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BPS - 5th Ed. Chapter 121 General Rules of Probability.

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Presentation on theme: "BPS - 5th Ed. Chapter 121 General Rules of Probability."— Presentation transcript:

1 BPS - 5th Ed. Chapter 121 General Rules of Probability

2 BPS - 5th Ed. Chapter 122 Probability Rules from Chapter 10

3 BPS - 5th Ed. Chapter 123 Venn Diagrams Two disjoint events: Two events that are not disjoint, and the event {A and B} consisting of the outcomes they have in common:

4 BPS - 5th Ed. Chapter 124 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) Multiplication Rule for Independent Events

5 BPS - 5th Ed. Chapter 125 Multiplication Rule for Independent Events Example u Suppose that about 20% of incoming male freshmen smoke. u Suppose these freshmen are randomly assigned in pairs to dorm rooms (assignments are independent). u 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?

6 BPS - 5th Ed. Chapter 126 Addition Rule: for Disjoint Events P(A or B) = P(A) + P(B)

7 BPS - 5th Ed. Chapter 127 General Addition Rule P(A or B) = P(A) + P(B)  P(A and B)

8 BPS - 5th Ed. Chapter 128 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) = What is the probability that a student is either an in- state student or a part-time student?

9 BPS - 5th Ed. Chapter 129 Other Students P(A or B)= P(A) + P(B)  P(A and B) =  0.20 = 0.90 All Students Part-time (B) 0.30 {A and B} 0.20 Case Study In-state (A) 0.80

10 BPS - 5th Ed. Chapter 1210 Other Students All Students Part-time (B) 0.30 Case Study {A and B} 0.20 In-state (A) 0.80 In-state, but not part-time (A but not B): 0.80  0.20 = 0.60

11 BPS - 5th Ed. Chapter 1211 u The probability of one event occurring, given that another event has occurred is called a conditional probability. u 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 Conditional Probability

12 BPS - 5th Ed. Chapter 1212 Conditional Probability When P(A) > 0, the conditional probability of B given A is

13 BPS - 5th Ed. Chapter 1213 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) = Given that a student is in-state (A), what is the probability that the student is part-time (B)?

14 BPS - 5th Ed. Chapter 1214 General Multiplication Rule P(A and B) = P(A)  P(B|A) or P(A and B) = P(B)  P(A|B) 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:

15 BPS - 5th Ed. Chapter 1215 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) = What is the probability that a student smokes and is a freshman?

16 BPS - 5th Ed. Chapter 1216 Case Study Student Demographics P(A) = 0.25, P(B|A) = % of all students are freshmen smokers. P(A and B)= P(A)  P(B|A) = 0.25  0.20 = 0.05

17 BPS - 5th Ed. Chapter 1217 Independent Events u 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)

18 BPS - 5th Ed. Chapter 1218 Tree Diagrams u Useful for solving probability problems that involve several stages u 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)

19 BPS - 5th Ed. Chapter 1219 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?

20 BPS - 5th Ed. Chapter 1220 Case Study Binge Drinking and Accidents P(Accident) = P(A) = = P(A and B) = P(B)  P(A|B) = (0.30)(0.18) Accident No accident Accident No accident Binge drinker Non-binge drinker


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