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

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.

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.

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

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)

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

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

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)

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

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

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

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

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

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

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 Probability Trees

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

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

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

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

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

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

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) = (continued on next slide) Dependent and Independent Events

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 So, the probability that the student is assigned an apartment is = 0.59.

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

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

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