Mathematics 102 Solutions for HWK 22 SECTION 7.6 P (R 1 )=0.05

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1 Mathematics 102 Solutions for HWK 22 SECTION 7.6 p 368 Problem 3. Assume R 1, R 2, and R 3 are mutually exclusive events and we have P (R 1 )0.05 P (R 2 )0.6 P (R 3 )0.35 P (Q R 1 )0.4 P (Q R 2 )0.3 P (Q R 3 )0.6 Find P (R 1 Q). Solution. Notice that the three events R 1,R 2,R 3 are not only mutually exclusive (as given) but also exhaustive (since their probabilities add to 1). In other words, whatever the outcome is, one of the three events R 1,R 2,R 3 must occur. We have exactly the setup described in Bayes Theorem. Aschematic diagram of the sample space might look like this: Page 1 of 10 A. Sontag April 22, 2004

2 To begin applying Bayes Theorem, use the given data to set up a tree diagram as follows. Recall that P (R 1 Q) P (R 1 Q). Bayes theorem lets us find the numerator and denominator P (Q) we need by using branches of the tree that end in Q. Inparticular, the numerator we need is the probability for the top branch in the tree, the branch showing that R 1 occurs and Q occurs. The denominator is the sum of the probabilities for the three (red) branches that end in Q. So P (R 1 Q) P (R 1 Q) P (Q) (0.5)(0.4) (0.5)(0.4)+(0.6)(0.3)+(0.35)(0.6) p 368 Problem 6. Under the same assumptions as in Problem 3, find P (R 1 Q). Solution. Using the result from Problem 3, we have P (R 1 Q) 1 P (R 1 Q) Page 2 of 10 A. Sontag April 22, 2004

3 p 368 Problem 8. Suppose you have three jars with the following contents: 2 black balls and 1 white ball in the first, 1 black ball and 2 white balls in the second, and 1 black ball and 1 white ball in the third. One jar is to be selected, and then 1 ball is to be drawn (at random) from the selected jar. If the probabilities of selecting the first, second, or third jar are 1 2, 1 3, and 1 6, respectively, find the probabilities that if a white ball is drawn, it came from the third jar. (In other words, find the probability that the ball came from the third jar, given that the ball is white.) Solution. Begin with a tree diagram: Let W be the event that the ball is white. Let J1,J2,J3bethe events that the ball comes from jar 1, jar 2, jar 3, respectively. Using the three branches that end in white, we find Also, P (W )( 1 2 )(1 3 )+(1 3 )(2 3 )+(1 6 )(1 2 ) P (J3 W )( 1 6 )(1 2 ) Page 3 of 10 A. Sontag April 22, 2004

4 Therefore P (J3 W ) P (J3 W ) P (W ) p 368 Problem 9. Employment Test. A manufacturing firm finds that 70% of its new hires turn out to be good workers and 30% become poor workers. All current workers are given a reasoning test. Of the good workers, 80% pass it; 40% of the poor workers pass it. Assume that these figures will hold true in the future. if the company makes the test part of its hiring procedure and only hires people who meet the previous requirements and also pass the test, what percent of the new hires will turn out to be good workers? Solution. For a potential new hire chosen at random we have the following tree diagram: Let G be the event that the new hire turns out to be a good worker. Let T be the event that the new hire passes the test. From the tree we find P (T )(0.7)(0.8) + (0.3)(0.4) Moreover P (G T )(0.7)(0.8) So the probability that a potential new hire turns out to be a good worker, given that s/he passed the test, is P (G T ) P (G T ) P (T ) Thus one could expect approximately 82.35% of the new hires to be good workers. Page 4 of 10 A. Sontag April 22, 2004

5 p 368 Problem 16. Appliance Reliability. Companies A, B, and C produce 15%, 40%, and 45%, respectively of the major appliances sold in a certain area. In that area, 1% of the company A appliances, 1.5% of the company B appliances, and 2% of the company C appliances need service within the first year. Suppose a defective appliance is chosen at random; find the probability that it was manufactured by Company A. Solution. It appears that defective should be interpreted as needs service within the first year. So we have the following tree diagram, for a randomly chosen appliance sold in the specified area. Let D be the event that the randomly chosen appliance is defective (i.e. needs service within a year), and let A be the event that it was manufactured by Company A. Using the three branches that yield a defective outcome, find P (D) (.15)(.01) + (.40)(.015) + (.45)(.02) Also, P (A D) (.15)(.01) Therefore the required probability is P (A D) P (A D) P (D) Page 5 of 10 A. Sontag April 22, 2004

6 p 368 Problem 17. Same as Problem 16, except for Company B. Solution. We already computed that P (D) Consulting the tree again, see that P (B D) (.40)(.015).006. Therefore the probability we want to find is given by P (B D) P (B D) P (D) p 368 Problem 20. Shipping Errors. Three shipping terminals are operated by Krag Corporation, namely a land terminal, an air terminal, and a sea terminal. The land terminal handles 50% of Krag s cargo, the air terminal handles 40%, and the sea terminal handles the remaining 10%. For the cargo shipments handled by the land terminal, 2% of the shipping documents contain errors. The corresponding figures for the air and sea terminals are 4% and 14%, respectively. Krag s internal auditor randomly selects one set of shipping documents, ascertaining that the set selected contains an error. Which of the figures (a) 0.02, (b) 0.10, (c) 0.25, (d) 0.50 gives the probability that the error occurred in the land terminal? Solution. For arandomly chosen shipping document, the data in the problem gives us the following tree. If we let L denote the event that the document came from the land terminal and E the event that it contains an error, then we need to find P (L E). Using the tree diagram to apply Bayes Page 6 of 10 A. Sontag April 22, 2004

7 Theorem, we find P (L E) P (L E) P (E) (.5)(.02) (.5)(.02)+(.4)(.04)+(.1)(.14) so the correct answer would be (c). p 368 Problem 22. Toxemia Test. A magazine article described a new test for toxemia, a disease that affects pregnant women. To perform this test, the woman lies on her left side and then rolls onto her back. The test is considered positive if there is a 20 millimeter rise in her blood pressure within one minute. The article gives the following probabilities, where T represents having toxemia at some time during the pregnancy, and N represents a negative test. P (T N) 0.90 P (T N )0.75 Assume that P (N )0.11, and find (a) P (N T ); (b) P (N T ). Solution. We are given enough information to make the following tree (for a randomly selected pregnant woman). The probability that the woman has toxemia at some point is given by P (T )(.89)(.1) + (.11)(.75) Page 7 of 10 A. Sontag April 22, 2004

8 For the probability that a woman tests negative and has toxemia at some time, we have P (N T )(.89)(.1).089. Part (a) asks for the probability of a negative test given some toxemia, which is P (N T ) P (N T ) P (T ) Part (b) asks for the probability of a positive test given toxemia, which is P (N T ) P (N T ) P (T ) (.11)(.75) It looks to me that this isn t a terribly good test. Page 8 of 10 A. Sontag April 22, 2004

9 p 368 Problem 32. Seat Belt Effectiveness. A federal study showed that in 1990, 49% of all those involved in a fatal car crash wore seat belts. Of those in a fatal crash who wore seat belts, 44% were injured and 27% were killed. For those not wearing seat belts, the comparable figures were 41% and 50%, respectively. Find the probability that a randomly selected person who was killed in a car crash was wearing a seat belt. Solution. The data supplied gives us the following tree (for those involved in a fatal crash). Let K denote the event that a randomly selected person involved in a fatal crash was killed, and let B denote the event s/he was wearing a seat belt. Then we find P (B K) P (B K) P (K) (.49)(.27) (.49)(.27)+(.51)(.5) p 368 Problem 33. Using the same data as in Problem 32, find the probability that a randomly selected person who was unharmed in a fatal crash was not wearing a seat belt. Solution. Let U denote the event that a randomly selected person involved in a fatal crash was unharmed. This time we re looking for P (B U), which is P (B U) P (B U) P (U) (.51)(.09) (.49)(.29) + (.51)(.09) Page 9 of 10 A. Sontag April 22, 2004