1. Section 1.1, Slide 1 Copyright © 2014, 2010, 2007 Pearson Education, Inc. Section 13.3, Slide 1 13 Probability What Are the Chances?

2. Copyright © 2014, 2010, 2007 Pearson Education, Inc. Section 13.3, Slide 2 Conditional Probability and Intersection of Events 13.3 Understand how to compute conditional probability. Calculate the probability of the intersection of two events.

3. Copyright © 2014, 2010, 2007 Pearson Education, Inc. Section 13.3, Slide 3 Conditional Probability and Intersection of Events 13.3 Use probability trees to compute conditional probabilities. Understand the difference between dependent and independent events.

4. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.3, Slide 4 Conditional Probability

5. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.3, Slide 5 Example: Assume that we roll two dice and the total showing is greater than nine. What is the probability that the total is odd? Solution: This sample space has 36 equally likely outcomes. We will let G be the event “we roll a total greater than nine” and let O be the event “the total is odd.” Therefore, G = {(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)}. Conditional Probability (continued on next slide)

6. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.3, Slide 6 We now seek all pairs that give an odd total – the diagram below shows that there are two. Conditional Probability

7. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.3, Slide 7 Conditional Probability

8. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.3, Slide 8 Example: A survey of college graduates compared starting salaries to majors. If we select a graduate who was offered between $40,001 and $45,000, what is the probability that the student has a degree in the health fields? Conditional Probability (continued on next slide)

9. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.3, Slide 9 Solution: Let R be the event “graduate received a starting salary between $40,001 and $45,000” and H the event “student has a degree in the health fields.” We want to find P(H | R). Conditional Probability

10. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.3, Slide 10 The Intersection of Events

11. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.3, Slide 11 Example: Assume your professor has written questions on 10 assigned readings on cards and you are to randomly select two cards and write an essay on them. If you have read 8 of the 10 readings, what is your probability of getting two questions that you can answer? Solution: Let A be “you can answer the first question;” and B be “you can answer the second question.” (continued on next slide) The Intersection of Events

12. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.3, Slide 12 We need to calculate We may compute the following probabilities: The Intersection of Events

13. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.3, Slide 13 Probability Trees

14. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.3, Slide 14 Example: Again assume your professor will randomly ask questions on 2 of 10 assigned readings. If you have read 8 of the 10 readings, what is the probability that you will get questions on two readings that you have not done? Solution: Let A be “you can answer a question;” and N be “you cannot answer a question.” We will continue with a probability tree. (continued on next slide) Probability Trees

15. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.3, Slide 15 We have and. From below we see that the probability of receiving questions on the readings you haven’t done is 0.02. Probability Trees

16. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.3, Slide 16 Dependent and Independent Events

17. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.3, Slide 17 Example: Assume we roll a red and a green die. Are the events F, “a five shows on the red die,” and G, “the total showing on the dice is greater than 10,” independent or dependent? Solution: The three outcomes (5, 6), (6, 5), and (6, 6) give a total greater than 10, so We have F = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}, and G ∩ F = {(5, 6)}. (continued on next slide) Dependent and Independent Events

18. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.3, Slide 18 We may now compute Since the events are dependent. Dependent and Independent Events

19. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.3, Slide 19 Example: To get a dorm room, a student will have to draw a card that has the name of one dormitory, X, Y, or Z, and also has a two-person room number or an apartment number. 30% of the available spaces are in X, 50% in Y, and 20% of the spaces are in Z. Half of the available spaces in X are in rooms, 40% of Y’s spaces are in rooms, and 30% of the spaces in Z are in rooms. (continued on next slide) Dependent and Independent Events

20. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.3, Slide 20 a) Draw a probability tree to describe this situation. b) Given that a student selects a card for dormitory Y, what is the probability that she will be assigned to an apartment? c) What is the probability that a student will be assigned an apartment in one of the three dormitories? (continued on next slide) Dependent and Independent Events

21. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.3, Slide 21 (a) We will think of the assignments happening in two stages. First, the student is assigned a dormitory, and then she is assigned either a room or an apartment. (continued on next slide) Dependent and Independent Events

22. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.3, Slide 22 (b) Looking at the branch that is highlighted in red in, we see that P(Apartment|Y) = 0.60. (continued on next slide) Dependent and Independent Events

23. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.3, Slide 23 (c) We see that the event “A student is assigned an apartment” is the union of three mutually disjoint Dependent and Independent Events A student is assigned an apartment. subevents with probabilities 0.15, 0.30, and 0.14. So, the probability that the student is assigned an apartment is 0.15 + 0.30 + 0.14 = 0.59.

24. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.3, Slide 24 Example: It is estimated that 2% of the employees at a company use a certain drug, and the company is giving a drug test that is 99% accurate in identifying the drug users. What is the probability that if an employee is identified by this test as a drug user, the person is innocent? Solution: Let D be the event “the person is a drug user” and let T be the event “the person tests positive for the drug.” We wish to compute (continued on next slide) Dependent and Independent Events

25. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.3, Slide 25 We create a probability tree for this situation and compute the probabilities shown below. (continued on next slide) Dependent and Independent Events

26. Copyright © 2014, 2010, 2007 Pearson Education, Inc.Section 13.3, Slide 26 We may now compute the desired probability. Dependent and Independent Events