chance of event A occurring in this new sample space
In other words, if B occurred, then what is the chance of A occurring

29. Example: Newspaper Subscribers
Of the households that subscribe to the Atlantic Journal, what is the chance that they also subscribe to the Beacon News?
Want P(B|A), where

30. Independence of Events
Two events A and B are said to be independent if and only if: P(A|B) = P(A)
This is equivalently to P(B|A) = P(B)

31. Example: Newspaper Subscribers
Of the Atlantic Journal subscribers, what is the chance that they also subscribe to the Beacon News?
If independent, the P(B|A) = P(A)
Is P(B|A) = P(A)?
Know that P(A) = 0.65
Just calculated that P(B|A) =
0.65 ≠ , so P(B|A) ≠ P(A)
A is not independent of B
A and B are said to be dependent

32. The Multiplication Rule
The joint probability that A and B (the intersection of A and B) will occur is P(A∩B) = P(A) • P(B|A) = P(B) • P(A|B)
If A and B are independent, then the probability that A and B will occur is: P(A∩B) = P(A) • P(B) = P(B) • P(A)

33. Example: Genders of Two Children
B is the outcome that child is boy
G is the outcome that child is girl
Sample space S = {BB, BG, GB, GG}
If B and G are equally likely, then P(B) = P(G) = ½ and
P(BB) = P(BG) = P(GB) = P(GG) = ¼

34. Example: Genders of Two Children Continued
Of two children, what is the probability of having a girl first and then a boy second?
Want P(G first and B second)?
Want P(G∩B)
P(G∩B) = P(G)  P(B|G)
But gender of siblings is independent
So P(B|G) = P(B)
Then P(G∩B) = P(G)  P(B) = ½  ½ = ¼
Consistent with the tree diagram

35. Contingency Tables

36. EXAMPLE 4.16 The AccuRatings Case: Estimating Radio Station Share by Daypart
5,528 L.A. residents sampled
2,827 of residents sampled listen during some portion of the 6-10 a.m. daypart
Of those, 201 prefer KIIS
KIIS Share for 6-10 a.m. daypart: P(KIIS|6-10 a.m.) = P(KIIS  6-10 a.m.) / P(6-10 a.m.) = (201/5528)  (2827/5528) = 201 / 2827 =

37. Example 4.16 Continued