Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide 15- 1

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Chapter 15 Probability Rules!

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Recall That… For any random phenomenon, each trial generates an outcome. An event is any set or collection of outcomes. The collection of all possible outcomes is called the sample space, denoted S.

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Events When outcomes are equally likely, probabilities for events are easy to find just by counting. When the k possible outcomes are equally likely, each has a probability of 1/k. For any event A that is made up of equally likely outcomes,

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide The First Three Rules for Working with Probability Rules Make a picture.

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Picturing Probabilities The most common kind of picture to make is called a Venn diagram. We will see Venn diagrams in practice shortly…

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide The General Addition Rule When two events A and B are disjoint, we can use the addition rule for disjoint events from Chapter 14: P(A or B) = P(A) + P(B) However, when our events are not disjoint, this earlier addition rule will double count the probability of both A and B occurring. Thus, we need the General Addition Rule.

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide The General Addition Rule (cont.) General Addition Rule: For any two events A and B, P(A or B) = P(A) + P(B) – P(A and B) The following Venn diagram shows a situation in which we would use the general addition rule:

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Example: Traffic Stops Police report that 78% of drivers stopped on suspicion of drunk driving are given a breath test, 36% a blood test, and 22% both tests. What is the probability that a randomly selected DWI suspect is given: a. A test? b. A blood test or a breath test, but not both? c. Neither test?

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Example continued…. 1. Draw a picture (Venn Diagram) 2. Figure out what you want to know in words. 3. Translate words to equations. P(A)=.78P(B)=.36 P(A∩B)=.22

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Example continued…. What is the probability that the suspect is given a test? P(A or B)=P(A) + P(B) – P(A∩B) =.92 What is the probability that the suspect gets either a blood test or a breath test but NOT both? P(A and not B or B and not A)=P(A)P(not B) + P(B)P(not A) =.70 What is the probability that the suspect gets neither test? P(neither test) = 1 – P(either test) = 1 – P(A or B) = 1 –.92 = 0.08 Interpret your result in context!!

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Example Two psychologists surveyed 478 children in grades 4, 5, and 6 in elementary schools in Michigan. They stratified their sample, drawing one third each from rural, suburban, and urban schools. They asked the students whether their primary goal was to get good grades, to be popular, or to be good at sports. The results are shown in the table below. 1. What is the probability of randomly picking a girl? 2. What is the probability of randomly picking a student whose goal is to be popular? 3. What is the probability of picking a student who is a boy and wants to be good at sports? 4. What is the probability of picking a student who is a girl or wants to get good grades? GradesPopularSports Boy Girl

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide It Depends… Back in Chapter 3, we looked at contingency tables and talked about conditional distributions. When we want the probability of an event from a conditional distribution, we write P(B|A) and pronounce it “the probability of B given A.” A probability that takes into account a given condition is called a conditional probability.

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide It Depends… (cont.) To find the probability of the event B given the event A, we restrict our attention to the outcomes in A. We then find the fraction of those outcomes B that also occurred. Note: P(A) cannot equal 0, since we know that A has occurred.

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Example: Girls and Sports In a recent study it was found that the probability that a randomly selected student is a girl is.51 and is a girl and plays sports is.10. If the student is female, what is the probability that she plays sports?

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Example: Boys and Sports The probability that a randomly selected student plays sports if they are male is.31. What is the probability that the student is male and plays sports if the probability that they are male is.49?

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Probabilities from two way tables StuStaffTotal American European Asian Total ) What is the probability that the driver is a student?

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Probabilities from two way tables StuStaffTotal American European Asian Total ) What is the probability that the driver drives a European car?

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Probabilities from two way tables StuStaffTotal American European Asian Total ) What is the probability that the driver drives an American or Asian car? Disjoint?

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Probabilities from two way tables StuStaffTotal American European Asian Total ) What is the probability that the driver is staff or drives an Asian car? Disjoint?

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Probabilities from two way tables StuStaffTotal American European Asian Total ) What is the probability that the driver is staff and drives an Asian car?

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Probabilities from two way tables StuStaffTotal American European Asian Total ) If the driver is a student, what is the probability that they drive an American car? Condition

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Probabilities from two way tables StuStaffTotal American European Asian Total ) What is the probability that the driver is a student if the driver drives a European car? Condition

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Example: Conditional Probability After School Activities At George Washing HS, after school activities can be classified into three types: athletic, fine arts, and other. The following table gives the number of students participating in these types of activities by grade: 9th10th11th12thTotal Athletics Fine Arts Other Total

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Example continued… Is it true from the table that: There are th graders participating in athletic? The number of senior participating in fine arts activities is 125? There are 435 students in fine arts activities? GWHS has 410 juniors? The total number of students is 1600?

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Example continued… What is the probability that a randomly selected student is a senior athlete? P(senior athlete)= What is the probability that the selected student is an athlete, given that the student is a senior? P(athlete|senior)=

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Example: GFI Switches A GFI (ground fault interrupt) switch turns off power to a system in the event of an electrical malfunction. A spa manufacturer currently has 25 spas in stock, each equipped with a single GFI sw9itch. Two different companies supply the switches and some of the switches are defected as summarized in the table: NondefectiveDefectiveTotal Company Company Total187

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Example continued… Given: E = event that GFI switch in selected spa is from company 1 D = event that GFI switch in selected spa is defective Find: P(E) P(D) P(E and D) = P(E∩D)

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Example continued… P(E)=15/25 =.6 P(D) = 7/25 =.28 P(E and D) = P(E∩D) = 5/25 =.2 Now suppose that testing reveals a defective switch. How likely is it that the switch came from the first company? P(company 1 given defective switch)=P(E|D) =5/7 =.714

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide The General Multiplication Rule When two events A and B are independent, we can use the multiplication rule for independent events from Chapter 14: P(A and B) = P(A) x P(B) However, when our events are not independent, this earlier multiplication rule does not work. Thus, we need the General Multiplication Rule.

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide The General Multiplication Rule (cont.) We encountered the general multiplication rule in the form of conditional probability. Rearranging the equation in the definition for conditional probability, we get the General Multiplication Rule: For any two events A and B, P(A and B) = P(A) x P(B|A) or P(A and B) = P(B) x P(A|B)

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Independence Independence of two events means that the outcome of one event does not influence the probability of the other. With our new notation for conditional probabilities, we can now formalize this definition: Events A and B are independent whenever P(B|A) = P(B). (Equivalently, events A and B are independent whenever P(A|B) = P(A).)

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Independent ≠ Disjoint Disjoint events cannot be independent! Well, why not? Since we know that disjoint events have no outcomes in common, knowing that one occurred means the other didn’t. Thus, the probability of the second occurring changed based on our knowledge that the first occurred. It follows, then, that the two events are not independent. A common error is to treat disjoint events as if they were independent, and apply the Multiplication Rule for independent events—don’t make that mistake.

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Depending on Independence It’s much easier to think about independent events than to deal with conditional probabilities. It seems that most people’s natural intuition for probabilities breaks down when it comes to conditional probabilities. Don’t fall into this trap: whenever you see probabilities multiplied together, stop and ask whether you think they are really independent.

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Drawing Without Replacement Sampling without replacement means that once one individual is drawn it doesn’t go back into the pool. We often sample without replacement, which doesn’t matter too much when we are dealing with a large population. However, when drawing from a small population, we need to take note and adjust probabilities accordingly. Drawing without replacement is just another instance of working with conditional probabilities.

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Tree Diagrams A tree diagram helps us think through conditional probabilities by showing sequences of events as paths that look like branches of a tree. Making a tree diagram for situations with conditional probabilities is consistent with our “make a picture” mantra.

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Tree Diagrams (cont.) Figure 15.4 is a nice example of a tree diagram and shows how we multiply the probabilities of the branches together:

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Example: Management has determined that customers return 12% of the items assembled by inexperienced employees, whereas only 3% of the items assembled by experienced employees are returned. Due to turnover and absenteeism at an assembly plant, inexperienced employees assemble 20% of the items. Construct a tree diagram or a chart for this data. What is the probability that an item is returned? If an item is returned, what is the probability that an inexperienced employee assembled it? P(returned) = 4.8/100 = P(inexperienced|returned) = 2.4/4.8 = 0.5 ReturnedNot returned Total Experienced Inexperienced Total

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Reversing the Conditioning Reversing the conditioning of two events is rarely intuitive. Suppose we want to know P(A|B), but we know only P(A), P(B), and P(B|A). We also know P(A and B), since P(A and B) = P(A) x P(B|A) From this information, we can find P(A|B):

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide Reversing the Conditioning (cont.) When we reverse the probability from the conditional probability that you’re originally given, you are actually using Bayes’s Rule.

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide What Can Go Wrong? Don’t use a simple probability rule where a general rule is appropriate: Don’t assume that two events are independent or disjoint without checking that they are. Don’t find probabilities for samples drawn without replacement as if they had been drawn with replacement. Don’t reverse conditioning naively. Don’t confuse “disjoint” with “independent.”

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide What have we learned? The probability rules from Chapter 14 only work in special cases—when events are disjoint or independent. We now know the General Addition Rule and General Multiplication Rule. We also know about conditional probabilities and that reversing the conditioning can give surprising results.

Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison-Wesley Slide What have we learned? (cont.) Venn diagrams, tables, and tree diagrams help organize our thinking about probabilities. We now know more about independence—a sound understanding of independence will be important throughout the rest of this course.