1. Chapter Probability © 2010 Pearson Prentice Hall. All rights reserved 3 5

2. Section 5.2 Probability Rules 5-2 © 2010 Pearson Prentice Hall. All rights reserved

3. 5-3 © 2010 Pearson Prentice Hall. All rights reserved

4. Two events are disjoint if they have no outcomes in common. Another name for disjoint events is mutually exclusive events. 5-4 © 2010 Pearson Prentice Hall. All rights reserved

5. We often draw pictures of events using Venn diagrams. These pictures represent events as circles enclosed in a rectangle. The rectangle represents the sample space, and each circle represents an event. For example, suppose we randomly select a chip from a bag where each chip in the bag is labeled 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. Let E represent the event “choose a number less than or equal to 2,” and let F represent the event “choose a number greater than or equal to 8.” These events are disjoint as shown in the figure. 5-5 © 2010 Pearson Prentice Hall. All rights reserved

6. 5-6 © 2010 Pearson Prentice Hall. All rights reserved

7. The probability model to the right shows the distribution of the number of rooms in housing units in the United States. Number of Rooms in Housing Unit Probability One0.010 Two0.032 Three0.093 Four0.176 Five0.219 Six0.189 Seven0.122 Eight0.079 Nine or more0.080 Source: American Community Survey, U.S. Census Bureau EXAMPLE The Addition Rule for Disjoint Events (a) Verify that this is a probability model. All probabilities are between 0 and 1, inclusive. 0.010 + 0.032 + … + 0.080 = 1 5-7 © 2010 Pearson Prentice Hall. All rights reserved

8. Number of Rooms in Housing Unit Probability One0.010 Two0.032 Three0.093 Four0.176 Five0.219 Six0.189 Seven0.122 Eight0.079 Nine or more0.080 (b) What is the probability a randomly selected housing unit has two or three rooms? P(two or three) = P(two) + P(three) = 0.032 + 0.093 = 0.125 5-8 © 2010 Pearson Prentice Hall. All rights reserved

9. (c) What is the probability a randomly selected housing unit has one or two or three rooms? Number of Rooms in Housing Unit Probability One0.010 Two0.032 Three0.093 Four0.176 Five0.219 Six0.189 Seven0.122 Eight0.079 Nine or more0.080 P(one or two or three) = P(one) + P(two) + P(three) = 0.010 + 0.032 + 0.093 = 0.135 5-9 © 2010 Pearson Prentice Hall. All rights reserved

10. 5-10 © 2010 Pearson Prentice Hall. All rights reserved

11. 5-11 © 2010 Pearson Prentice Hall. All rights reserved

12. 5-12 © 2010 Pearson Prentice Hall. All rights reserved Suppose that a pair of dice are thrown. Let E = “the first die is a two” and let F = “the sum of the dice is less than or equal to 5”. Find P(E or F) using the General Addition Rule. EXAMPLEIllustrating the General Addition Rule

13. 5-13 © 2010 Pearson Prentice Hall. All rights reserved

14. 5-14 © 2010 Pearson Prentice Hall. All rights reserved

15. Complement of an Event Let S denote the sample space of a probability experiment and let E denote an event. The complement of E, denoted E C, is all outcomes in the sample space S that are not outcomes in the event E. 5-15 © 2010 Pearson Prentice Hall. All rights reserved

16. Complement Rule If E represents any event and E C represents the complement of E, then P(E C ) = 1 – P(E) 5-16 © 2010 Pearson Prentice Hall. All rights reserved

17. According to the American Veterinary Medical Association, 31.6% of American households own a dog. What is the probability that a randomly selected household does not own a dog? P(do not own a dog) = 1 – P(own a dog) = 1 – 0.316 = 0.684 EXAMPLE Illustrating the Complement Rule 5-17 © 2010 Pearson Prentice Hall. All rights reserved

18. The data to the right represent the travel time to work for residents of Hartford County, CT. (a) What is the probability a randomly selected resident has a travel time of 90 or more minutes? EXAMPLE Computing Probabilities Using Complements Source: United States Census Bureau There are a total of 24,358 + 39,112 + … + 4,895 = 393,186 residents in Hartford County, CT. The probability a randomly selected resident will have a commute time of “90 or more minutes” is 5-18 © 2010 Pearson Prentice Hall. All rights reserved

19. (b) Compute the probability that a randomly selected resident of Hartford County, CT will have a commute time less than 90 minutes. P(less than 90 minutes) = 1 – P(90 minutes or more) = 1 – 0.012 = 0.988 5-19 © 2010 Pearson Prentice Hall. All rights reserved