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Chapter 11

Chapter 11. General Rules of Probability. Probability Rules from Chapter 9. Two events that are not disjoint, and the event {A and B} consisting of the outcomes they have in common:. Two disjoint events:. Venn Diagrams. Multiplication Rule for Independent Events.

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Chapter 11

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  1. Chapter 11 General Rules of Probability Chapter 11

  2. Probability Rules from Chapter 9 Chapter 11

  3. Two events that are not disjoint, and the event {A and B} consisting of the outcomes they have in common: Two disjoint events: Venn Diagrams Chapter 11

  4. Multiplication Rulefor Independent Events If two events A and B do not influence each other, and if knowledge about one does not change the probability of the other, the events are said to be independent of each other. If two events are independent, the probability that they both happen is found by multiplying their individual probabilities: P(A and B) = P(A)  P(B) Chapter 11

  5. Multiplication Rulefor Independent EventsExample • Suppose that about 20% of incoming male freshmen smoke. • Suppose these freshmen are randomly assigned in pairs to dorm rooms (assignments are independent). • The probability of a match (both smokers or both non-smokers): • both are smokers: 0.04 = (0.20)(0.20) • neither is a smoker: 0.64 = (0.80)(0.80) • only one is a smoker: ? } 68% 32% (100%  68%) What if pairs are self-selected? Chapter 11

  6. Addition Rule: for Disjoint Events P(A or B) = P(A) + P(B) Chapter 11

  7. General Addition Rule P(A or B) = P(A) + P(B) P(A and B) Chapter 11

  8. Case Study Student Demographics At a certain university, 80% of the students were in-state students (event A), 30% of the students were part-time students (event B), and 20% of the students were both in-state and part-time students (event {A and B}). So we have that P(A) = 0.80, P(B) = 0.30, and P(A and B) = 0.20. What is the probability that a student is either an in-state student or a part-time student? Chapter 11

  9. All Students Part-time (B) 0.30 {A and B} In-state (A) 0.20 0.80 Case Study Other Students P(A or B) = P(A) + P(B)  P(A and B) = 0.80 + 0.30  0.20 = 0.90 Chapter 11

  10. All Students Part-time (B) 0.30 {A and B} In-state (A) 0.20 0.80 In-state, but notpart-time (A but not B):0.80  0.20 = 0.60 Case Study Other Students Chapter 11

  11. Conditional Probability • The probability of one event occurring, given that another event has occurred is called a conditional probability. • The conditional probability of B given A is denoted by P(B|A) • the proportion of all occurrences of A for which B also occurs Chapter 11

  12. Conditional Probability When P(A) > 0, the conditional probability of B given A is Chapter 11

  13. Case Study Student Demographics In-state (event A): P(A) = 0.80 Part-time (event B): P(B) = 0.30 Both in-state and part-time: P(A and B) = 0.20. Given that a student is in-state (A), what is the probability that the student is part-time (B)? Chapter 11

  14. General Multiplication Rule For ANY two events, the probability that they both happen is found by multiplying the probability of one of the events by the conditional probability of the remaining event given that the other occurs: P(A and B) = P(A)  P(B|A) or P(A and B) = P(B)  P(A|B) Chapter 11

  15. Case Study Student Demographics At a certain university, 20% of freshmen smoke, and 25% of all students are freshmen. Let A be the event that a student is a freshman, and let B be the event that a student smokes. So we have that P(A) = 0.25, and P(B|A) = 0.20. What is the probability that a student smokes and is a freshman? Chapter 11

  16. Case Study Student Demographics P(A) = 0.25 , P(B|A) = 0.20 P(A and B) = P(A)  P(B|A) = 0.25  0.20 = 0.05 5% of all students are freshmen smokers. Chapter 11

  17. Independent Events • Two events A and B that both have positive probability are independent if P(B|A) = P(B) • General Multiplication Rule: P(A and B) = P(A) P(B|A) • Multiplication Rule for independent events: P(A and B) = P(A) P(B) Chapter 11

  18. Tree Diagrams • Useful for solving probability problems that involve several stages • Often combine several of the basic probability rules to solve a more complex problem • probability of reaching the end of any complete “branch” is the product of the probabilities on the segments of the branch (multiplication rule) • probability of an event is found by adding the probabilities of all branches that are part of the event (addition rule) Chapter 11

  19. Case Study Binge Drinking and Accidents At a certain college, 30% of the students engage in binge drinking. Among college-aged binge drinkers, 18% have been involved in an alcohol-related automobile accident, while only 9% of non-binge drinkers of the same age have been involved in such accidents. Let event A = {accident related to alcohol}. Let event B = {binge drinker}. So we have P(A|B)=0.18, P(A|’not B’)=0.09, & P(B)=0.30 . What is the probability that a randomly selected student has been involved in an alcohol-related automobile accident? Chapter 11

  20. P(A and B) = P(B)P(A|B) = (0.30)(0.18) 0.18 Accident Bingedrinker 0.82 0.30 No accident 0.246 0.70 0.09 Accident 0.063 Non-binge drinker 0.91 No accident 0.637 Case Study Binge Drinking and Accidents 0.054 P(Accident) = P(A) = 0.054 + 0.063 = 0.117 Chapter 11

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