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MATH 165 - Ch. 13 – Homework Name (Print) Student ID # Independent and the Multiplication Rule (1) Lost Internet sites Internet sites often vanish or move, so that references to them can’t be followed. In fact, 13 % of Internet sites referenced in major scientific journals are lost within two years after publication. If a paper contains seven Internet references, what is the probability that all seven are still good two years later? What specific assumptions did you make to calculate this probability? P(all seven are still good two years later)= [P(one still good two years later)]7 = (1 − 0.13)7 = (0.87)7 = 0.377254794 ≈ 0.3773 The General Addition Rule (2) Photo and Video Sharing Photos and videos have become an important part of the online social experience, with more than half of Internet users posting photos or videos online that they have taken themselves. Let A be the event an Internet user posts photos that they have taken themselves, and B be the event an Internet user posts videos that they have taken themselves. Pew Research Center finds that P(A) = 0.52, P(B) = 0.26, and P(A or B) = 0.54. (a) Make a Venn diagram similar to Figure 13.4 showing the events {A and B}, {A and not B}, {B and not A}, and {neither A nor B}. (b) Describe each of these events in words. (c) Find the probabilities of all four events and add the probabilities to your Venn diagram. P(A or B) = P(A) + P(B) − P(A and B) = 0.52 + 0.26 − P(A and B) = 0.54 P(A and B) = P(A) + P(B) − P(A or B) P(A and B) = 0.52 + 0.26 − 0.54 = P(a user who posts both) P(A ∩ B) = 0.24 P(A and not B) = 0.52 − 0.24 = P(a user who posts photos but not videos) P(A ∩ not B) = 0.28 P(B and not A) = 0.26 − 0.24 = P(a user who posts videos but not photos) P(B ∩ not A) = 0.02 P(neither A nor B) = 1 − 0.54 = P(a user who posts neither photos nor videos) P(neither A nor B) = 0.46 Online Videos No Videos Total —————————————————————————– Photos 0.24 0.28 0.52 No Photos 0.02 0.46 0.48 —————————————————————————Total 0.26 0.74 1 Conditional Probability (3) Photo and Video Sharing In the setting of the question # 2, what is the conditional probability that an Internet user posts photos that they have taken themselves, given that they post video that they have taken themselves? P (A ∩ B) = 0.24 P (a user who posts phot| a user who posts video) = = 0.24 0.26 = 0.923076923 ≈ 0.9231 P (a user who posts both) P (a user who posts video) (4) Computer games Here is the distribution of computer games sold by type of game: Game type Probability ——————————————————————————————————————– Strategy 0.354 Role playing 0.139 Family Entertain. Shooters Children’s 0.127 0.109 0.057 Other 0.214 What is the conditional probability that a computer game is a role-playing game, given that it is not a strategy game? P (B | A) = P (A and B) P (A) P (role playing game | not strategy game) = = P (not strategy and role playing) P (not strategy) 0.139 1 − 0.354 = 0.215170278 ≈ 0.2152 Check Your Skills (5) An instant lottery game gives you probability 0.02 of winning on any one play. Plays are independent of each other. If you play three times, the probability that you win on none of your plays is about P(no winning on three lotteries) = [P(no winning on each lottery)]3 = (1 − 0.02)3 = (0.98)3 = 0.941192 ≈ 0.9412 (6) The probability that you win on one or more of your three plays of the game in previous exercise is about P(winning on one or more of three lotteries) = 1− P(no winning on three lotteries) = 1−0.941192 (from the previous exercise) = 0.058808 ≈ 0.0588 (7) An athlete suspected of having used steroids is given two tests that operate independently of each other. Test A has probability 0.9 of being positive if steroids have been used. Test B has probability 0.8 of being positive if steroids have been used. What is the probability that at least one test is positive if steroids have been used? P(at least one of them is positive) = 1− P(both negative) P(both negative) = (1 − 0.9) · (1 − 0.8) = 0.02 P(at least one of them is positive) = 1− P(both negative) = 1 − 0.02 = 0.98 (8) What is the distribution of doctorates conferred by field and sex? Here are the counts from the most popular fields in 2012. The physical sciences include mathematics and computer and information sciences; the life sciences include agricultural sciences/natural resources, biological/biomedical sciences, and health sciences; and the social sciences include psychology. Degree Male Female Total ——————————————————————————————————————– Engineering 6, 527 1, 883 8, 410 Physical sciences 6, 393 2, 551 8, 944 Life sciences Social sciences Education 5, 331 3, 488 1, 501 6, 698 4, 861 3, 297 12, 029 8, 349 4, 798 Humanities 2, 654 2, 847 5, 501 Other 1, 496 1, 425 2, 921 ———————————————————————————————– Total 27, 390 23, 562 50, 952 (a) Choose a doctoral recipient at random from this group. The probability that the recipient is male is about P(the recipient is male) = 27, 390 = 0.537564766 ≈ 0.5376 50, 952 (b) The conditional probability that the recipient is male, given that the degree is in education, is about P(the recipient is male | in Education) = = P (Eduation and male) P (Eduation) 1, 501 = 0.312838682 ≈ 0.3128 4, 798 (c) The conditional probability that the degree is in education, given that the recipient is female, is about P(the recipient is in Education | female) = = P (f emale and Eduation) P (f emale) 3, 297 = 0.139928698 ≈ 0.1399 23, 562 (d) Let A be the event that the degree is in engineering and B the event that the recipient is a female. The proportion of engineering doctorates conferred on females is expressed in probability notation as A: The degree is in Engineering B: The recipient is a female P (The proportion of engineering doctorates conferred on female) = P (Engineering and female) P (Engineering) = P (A ∩ B) P (A) = P (B|A) Answer is (c). (9) Choose an American adult at random. The probability that you choose a woman is 0.52. The probability that the person you choose has never married is 0.25. The probability that you choose a woman who has never married is 0.11. The probability that the person you choose is either a woman or never married (or both) is therefore about Adult Never Married Married Total —————————————————————————– Woman 0.11 0.41 0.52 Man 0.14 0.34 0.48 —————————————————————————Total 0.25 0.75 1 P (woman or never married) = P (woman) + P (never married ) − P (woman and never married) = 0.52 + 0.25 − 0.11 = 0.66 (10) Universal blood donors People with type O-negative blood are referred to as universal donors, although if you give type O-negative blood to any patient you run the risk of a transfusion reaction due to certain antibodies present in the blood. However, any patient can receive a transfusion of O-negative red blood cells. Only 7.2 % of the American population have O-negative blood. If 10 people appear at random to give blood, what is the probability that at least one of them is a universal donor? P(at least one of them is a universal donor) = 1− P(All 10 people are not universal donor) P(the person is not universal donor) = 1− P(the person is a universal donor) = 1 − 0.072 = 0.928 P(at least one of them is a universal donor) = 1− P(All 10 people are not universal donor) = 1 − (0.928)10 = 1 − 0.473674234 = 0.526325765 ≈ 0.5263