midterm1-practice

midterm1-practice - KEEP Tms Midterm 1(Stat 126 October 5...

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Unformatted text preview: KEEP Tms. Midterm 1 (Stat 126) October 5, 2005 The midterm has 10 problems. Each problem is worth 3 points. Show all the work and write in the space provided. If more space is needed, write on the back of the pages. Problem 1. In how many ways can 10 people be seated in a row if there are 5 men and 5 women and no 2 men or 2 women 'can sit next to each other? Solution: '11":th (M=f"lon\W§V\/omm) 921% MWMWMw...Mw or WMWMWM..-WM 'T 1". \‘ MogP.ss'.L.-ms: 5'35! 51-5, TotoQ: g.sI.-sl_ 7 Problem 2 A dance class consists of 18 students, 10 women and 8 men. If 4 men and , 4 women are to be chosen and then paired off, how many results are possible? <2: 4:)" T T To hclvosz‘l #5 4 41 ‘ woo rmx‘ Mmﬂ W Solution: Problem 3. How many ways are there to distribute n distinct balls into 1" distinct urns? 7 (An urn can contain any number of n balls.) Solution: Problem 4. A die is rolled continually until a 6 appears, at which point the experiment stops. What is the sample space of this experiment? Let E, be the eVent that n rolls are necessary to complete the experiment. What outcomes of the sample space are contained in En? What is ( °° En)"'? ' n=1 Solution: 5 =’l (zl,x1‘-_-,x,,,,_,'é,l,3<;= 1’213i4'5—(m7’4‘t 0“) ’j‘ (IL,9C1,9C3|.~) § ’t 6mm I], B, =‘l (11,9<z,-~’9cm_,’6) ,1: =1,2.3,‘4l3’l Wu 7, n. (0‘75 (11,12 (CD) EX : 4 6 W WW4” “:1 Problem ,5. A total of 28‘ percent of American males smoke cigarettes, 7 percent smoke cigars, and 5 percent smoke both cigars and cigarettes. What percentage of American males smoke cigars but not cigarettes, or cigarettes but not cigars? Solution: C zgray‘iombvicw maklcmoldig (Ago-{33 P(C) :. 0e07- 34 Wk ma. gametes W?) = mg PMOR) = 005 \Nom’cf ' PMCns‘mm‘n m) = Psz“) + Prc°n m = 19(6) — Wm?) + 19(2) — Pan {2) = 0.0? + 0.;28 — at 0.03 = 0325 Problem 6. A forrest contains 15 elks, of which 3 are captured, tagged, and then released. A certain time later 5 of the 15 elks are captured. What is the probability that 2 of these 5 have been tagged? Solution: ch°°W’%2 Gull—Oil 3+0? “lionsvﬂg 3 01.4"“? 42 um‘l’W 4 (3H?) (4:) 5* ”wwlwlam W‘ ”W61 4?“ Problem 7. A fair coin is flipped twice. Let E be the event that both tosses land on Heads, and F be the event that at least one toss lands on Heads. Compute P(E|F). Solution: 3 £4 HH,HTITH,TT§ E :4 HH‘y' ‘F:),HT,TH,’HH§ 1‘ P EnFl P(Hm 4 ”47,1 P(EIF) : ( :. - 3— -— 3 WP) P(HTfl'HIHHl —- Problem 8. A laboratory test is 90 percent effective in detecting a certain disease when it is present. The test is also “false positive” for 2 percent of the healthy persons tested. If 1 percent of the population actually has the disease, what is the probability that a . person has the disease given that the test result is negative? Solution: D :& plasma 71w.) dismsef] , P(TlDl = 0.9 T sAhs‘t l5 Ywiﬁjﬁb (Pl T lDC) C 002 WD) = 001 wont: _ ' PtT‘lp) Hp) P(T°]DlP{Dl Pl Dch) = c ’- c c c. HT) P(T [13mm + HT in) P(D ‘ 0.1. tan/1 I: “M = (2004.02 0-10.01 +Oi93. 0.99 P(F) :P(}:16\p,a.d-y) P‘Lb,MT)—+ VLF lmn’ma.) lylnwedgb > WPWbm) meme) +¥(FJQ;L.aAw) F/MhU‘o,) = P(Flueoay) = WFlE). Problem 9. Let E be the event that a randomly selected U.S. student goes to UN C, and F be the event that a randomly selected U.S. student takes a basic course on probability theory. If you had to construct a mathematical model for events E and F, would you assume that they were independent events? Explain your reasoning. Solution: E clg‘geo-h) UNC‘I IF 74% Wit/kw} Ben answers “ﬂexible at long 0a M are MSonq‘ohaoramd for; 9352421“ Imam ct Pl IE)< PIF) becwe x4: Mﬂharmqu,sk an 6' 2m; (Meal? +0 Lg {FumUNC uOl'ud’l .3 abodar‘h' Collage. m3 ‘3 5a,»; (M7UOL‘é2la w Misha M (ALUaJl 01+} CORRQ4 heme q smadlu leafy-lien s} WE hung) m‘babd/Cl’v _ + :mloq—hlahwnimla 42'5“; “1 a. nerd» owl?) (play. Wmabo lurid oKnrscmrice 03qu “ed one Vh [-(Alm MW We go; no 53)qu Problem 10. Let E, F and be 3 events with P(E) = . , (F) = .3 and P(G) = .1. Oxbow}- Compute P(E U Fc U G) when E, F and G are independent. +111 :40 Solution: Om “0‘6: W‘EUF‘U (x) = WEN P(F‘) + We) —- P(EDF‘I— mgr: ca- P(F‘ne) + WED renew = P(E) + PM + New — acme) -P(ElP((sl — l’LP‘) we @mu + P(E) Ptp‘me Second wcua/ P(EUF‘UG) = 2L— P((EUF‘U6l“)-‘= L— P/E‘CIFne‘) =. /L- mecmﬂws‘) = 1—.8-.3-.9 = ORM J (hour-"chant; ...
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