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### solutions03

Course: MATH 302, Spring 2011
School: UBC
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302.102 Math Fall 2010 Solutions to Assignment #3 Let A be the event that a randomly selected person is a man so that Ac is the event that a randomly selected person is a woman. Let B be the event that a person is colour blind. We are told that P {B | A} = 0.05 and P {B | Ac } = 0.0025. (a) If men and women each make up the same proportion of the population, then P {A} = 0.5 and P {Ac } = 0.5, so that Bayes Rule...

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302.102 Math Fall 2010 Solutions to Assignment #3 Let A be the event that a randomly selected person is a man so that Ac is the event that a randomly selected person is a woman. Let B be the event that a person is colour blind. We are told that P {B | A} = 0.05 and P {B | Ac } = 0.0025. (a) If men and women each make up the same proportion of the population, then P {A} = 0.5 and P {Ac } = 0.5, so that Bayes Rule tells us that P {A | B } = P {B | A} P {A} P {B | A} P {A} (0.05)(0.5) = = c } P {Ac } P {B } P {B | A} P {A} + P {B | A (0.05)(0.5) + (0.0025)(0.5) 20 . = = 0.952381. 21 1. (b) If there are twice as many women as men in this population, then P {A} = 1/3, P {Ac } = 2/3, and Bayes Rule tells us that P {A | B } = P {B | A} P {A} P {B | A} P {A} (0.05)(1/3) = = c } P {Ac } P {B } P {B | A} P {A} + P {B | A (0.05)(1/3) + (0.0025)(2/3) 10 . = = 0.909091. 11 Let A be the event that a randomly selected chip is good so that Ac is the event that a randomly selected chip is bad. Let B be the event that a chip passes the cheap chip test. We are told that P {A} = 0.8 and P {Ac } = 0.2. Since all good chips pass the cheap chip test, P {B | A} = 1, but since 10% of bad chips also pass the cheap chip test, we have P {B | Ac } = 0.1. (a) Using Bayes Rule we nd P {A | B } = P {B | A} P {A} (1)(0.8) P {B | A} P {A} = = c } P {Ac } P {B } P {B | A} P {A} + P {B | A (1)(0.8) + (0.1)(0.2) 40 . = = 0.975609. 41 2. (b) If the company sells all chips which pass the cheap chip test, then the percentage of chips sold that are bad is simply 1 40 P {Ac | B } = 1 P {A | B } = 1 =. 41 41 Let A be the event that a randomly selected person has the disease so that Ac is the event that a randomly selected person does not have the disease. Let B be the event that that the laboratory test on the blood sample returns a positive result. We are told that P {A} = 0.01 so that P {Ac } = 0.99. We are also told that P {B | A} = 0.95 and P {B | Ac } = 0.02. Using Bayes Rule we nd P {A | B } = P {B | A} P {A} (0.95)(0.01) P {B | A} P {A} = = c } P {Ac } P {B } P {B | A} P {A} + P {B | A (0.95)(0.01) + (0.02)(0.99) 95 . = = 0.324232. 293 3. 4. The key to solving this problem is to realize that the event A from the previous problem needs to change. No longer is the patient who walks into the doctors oce randomly selected from the population. Hence, we take A to be the event that the patient has the disease. The doctors opinion is that P {A} = 0.3. If the blood test result is positive, then Bayes Rule implies P {A | B } = P {B | A} P {A} P {B | A} P {A} (0.95)(0.30) = = c } P {Ac } P {B } P {B | A} P {A} + P {B | A (0.95)(0.30) + (0.02)(0.70) 285 . = 0.953177. = 299 5. For j = 1, 2, . . ., let Aj be the event that a 3 appears on roll j , let Bj be the event that a 1 or a 6 appears on roll j , let Cj be the event that a 2 appears on roll j , let Dj be the event that a 4 appears on roll j , and let Ej be the event that a 5 appears on roll j . If W denotes the event that Player A wins, then the Law of Total Probability implies P {W } = P {W | A1 } P {A1 } + P {W | B1 }P {B1 } + P {W | C1 } P {C1 } + {W P | D1 } P {D1 } + P {W | E1 } P {E1 } Since Player A wins immediately if a 3 appears on the rst roll, P {W | A1 } = 1, and since Player A loses immediately if a 1 or a 6 appears on the rst roll, P {W | B1 } = 0. Furthermore, since the die is fair, P {A1 } = P {C1 } = P {D1 } = P {E1 } = 1/6. Therefore, P {W } = 11 1 1 + P {W | C1 } + P {W | D1 } + P {W | E1 } . 66 6 6 The next observation is that P {W | C1 } = P {W | D1 } = P {W | E1 }. If Player A rolls a 2 initially, then the only way for Player A to win is if a 2 is rolled before a 3 on subsequent rolls. Similarly, if Player A rolls a 4 initially, then the only way for Player A to win is if a 4 is rolled before a 3 on subsequent rolls, and if Player A rolls a 5 initially, then the only way for Player A to win is if a 5 is rolled before a 3 on subsequent rolls. Since the die is fair, all of these events have the same probability. Thus, 11 P {W } = + P {W | C1 } . 62 The remaining step is to compute P {W | C1 }. It is reasonable to guess that P {W | C1 } = 1/2. This is because if a 2 is rolled initially, then the only way for Player A to win is if a 2 appears before a 3 on subsequent rolls. Since only a 2 or 3 matter (all other numbers cause a re-roll), and since 2 and 3 are both equally likely, we must have P {W | C1 } = 1/2. Alternatively, we can sum up an innite series as follows. (continued) Let Fj = Bj Dj Ej be the event that 1, 4, 5, or 6 appears on roll j . Then, using the fact that the results of subsequent rolls are independent, we nd P {W | C1 } = P {C2 } + P {F2 } P {C3 } + P {F2 } P {F3 } P {C4 } + = = = 1 41 ++ 6 66 2 1 1+ + 6 3 4 6 2 3 2 2 1 + 6 + 1 1 6 1 2/3 1 =. 2 P {W } = 5 1 11 +=, 6 22 12 5 7 =. 12 12 This implies that and so the probability that Player A loses is P {W c } = 1 6. (a) Since 1 x2 dx = x1 1 = 0 (1) = 1 we see that taking c = 1 makes f a legitimate probability density. (b) Since 1 x1 dx = ln(x) 1 =0= we see that there is no such c that makes f a legitimate probability density. (c) Using integration-by-parts twice (see Prerequisite Review Handout) gives x2 ex dx = x2 ex 2xex 2ex . Therefore, since 1 1 1 xe 2 x dx = x e 2 x 2xe x 2e x 1 = [e1 2e1 2e1 ] [e1 +2e1 2e1 ] = e 5e1 , we see that taking c = (e 5e1 )1 = e/(e2 5) makes f a legitimate probability density. (d) As in (c), using integration-by-parts twice gives 0 x2 ex dx = x2 ex 2xex 2ex 0 = 0 (2) = 2 and so we see that taking c = 1/2 makes f a legitimate probability density. (e) Since 0 0 xex dx = [xex ex ] = (0 1) (0 0) = 1 by see that taking c = 1 makes f a legitimate probability density. Note that xex 0 for x 0. This means that if we multiply xex by a negative number it will always be non-negative. Hence, f (x) = xex for x 0 is a non-negative function that integrates to 1. (f ) Note that the function xex assumes both positive and negative values when 1 x 1. This means that there is no single value of c that could be multiplied by xex to make it strictly non-negative. Hence, there is no possible value of c that makes f a legitimate probability density.
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6.tmi Sastitmi: dag: Fallstjrn Persnufornfn Orar Eignarsambnd:Sagnirnareiga,hafa,vera me EignarfornfnOrar Frumlag+sgn+andlag 1)Konanhjlparstelpunni nf.+gf.gf. nf.+gf.gf.Venjulegorar 2)Stelpanhjlparkonunni Andlag+sgn+frumlagfugorar 3)Stelpunnihjl
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10.tmi Sastitmi: dag: kk.no.oglo.nendingarnf.et. Meiraumlo.nendingar(tvkv) Brottfallsrhljsrstofnitvkvrano.og lo.Tvkvlo.kk.nendingar nf.et.kk. kvk. et.nf.viturmaurviturkona gulpeysa et.nf.gulurdiskur ertilbrigiafurendingunni Sama beygingardmi og gulur,
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11. tmi Sasti tmi Brottfall srhljs r stofni tvkvra no. og lo. dag Sagnir Nt sagnaUm sagnir Sgn (ft.: sagnir) = sagnor (so.) Sagnbeyging Mismunandi endingar Innskotsstafur j milli stofns og endinga sumum myndum N srhljavxl: B-vxlNafnhttur Nafnhttu
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19. tmi Sasti tmi: reglulegar veikar sagnir Httarsagnir dag: Staaratviksor Samandregnar myndir spurnarsetningum OrarStaaratviksor Atviksor (ao.) beygjast ekki. Staaratviksor tkna: dvl sta (rest at a place), hreyfingu til staar (movement to a place)
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2.tmi Sastitmi dag Inngangur Upprifjunfrhaustmisseri. BeygingnafnoraII: KarlkynKk:gestur Et.nf.gestur f.gest gf.gesti ef.gests Ft.nf.gestir f.gesti gf.gestum ef.gesta hestur hest hesti hests hestar hesta hestum hestaKarlkynsor Algeng karlkynsnafnor
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3.tmi Sastitmi: dag: Njarbeygingarkk.:kkII Beygingnafnora Meiraumkarlkyn:kk.IIogkk.III Kvenkyn:kvk.IISrhljavxlstofni 1) Bvxl:,oy 2)Flknarihljavxl: a)ea a&gt;=Avxl a&gt;e=Bvxl Klofning(breaking)i&gt;j,i&gt;ja b)jija Alltafigf.et.ogaref.et.Hljavxl:jija nf. f
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4.tmi Sastitmi: dag: Njarbeygingarkk.ogkvk. Beygingnafnora: Stigbreytinglsingarora. Meiraumkvenkyn:kvk.III HvorugkynKvk.IIIkindskei mynd Et.nf.kind skei mynd f.kind skei mynd gf.kind skeiar myndar ef.kindar myndir Ft.nf.kindur skeiar skeiar myndir f
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5.tmi Sastitmi: dag: Njarbeygingarkvk.oghvk. Aeinsumstigbreytingulsingarora. framumstigbreytingulsingarora.Stigbreyting Srstkmyndaflo.ernotutilabera saman. Stofnlo.+viskeyti+endingar Viskeyti: Mst:(a)r Est:(a)st Mistigogefstastig Srstakarendingarf
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6.tmi Sastitmi: dag: Stigbreyting. Notkungreinis. bendingarfornfn.Greinir Nafnor geta veri kvein (indefinite) ea kvein(definite). kvein no. hafa viskeyttan greini inn. Greinirinn er settur aftan vi beygingarendinguno.:hesturinn. kveinno.hafaengangrein
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7.tmi Sastitmi: dag: Notkungreinis bendingarfornfn(fn.) Notkunveikrarbeygingarlo.bendingarfornafnis kk. Et.nf. s f. ann gf.eim eirri ef. ess Ft.nf. eir r f. gf. eim ef. eirra kvk. hvk. s a a v eirrar ess au r au eim eim eirraeirrabendingarfornafniess
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8.tmi Sastitmi: dag: bendingarfornfn Notkunveikrarbeygingarlsingarora Meiraumnotkunveikrarbeygingarlsingarora Eintluogfleirtlunafnor persnulegarsagnirNotkunveikrarbeygingarlo. 1) egar lo. stendur me no. me greini: Rauibllinnerbilaur Annahittigmlukonu
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9.tmi Sastitmi: Notkunveikrarbeygingarlsingarora Eintluogfleirtlunafnor persnulegarsagnir Aukafallsliir dag:Persnulegarsagnir Frumlagstendurnf. Beygingasamrmi:Sagnirlagasigafrumlagi tluog persnu. Andlgstandaaukafalli(f.,gf,eaef.): Beygingasamrmi:Lo.
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10.tmi Sastitmi: dag: persnulegarsagnir Aukafallsliir Meiraumaukafallslii Notkuna NotkunsjlfurAukafallsliir Falloraukafallinfallvalds,.e.n sagnareaforsetningarsemstrir aukafallinu. Aukafallsliirmerkjatma,staea magn(quantity,number,measurement): tmali
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SE201G: Mlfri II 1. mars 201113. tmi KVEIN FORNFN bir, sumirINNGANGUR Annar, feinir, enginn, neinn, mis, bir, srhver, hvorugur, sumir, hver og einn, hvor og nokkur, einhver. Sj lka: allur, hvor tveggjaFORNFN UM TVO/ FORNFN UM RJ + HEILD &gt;2: allirH
Uni. Iceland - HUMANITIES - ÍSE202G
SE201G: Mlfri II 3. mars 201114. tmi KVEIN FORNFN hvorugur, annarFORNFN UM TVO/ FORNFN UM RJ + HEILD &gt;2: allirHLUTIsumir/nokkrir einn/hinn annar hver enginnNEITUN 2: bir 2annar/hinn annar hvor 1-1 1: einn/einhverhvorugur 1-1 enginnhvorugur, bls
Uni. Iceland - HUMANITIES - ÍSE202G
SE201G: Mlfri II 8. mars 201115. tmiGagnkvmi (reciprocity): gera e- vi hvort ea hvert annaannar, bls. 169 annar annan rum annars nnur anna ara anna annarri ru annarrar annars arar arar nnur nnur arir ararum annarrahver, bls. 111 hver hvern hverjum
Uni. Iceland - HUMANITIES - ÍSE202G
SE201G: Mlfri II 10. mars 201116. tmiSagnir sem enda st/ mimyndarsagnir (middle voice verbs)Myndun - merking Engin tengsl (): anda andast; fara farast; lta ltast; taka takast; ykja ykjast Sgn ekki til: Pll feraist miki. ferast (&lt;- fer), ekki *fera st
Uni. Iceland - HUMANITIES - ÍSE202G
SE201G: Mlfri II 15. mars 201117. tmiMyndun og beyging vitengingarhttar I, bls. 210-212.Myndun vitengingarhttar I Vitengingarhttur I = vh.I er myndaur af nafnhtti. Nafnhttur kalla heyra telja bja fara ba 1. p. et. vh.I kalli heyri telji bji fari bi
Uni. Iceland - HUMANITIES - ÍSE202G
SE201G: Mlfri II 17. mars 201118. tmiMyndun og beyging vitengingarhttar II, bls. 210-212.Vitengingarhttur II Vitengingarhttur II er notaur aalsetningum kurteisisskyni: Gtir veri svo vnn a hjlpa mr? tti g a hjlpa honum? Mtti g bija ig um asto? Vildir v
Uni. Iceland - HUMANITIES - ÍSE202G
SE201G: Mlfri II 22. mars 201119. tmiNotkun vitengingarhttar aalsetningum bls. 187-190.Myndun vitengingarhttar I Vitengingarhttur I = vh.I er myndaur af nafnhtti. Nafnhttur kalla heyra telja bja fara ba 1. p. et. vh.I kalli heyri telji bji fari bi E
Uni. Iceland - HUMANITIES - ÍSE202G
SE201G: Mlfri II 24. mars 201120. tmiNotkun vitengingarhttar aukasetningum bls. 191-193.Bein ra bein ra Bein ra: Jn segir: Kalli er gur strkur. bein ra Jn segir a Kalli s gur strkur.Forsendur beinnar ru Vitengingarhttur aukasetningum: aalsetning me
Uni. Iceland - HUMANITIES - ÍSE202G
SE201G: Mlfri II 29. mars 201121. tmiMyndun og notkun lsingarhttar tarSterkar sagnir, 108-110 1. fl.: 2. fl.: bta ba beit bei bitum bium biti bEi brotibrjta brautbrutum 3. fl.: a. syNgja brega drekka b. detta hverfa o.fl.sng sungumdattduttumsuN
Uni. Iceland - HUMANITIES - ÍSE202G
SE201G: Mlfri II 31. mars 201122. tmiolmynd: myndun og notkunTilgangur olmyndar Tilgangur hennar er lsa verknainum og helst a fela gerandann sem er ALLATF hgt a bta vi. g las bkina. -&gt; (germynd; alltaf gerandi) Bkin (kvk.et.) var lesin (af X). olmynd
Uni. Iceland - HUMANITIES - ÍSE202G
SE201G: Mlfri II 5. aprl 201123. tmipersnuleg olmynd: myndun og notkunSgn strir gufalli X kastai teningnum (gf.et.). Teningnum var kasta (af X). kasta = lh.t.hvk.et. X kastai teningunum (gf.flt.). Teningunum var kasta (af X). a sem gerist: gufalli
Uni. Iceland - HUMANITIES - ÍSE202G
SE201G: Mlfri II 7. aprl 201124. tmiLsingarhttur tar sem lsingarorInnskot fr v sast Margar sagnir taka me sr tv fll. Dmi: gefa + gf. og f. beina andlagi= persnan er gf., beina andlagi f. leyna + f. og gf. beina andlagi= persnan er f., beina andlagi g
UC Irvine - PSYCH - 141H
ClinicalHealthLecture#21/6 A. B. QuiznextTue:GreenbergReading00:34C. A. B.II.Otherimportantconcepts: BiopsychosocialModel(GeorgeEngels) SystemsTheory(vonBertalanffy1968)&gt;postingonline i. Natureisorganizedintermsofahierarchyofunitsreflectingacontinuum
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Rutgers Business School INTRODUCTION TO FINANCIAL MANAGEMENT (33:390:300:02)SPRING 2011Class Meetings: Monday &amp; Thursday 10:20 11:40 a.m. Beck Hall 253 Professor: Office: Office Hours: Phone: E-Mail: Jin-Mo Kim, Ph.D. in Finance 121 Levin Monday 1:00 3:
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