#### Transcript Section 3.5-3.6 - Pendleton County Schools

Sta220 - Statistics Mr. Smith Room 310 Class #9 Section 3.5-3.6 Problem: To develop programs for business travelers staying at convention hotels, Hyatt Hotels Corp. commissioned a study on executives who play golf. The study revealed that 55% of the executives admitted that they had cheated at golf. Also, 20% of the executives admitted that they had cheated at golf and had lied in business. Given that an executive had cheated at golf, what is the probability that they executive also had lied in business? The event probabilities weβve been discussing give the relative frequencies of the occurrences of the events when the experiment is repeated a very large number of times. Such probabilities are often called unconditional probabilities. 3.5 Conditional Probability A probability that reflects such additional knowledge is called the conditional probability of the event. Example: Weβve seen that the probability of an even number (event A) on a toss of a fair die is ½. But suppose weβre given the information that on a particular throw of the die the results was a number less than or equal to 3 (event B). Would the probability of observing an even number on that throw of the die still be equal to ½? Reduced sample space for the die-toss experiment: given that event B has occurred Copyright © 2013 Pearson Education, Inc.. All rights reserved. To get the probability of event A given that event B occurs, we divide the probability of the part A that falls within the reduced sample space B, namely π(π΄ β© B), by the total probability of the reduced sample space, namely, P(B). P(A|B) = π π΄β©B π π΅ = π 2 π 1 + 2 +π 3 = 1 6 3 6 = 1 3 β’ Note: π π΅ π΄ = π π΄β©π΅ π π΄ [ We assume π(π΄) β 0] Problem: To develop programs for business travelers staying at convention hotels, Hyatt Hotels Corp. commissioned a study on executives who play golf. The study revealed that 55% of the executives admitted that they had cheated at golf. Also, 20% of the executives admitted that they had cheated at golf and had lied in business. Given that an executive had cheated at golf, what is the probability that they executive also had lied in business? Solution Let A: {Executive had cheated at golf} B: {Executive had lied in business} Then π(π΄) = .55 and π(π΄ β© π΅) = .20 We want P(B|A)=? π π΄β©π΅ .20 π π΅π΄ = = = .364 π π΄ .55 Thus, given that a certain executive had cheated at golf, the probability that the executive also had lied in business is .364. Problem: Many medical researchers have conducted experiments to examine the relationship between cigarette smoking and cancer. Consider an individual randomly selected from the adult male population. Let A represent the event that the individual smokes, and let π΄π denote the complement of A (the even that the individual does not smoke). Similarity, let B represent the even that the individual develops cancer, and let π΅π be the complement of that event. Then the four sample points associated with the experiment are shown in Figure 3.15. Use these sample point probabilities to examine the relationship between smoking and cancer. Solution A method in determining whether the given probabilities indicate that smoking and cancer are related is compare the conditional probability that an adult male acquires cancer given that he smokes with the conditional probability that an adult male acquires cancer given that he does not smoke [ i.e. compare π(π΅|π΄) with π(π΅|π΄πΆ ) So we are going to reduce the sample space A corresponding to adult male smokers. Two conditional probabilities: π π΅π΄ = π π΄β©π΅ π π΄ and π π΅π π΄ = π π΄β©π΅π π π΄ π π΄ = π π΄ β© B + π π΄ β© π΅πΆ = .05 + .20 = .25 So now π π΅π΄ = .80 π π΄β©π΅ π π΄ = .05 .25 = .20 and π π΅π π΄ = π π΄β©π΅ π π π΄ = .20 .25 = These two numbers represent the probabilities that an adult male smoker develops cancer and does not develop cancer, respectively. π π΅ π΄π π π΅π π΄π π π΄π β© π΅ .03 = = = .04 π π π΄ .75 π π΄π β© π΅π .72 = = = .96 π π π΄ .75 These are the conditional probabilities of an adult male nonsmoker developing cancer and not developing cancer. The conditional probability that an adult male smoker develops cancer (.20) is five times the probability that a nonsmoker develop cancer (.04). This relationship does not imply that smoking causes cancer, but it does suggest a pronounced link between smoking and cancer. 3.6 The Multiplicative Rule and Independent Events The probability of an intersection of two events can be calculated with the multiplicative rule. Problem: In a classic psychology study conducted in the early 1960s, Stanley Milgram performed a series of experiment in which a teacher is asked to shock a learner who is attempting to memorize word pairs whenever the learner gives the wrong answer. The shock levels increase with each successive wrong answer. {Unknown to the teacher, the shocks are not real.} Two events of interest are A: {The teacher βappliesβ a sever shock (450 volts).} B: {The learner protest verbally prior to receiving the shock.} A recent application of Milgramβs shock study revealed that P(B) = .5 and P(A|B) = .7. On the basis of the information, what is the probability that a learner will protest verbally and a teacher will apply a severe shock? Solution We want to calculate π(π΄ β© B). π π΄ β© B = π π΅ π π΄ π΅ = .5 .7 = .35 Thus, about 35% of the time the learner will give a verbal protest and the teacher will apply severe shock. Problem: A county welfare agency employs 10 welfare workers who interview prospective food stamp recipients. Periodically, the supervisor selects, at random, the forms completed by two workers and subsequently audits them for illegal deductions. Unknown to the supervisor, three of the workers have regularly been giving illegal deductions to applicants. What is the probability that both of the workers chosen have been giving illegal deductions? Solution Let N: Non-Illegal I: Illegal Copyright © 2013 Pearson Education, Inc.. All rights reserved. Tree diagram 6 1 πΌβ©I= = 90 15 Thus, there is a 1-in-15 chance that both workers chosen by the supervisor have been giving illegal dedication to food stamp recipients. We have showed that the probability of event A may be substantially altered by the knowledge that an event B has occurred. However, this will not always be the case; in some instances, the assumption that event B has occurred will not alter the probability of event A at all. Problem: Consider the experiment of tossing a fair die, and let A = {Observe an even number.} B = {Observe a number less than or equal to 4.} Are A and B independent events? Venn diagram for die-toss experiment Copyright © 2013 Pearson Education, Inc.. All rights reserved. Solution π(π΄) = 1/2 π(π΅) = 2/3 π(π΄ β© π΅) = 1/3 Assuming B has occurred, we see that the conditional probability that A given B is 1 π π΄β©π΅ 1 3 π π΄π΅ = = = = P(A) 2 π π΅ 2 3 Thus, assuming that the occurrence of event B does not alter the probability of observing an even number, the probability remains ½ . Therefore, the events A and B are independent. Problem An experiment results in one of five sample points with the following probabilities: π(πΈ1 ) = .22, π(πΈ2 ) = .31, π(πΈ3 ) = .15, π(πΈ4 ) = .22 πππ π(πΈ5 ) = .1. The following events have been defined: π΄: {πΈ1 , πΈ3 } π΅: {πΈ2 , πΈ3 , πΈ4 } πΆ: {πΈ1 , πΈ5 } Find each of the following probabilities: a) P(A) b) P(B) c) π(π΄ β© π΅) d) π π΄ π΅ e) π(π΅ βͺ πΆ) f) π(πΆ|π΅) g) Consider each pair of events A and B, A and C, and B and C. Are any of the pairs of events independent? Why? Solution A. P(A) = .22 +.15 = .37 B. P(B) = .31+.15 +.22 = .68 C. π(π΄ β© π΅) = .15 D. π π΄ π΅ = .15 .68 = .221 E. P(B\capC) = 0 F. P(C|B) = 0/.68 = 0 G. For any of the pairs of events to be independent, the following must hold: π(π΄ β© B) = π π΄ π(π΅) π(π΄)π(π΅) = .37(.68) = .2516 β π(π΄ β© B) = .15 π(π΄)π(πΆ) = .37(.32) = .1184 β π(π΄ β© C) = .22 π(π΅)π(πΆ) = .68(.32) = .2176 β π(π΅ β© C) = 0