Chapter 4 Probability

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

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

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

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

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

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%

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

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

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

A Tree Diagram of the Genders of Two Children

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

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

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

Example 4.7: AccuRatings Case Of 5528 residents sampled, 445 prefer KPWR Estimated Share P(KPWR) = 445 / 5528 = 0.0805 So the probability that any resident chosen at random prefers KPWR is 0.0805 Assuming 8,300,000 Los Angeles residents aged 12 or older: # Listeners = Population x Share so 8,300,000 x 0.0805 = 668,144

Example: AccuRatings Case Continued

Some Elementary Probability Rules Complement Union Intersection Addition Conditional probability Multiplication

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

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

Some Elementary Probability Rules

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

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

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

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

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

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

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

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

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

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)

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) = 0.3846 0.65 ≠ 0.3846, so P(B|A) ≠ P(A) A is not independent of B A and B are said to be dependent

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)

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

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

Contingency Tables

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 = 0.0711

Example 4.16 Continued