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Chapter 3

Chapter 3. Probability. Probability. 3.1 The Concept of Probability 3.2 Sample Spaces and Events 3.3 Some Elementary Probability Rules 3.4 Conditional Probability and Independence. Probability Concepts. An experiment is any process of observation with an uncertain outcome

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Chapter 3

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  1. Chapter 3 Probability

  2. Probability 3.1 The Concept of Probability 3.2 Sample Spaces and Events 3.3 Some Elementary Probability Rules 3.4 Conditional Probability and Independence

  3. Probability Concepts • An experimentis 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 • 1. 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 • 2. The probabilities of all the experimental outcomes must sum to 1

  5. Assigning Probabilities toExperimental Outcomes • Classical Method • For equally likely outcomes • Relative frequency • In the long run • Subjective • Assessment based on experience, expertise, or intuition

  6. Classical Method • 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. Relative Frequency Method • Let E be an outcome of an experiment • If the experiment 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 to calculate probabilities • Example: Of 1000 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. The Sample Space The sample space of an experiment is the set of all possible experimental outcomes Example 3.2: Genders of Two Children Let: B be the outcome that child is boy G be the outcome that child is girl Sample space S: 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

  9. Example: Computing Probabilities Example 3.4: Genders of Two Children Events 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 Note: Experimental Outcomes: BB, BG, GB, GG All outcomes equally likely: P(BB) = … = P(GG) = ¼

  10. 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

  11. Example: 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  Share = 8,300,000 0.08 = 668,100

  12. Some Elementary Probability Rules The complement of an event A is the set of all sample space outcomes not in A Further, Union of A and B, Elementary events that belong to either A or B (or both) Intersection of A and B, Elementary events that belong to both A and B These figures are “Venn diagrams”

  13. The Addition Rule The probability that A or B (the union of A and B) will occur is where the “joint” probability of A and B both occurring together A and B are mutually exclusive if they have no sample space outcomes in common, or equivalently, if If A and B are mutually exclusive, then

  14. 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

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

  16. Example: Newspaper Subscribers #3 • Refer to the contingency table in Table 3.3 for all probabilities • For example, the chance that a household does not subscribe to either newspaper • Want , so from middle row and middle column of Table 3.3

  17. 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

  18. Conditional Probability • Interpretation: Restrict the sample space to just event B. The conditional probability is the chance of event A occurring in this new sample space • If A occurred, then what is the chance of B occurring? The probability of an event A, given that the event B has occurred, is called the “conditional probability of A given B” and is denoted as Further, Note: P(B) ≠ 0

  19. 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

  20. Independence of Events Two events A and B are said to be independent if and only if: P(A|B) = P(A) or, equivalently, P(B|A) = P(B)

  21. 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

  22. The Multiplication Rule The joint probability that A and B (the intersection of A and B) will occur is If A and B are independent, then the probability that A and B (the intersection of A and B) will occur is

  23. Example: Genders of Two Children Recall: B is the outcome that child is boy G is the outcome that child is girl Sample space S: 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

  24. 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

  25. Contingency Tables

  26. Example: AccuRatings Case Example 3.16: Estimating Radio Station Share by Daypart 5528 L.A. residents sampled 2827 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

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