EE351K_HW5-A Soln Fall 2009

EE351K_HW5-A Soln Fall 2009 - EE351K Solution to HW 5-A...

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Unformatted text preview: EE351K Solution to HW 5-A Prqb‘lemJ' ' text messaging is 0.2, the probability the customer will order unlimited Internet access is 0.3, and the probability, the customer will order unlimited text messaging or unlimited Internet access or both is‘ Before-the next, ‘ (a). both unlimited text messaging and mflimited Internet access (b) neither unlimited text messaging nor-unlimited Internet access. Also answer the following: . ‘ - 1 TI ~ - . I ’ - -_ . customer enters the store, calculate the probability that he/she will'order: ~ I .4 I .Internetaccess; - i f - . ' Dallmé, EQQ‘LCK‘I—Tfijl Dr‘EJJ—hs' ,uw\¢_wt~4«cl"4‘e—z«+ MAIL$$€I~SIV~I ! ‘ , '\\ . (\r 'l h ' , r‘ . ructmevtsdr‘ QLQ£S g 3 ,__, . zuzwr'flaj Endww bucks 'C 4‘ exifizofzvsh ggucaocts ‘ as (temps, (a) 'gTREf‘CfiltépaC‘TfD-‘TB- \> L33} .4- Ptfl ;__ PC 3:. (1+5 (RV/iowz‘t" ' .—I — t A. V l - t , 2L ?C .err. -Ovll/ (4.311 dQ-lfipwgtgg {44.1. pmhagitéletas -T2M¢.:tt £W3“.;§i¢de—wfi gogustt- J ' Ptzn'fll‘gclx' 'PCfn$§:-O.(¢... . i ' ' . , I I “(.103 F8550? [_ 9 LG) Pcfm 2‘ » w amt At a cell phone store past observationlindicates that the probability that a new customer-will order unlimited- (0) If‘the customer orders unlimited texturessaging', what is the probability he/she- will not order unlimited __..=.._.,. .._...... .i....;_._.r... _\._._.__._.,__ _,_.l .....-.._.....,i_-. . . .. . .. . ..-.._ ..~~........_.. . ._.... _.. ._...._........_.. _,_..___._ ._._._.... m-.. . -_...__.._. Mfr—1.... __..-........._,. _ T 4 Pmblema' Consider three events A, B, and c', where; P(C)_=9.5', P(CuA).='o.6, P(AhEnC)=_o.os, P(A)=o.7,' ‘ P(An~1§)=0.25., P(An§né)=o.1, P(Bn2§)=o.4 . .. Calculating the following probabilities: ' ' - _ _(a) P(AUB) . - (b) P((AUC)nB) - I (c). chA) (d) P(AUBUC|AhJ§) I ' (e) HEIAnB) "k .__..___ ___.. _- _._ ___... ,__. ._____._L__ 9 ‘4 53.0.UJqf‘ é \ea 3- {Mk'cc/ fiwzdl'sxrtwc; nag _Q‘I£Vc_i§§ a \Q g V . ‘ a . I , ~ . q W MAmamme} wake (5" d\;g&o~\v_—\~ flgzw+S~ “S. & Mona“ va '. ' ' A“ 0 Again“ +109— 7 W 'Corwsc-=+?O UK ed {Uszvx w~m go.“ 'szlr}¢~wu (ML 44“ g Pro.§2csb\l\IJx-\‘a_1 agJothLJr-p'cx LDC—H4 agau' 5mg a? gxgd‘awcf- QUQLs/LITE J v ’ Q1$ 'lkV\9\\‘Ccér9.¢/°. e—-—-—-~.r.n.-_.mm_.mu=—.._..._-.._,_i .___i. .- _ i.._____._.=i_~. .. . i 1 ~ i i; I 1‘ l E ’ . . was P.(A'U\3§=.C9i2 +0"; +uiaj+ar05+a05z 5):“; . // / // 3 . _ ' -' gig/f n L.) 2 01.1 +a‘1; ‘l—0‘ x ‘ . I..AU‘\3 ! '//' //‘ ‘ ‘ “ i . I . . I _ {will i/('[/’(‘(‘\ E7 .| 2’:— k, 4 ‘ _ , :1 ' Au; ‘ —. ~ , —— - g L ‘3 .(AUGBA Es ‘ . a 3 (.3 Pam/H: PCZnM "Facflatlfl (wk 7 “we. : ‘91!» 0.7a 0727"”3“. Vii-$4M n\ ij-al é/ I: V ’ " I =3 - .E - ' A -\ Low W l-‘rVEUL‘ Ana; . PC (Awaoéhmflgy) I cflA i, ’ . ' a . -=. AflPJAM'QI-L+glosfl@'zg . j -”m N PC An\3_3.~ W. fl —-._____._______—-__r__:_\q. ! 4// / , ” _' / - PC-Afl%3 a.?.-I-o,os O 25 "‘ * ‘ / / K” n l // = ' I ' ' '. ' . . f . . (Auauc) Ana” fins . B -'r '_ x Ana)... P J ' l g . . ' “I CBfllAmamg p{@4\ _ I .‘ - I 1 : .. M ‘Fc'q‘ngx a V 005 l 4 V: r : Er til—A . -' // ' ,_ .- I I - ' " i I . I i v ' lf ... M'fl (,,_._\{’___. fl ' ‘ -. I 3/6 I "—'—"——‘-'“-"‘f@———“m7w—mfl j*—4'5-———“——“f—“— *‘ ‘ ‘ "_”_ "T :__.__.:.._i______;*__________;‘ ~___'_;w____;.___~.._-_.____._.' ,. . -' I i--;' ' j l WT“ l -i._._. _-- __ _ .__i .m, Problem.3 Consider the system below where eaclf subsystem is reliability of 0:70, find the reliability pf the qverall system” 9.: (49.73559) ,2 0,44% R: @ri—Bxpfiwoca—31' 0 . 5433 A Q '. ‘elé‘le—‘aé>(x—,0 a ..._, .4? $04479 °( £9 .. we: g-ql‘iqudgdlim‘afl 44335“ (734739 1 -———————-45/6«—~— ‘represented by e'switch.-If each subsystem has 5.. ii I. l i ! r a _______ i I A: . Problem.4 I l i . . 52 cards, with four Aces per deck. Each person independently and randomly draws one card from their respective 1: ' ’ deck. How large must n (the number of people participating in this expen '1' A . one Ace being drawn is 0.97 p . I. I _ I igmMZkr4/'MFI - ‘0le «AL: hujkmzzk L_=IK1!3j4)"u-,w a. n) 9 Thus 2.. Nada «PH/VD U. ‘ ' ~' ' . ‘ ‘ ' I A "film {ngdrk :Ab-AL/fibnwf Aw'ckrg immpgwdemffi—J: I \ C/L13:\>L‘A2_\=I\>(Az}3 auuA FC_AV\B:__::;_ .‘VW—‘x .c'vPL‘ Na? \aCIJV‘ié-‘j'DC'x/L’tp, . I . .‘ bz-Hhe. e...\)2~.-}L“.: I “urge. £44“ Mfiosuh draw (LL-wk I J ~“‘“‘”““~‘*‘I “i? .om-.~ma~~+.w+ we Arie 1?: Male-As mam-m l‘“ Wink beam»; fi—kgnwfiudfl floured NET o‘vHOULE C1 ' _ - ~ 7A PW +P‘AL3+9CA55+.rM« PM»; 5449?. We .7; _Al~A'A?.C?;r>?—.mHAmg>AA:-j.. 0...? . fiJLL‘Q «LAKE Q‘C?_ 7+ (9" P L -.av ' — ,R '_ _ . F“. (3 m .Ach) : PL Am ALnAg Mn @3201 u ' . 3‘ h ' _“ dJ~u£§~ - A - A “30% PC will"): \2— .. "Pléwo AQQJS : - [52:73.13 ' Suppose you have 11 people, and each has their own deck of shuffled playing cards. Each de‘ck consiéts 6f ’ 'iment)-be so that the 'probability.of at least- ____________ _ j 2 ‘ ;_ _ _ ___§/_5__-_; ~~ Problems Consider three white boxes and two black boxes. Each box contains CDMA cell phone chips made by . !-, ‘1“ ‘ ' two different companies, Company A and Company B. In thewhite boxes, the ratio of chips made by Company A ‘ to Company B is 3 to l(i.e., there are three chips made by Company A for every one made by Company B). In the black boxes, the ratio of‘chips. made by Company A to Company B is 1 to 4(i.e., there is "1 chip made by Company A for every four chips made by Company B). While blindfolded you select a box at random and draw out a chip at ' random. Examination indicates that it is made by Company A. What is the probability that the chip was drawn from ' a white box? . '4 ' ' - ' - P $0M.) Jet 1M7 " LJD Rani-<9... PC sfleagimj - lobes; " _"\§C@IW\ .: . FCC?th 3. are .I- , .i g ' i ‘ . i a“??? .?.<:_w (GS) .1 P <@\ w) PM -- Mu..." - """’"““-—~<. ' M. _ ec‘cotmi WW) + P‘(@ 1 BR) flag I r 3 ' - -,<-~:~«=> + 1%C%béxéxf 4,f;'{; :— w_——‘._ ..__-.—.___~_J______ ._ __ t: 45 . W. ' .' E7 1‘“ \Mfl) i l j. i i a l i .I l i ll _mmn___ n_nn_nm_nwnwvml ...
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This note was uploaded on 01/31/2010 for the course EE 351k taught by Professor Bard during the Fall '07 term at University of Texas at Austin.

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EE351K_HW5-A Soln Fall 2009 - EE351K Solution to HW 5-A...

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