Basic Practice of Statistics - 3rd Edition Chapter 12 General Rules of Probability BPS - 5th Ed. Chapter 12 Chapter 11

Probability Rules from Chapter 10 BPS - 5th Ed. Chapter 12

Venn Diagrams Two disjoint events: Chapter 12 Two events that are not disjoint, and the event {A and B} consisting of the outcomes they have in common: Two disjoint events: BPS - 5th Ed. Chapter 12

Multiplication Rule for 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) BPS - 5th Ed. Chapter 12

Multiplication Rule for Independent Events Example 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? BPS - 5th Ed. Chapter 12

Addition Rule: for Disjoint Events P(A or B) = P(A) + P(B) BPS - 5th Ed. Chapter 12

P(A or B) = P(A) + P(B)  P(A and B) General Addition Rule P(A or B) = P(A) + P(B)  P(A and B) BPS - 5th Ed. Chapter 12

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? BPS - 5th Ed. Chapter 12

Case Study 0.30 0.20 0.80 All Students Other Students Part-time (B) 0.30 In-state (A) 0.80 {A and B} 0.20 P(A or B) = P(A) + P(B)  P(A and B) = 0.80 + 0.30  0.20 = 0.90 BPS - 5th Ed. Chapter 12

In-state, but not part-time (A but not B): 0.80  0.20 = 0.60 Case Study All Students Other Students Part-time (B) 0.30 In-state (A) 0.80 In-state, but not part-time (A but not B): 0.80  0.20 = 0.60 {A and B} 0.20 BPS - 5th Ed. Chapter 12

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 BPS - 5th Ed. Chapter 12

Conditional Probability When P(A) > 0, the conditional probability of B given A is BPS - 5th Ed. Chapter 12

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)? BPS - 5th Ed. Chapter 12

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) BPS - 5th Ed. Chapter 12

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? BPS - 5th Ed. Chapter 12

5% of all students are freshmen smokers. 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. BPS - 5th Ed. Chapter 12

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) BPS - 5th Ed. Chapter 12

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) BPS - 5th Ed. Chapter 12

Binge Drinking and Accidents 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? BPS - 5th Ed. Chapter 12

Binge Drinking and Accidents Case Study Binge Drinking and Accidents P(A and B) = P(B)P(A|B) = (0.30)(0.18) Accident No accident 0.09 0.82 0.18 0.91 0.054 Binge drinker Non-binge drinker 0.30 0.70 0.637 0.063 0.246 P(Accident) = P(A) = 0.054 + 0.063 = 0.117 BPS - 5th Ed. Chapter 12