Solution cap 2- Estatistica aplicada e probabilidade para engenheiros

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Sfl(tlort 2·1 
!H. tet a and bdeootea partsbm-eand below tJt.especificatio n. resp;x:ti\'ely. 
S = {aoo.oob.aba.abb.baa.bab.bba.bbbl 
2-;J. tet adenoteanacoeptable pD\\'et s-.:.pply. 
tet/. m.andcdenotea powers.:pply th.at OOss functioMI, minot, or oosmeticcrror. respccth\'!ly. 
S= {a.f,m.c} 
2·5-· tet y aodndenotea website ti'.at oontainssnddoes not contain banner ads. 
n.e sample space is the set of a lI possible SO!QUences of y and n of lengtlt :l4. An o:ampJe o-.:.tcome in t~.e sample spaoe isS = {yyJmynm•mtynynrumyynnyy}. 
2·7. Sis the sample spaceofloo possible two d ig it integ6S. 
2·11. S = {l.O,l.l,l.2, ... 14.0} 
2·lj.. S = {0,1.2 •... .) in milliseoonds. 
PRESS I 2 
, 2 3 • I ~ 8 
2·17. Lct caodbdeooteoonoect and bu;;y, respecthl!ly. Th.e:nS = {c. be. bbc.bbbc. bbbbc . .. .). 
b) 
c) 
A B 
c 
d) 
•l 
1~1. .a) l.ct S = tlt.i! oonncg.athoe integers from o to tf!.e largest integer that can bedispla~'ed by tlti'sca!e. 
l.d X deootethe wcight. 
.4 is tltet'\l:nt that X> u Sis thef!\"Cnt tf!.at X s 15 C is thef!\"Cnt tf!.at8 s X <12 
S = ~O.l , :1..3 ... .) 
b) s 
<:) n<X~l50rh:i,l3.14,l5~ 
d ) Xs uor~O.l,!!,. .• , ll} 
e) S 
f) .4. VCoontain; t.be values of X sn.ch thst: X ;e 8 
Ttms(.4 UC)' contains th.evalu.esof X sll:Cht~.s t: X< 8or ~0 . 1 . !!, .... 7~ 
g) " 
h) 8' oontains tf!.e\'alu.esof X sn.<:hthat X> 1.;. n.erefore. 8' (')Cis t.Jt.e empty set. n.ey ha\oe oooutoomes inoom.monor fl . 
i) 8 rtCis the C\~t 8 s X <12. Tlterefore,.4. U (B (')C) is thct'\'t'llt X~ 8or {8. 9, 10 .... } 
2·:!j. ldd .and odcoote a distorted bit and one that is oot distortcd(odeootesokay). resp.."-cth"C!y. 
a) 
dddd,dodd,oddd,oodd, 
S = dddo, dodo)oddo)oodo) 
ddod,dood,odod,oood, 
ddoo, dooo, odoo, oooo 
b) No, for V(ampleA.1(\~ = {dddd.dddo,ddOO.,ddoo} 
dddd,dodd, 
c) A,= dddo,dodo 
ddod,dood 
ddoo,dooo 
oddd,oodd, 
d) 
..t;= oddo,oodo) 
odod,oood, 
odoo,oooo 
e) A1(\4.21'\"*ll'\"4 = {dddd) 
0 '41!"\4>)U'43'"""') = ldddd.dcdd.dddo.oddd.ddod.oodd.ddooi 
2·::!..5· ld Pand N denote posith~.and ncg.ati\'e, res)X\.'1hdy. 
Tf!.esample spaoe is {PPP, PP!'l, Pfo.'P. NPP, PNN,:\'Pfo,',NNP,NNN). 
(a) A= lPPPI 
(b) 8 = {N]I.I]I.'} 
{<:) .lt r.B =~ 
{d) .4 UB = {PPP.NNN} 
:!·!!] . .a) A'r.B'=lO,B=l0,.4UB =92 
b) 
' 
...._, 
..... 
e 
G 
2·29 . .a) A'= {X I X" 7~-5) 
b) B'=<x lxss~.s) 
0 
<:) A()B:{x (.;.:l,s <X< 7-2-.S} 
d) AUB=~X ~X >O~ 
• 
• 
G 
., • 
~-.Jl . l.etgdeoote.a good board, rna board with miootdefects . .andj .a board with major defects. 
a) S= {gg.gm.g}.mg.mm.mj.Jg,Jm.JJ> 
b) S=(gg.gm.gJ,mg.mm.mJ.Jg.Jm~ 
c 
2-·.'}3. a) 
batch 2 
72.5 
b,jch 1 
b) 
bodch 2 
52.5 
tl&tCil1 
c) 
betcll 2 
52.5 
72.5 botch 1 
d) 
72.5 
~-!).5. From fl!.e multiplication rule. th.eanswer is .;x3x4x~ = 120. 
~-!)7. From fl!.emultiplicationrulc,3x4x-'x4 = 144. 
2·!)9. From the multiplication ruleancif:qU;Stion ~-!. fl!.eanswer is 5!5! = 14400. 
2-4l. a) Ftomtqt;:ation2-4. th-enumherofs.ample.sofs.i7.efive is[ 1 4 0 ] = 140 1 = 416965528 
5 5!1 35! 
b) Tht:rea.rt'lO '<NI)'Sofselectingoneooooonformingc~.ipandth;erearc [I 30] = IJO! = ) 1358880 ''<I)'S of selecting four 
4 4 !126! 
oonfonnir.gchips. T~.<refore. tt"'numberolsamplts thatoontaincxactlyone nonoonfonnir.gchipis I 0 x[ :30] = I I 3 588800. 
c) n~ numberofsamplesthatoontainat least one oonoonformir.gchip is tit<! total number of samples [ ~ 40 J minus tt~ numberofsamplts 
[
130] [140 ] [1 30) 140! 130! tt~toontainno nonoonfonnir.gchips 5 .Thatis 5 - 5 = 5 113 51
-
5
!1
251 
13072 1752. 
7! 
a) --= 2 lsequDOESarcpos.slble. 
2 !5 ! 
71 
b) ---·--= 2520sequell<:<Sarepossible. 
I !I ! I !I !I !2! 
c) 6! = 720 seqneooesar.: possible. 
1-45. a) from t.r.e multiplication rule,10 3 = 1000 prftixe:sare possible. 
b) From tr.e multiplication rule.8x:tx10 = 160 are possible. 
c) E'l:ryarrangement ofthree:digjtsselected from fl!.e tO d ig its results ina pos.slbJe:-prdix. 
10 I 0! P1 = -= 720prefixesarepossible. 7! 
i--47· a) TltetotaJ nu.mberofssmplcs pcliSSible is = = I 0,626. n.e Ol!mberofssmplcs iowltichexactlyonetankhas high [
24) 24! 
4 4!201 
viscosityis =--X--= 4 896.Therefore:, theprobabilitris = 0 .46 1 [6][18] 6! 18! 4896 I J 1!5! 3!15! 10626 
b) Thenumberofsamplesthatoootainootank \vithhig.hvisoosityis [' 8] = IS! 3060. n.erefore. thereque:stcdprobabiJityis 
4 4!14! 
I - 3060 0.712. 
10626 
c) Tt.enumberofsamples that mect the"'lwrcment.s ;s [6][4]['4] = _6_! x -4_! X.,..:-!4:.:1_ 
I I 2 1!5! 1!3! 2!12! 
2184 = 0.206. 
10626 
::!·.:;j.. a) A.r.B=t70 + 443""fl0=67j. 
b) .4' = ::!8 + j.6::}'"- .109""9$3- 39= 167~ 
~=0.0082 
1225 
c) AVB= 168.5 ""3733 ... 140$"" 2 '"-14 <>- ::!9 +46 - .l= 691.5 
d) AVB' = 1685 "{28- 0)-37~" {36;) -14) " {309- 09) " 1403 • (9~- 46)" (39- 3) = 8$9¢ 
e) A' r'IB' = ~8 -1 ~ 363- 14 ... .306-19 • 933-46'"" 39-3 = 1578 
::!·.5.5· a) P(A) = 04 
b) P(8) = 08 
c) P\K)=0.6 
d) P(AVB)= t 
e) P(.-\r'l 8) = 0:2 
2·57· a) l/U) 
b) .5/10 
::!·59· a) s = {1, ::!,3.4.5.6. 7.8) 
b) 1/8 
c) 6/8 
::!~1 . Thesamplesp<~ceis{O, ,.. 2., <>- $ .a:nd'"" 4). 
2 18 4 . Therefore. t~.eprobabilityis 
(a) Thef:\'t'llt t~.a ta cell has at Jeastoneofthe positive: nickel charged option; is {- !!., • j .and"" 4}. Tf!.e probabilityisOj.S • OS3 • 0.15 = 
Ollj. 
{b) The:t'\oen1 that a cell is ootcomposedof a positivt nickel charge greater t.l>.an ... 3 is <o. <>- 2-.a:nd .-. 3). Tile probabilityis0.17 .-. 0.35 ... 0.3;3 
= 0.8.:;. 
2~.3. jdigits bctwreno and9.sothe probabiJityofsnyt llree numbers is 1j(lo•lo•1o). 
3 
3. ~ 
3lettersA toZ.so the probability of aoytbroe numbers is t/(~6·2.6.::!6). Tlte: prob:lbilityyour lioens.e plate isc~..os.cn is tlt..eo{J/10 )•(l/ ::!6 ) = 5.7 x 10 
1~5. (3) Th.enumberofpossibleo:periments is4 .11. 4 x 3 • 4 x 3 x 3 =52 
{b) There are ,36 o:perimeots that use a ll three st eps. Tlte probabilitytlt.e best NSn!t uses aJI three steps is j6/52. = OJ>!)ij. 
(c) No. it will ootchange:. Withkamou.nts io tlte firststeptt.enumberofo:pctiments isk + jk 4 9k = 1jk. Tt:.e number of experiments thst 
oomplcteaJI three s teps i s'!).~ outofljk. Tlte probabilityis9/ l.3 = Ofl9::!j. 
1~. a) P{A) = 30/U)O = O.jO 
b) P(B) =n/ !OO=O.n 
c) P(..-\') = 1- o:;o = 0 .70 
d) P{..-\ r'l 8) = 12./100 = O.i-2 
e) P(AV8)=85/t00 =0.85 
o ft"-·v B) =92-/ 100 = 0 .92 
1~. a) Because ~:and E' a remuru.allyo:clushl.'f:\'f:ntssod EVE'= S 
t = P(S) = P(E V &') = P(E) • P(E'). TJ..crcfor<. P(f:) = t - P(8) 
b) Bo::.auscSand 0 .sre mutuaUyexcll!.'Sht: C\~ \\i tbS = S V0 
P(S) = P(S) • 1'(0). Tlt<rcfore. 1'(0) = o 
c) Now. 8 = AV(A'(') B)andU..eC\oentsAaod:A' r. Bare mutu:aJiyexch:sive. Tbetefore, P(8) = P(A)- P(A'r'l 8 ). Bec:.<luse P(A' r. 8) ~ 0. 
P(B), P(A). 
2·71.5 TotaJ number of pnssibletks.igns is9QO. Thessmplespaoeof a ll pnssibletks.igns t.hst may besrenoo fhoe visits. This space contains 
900 outcomes.. 
Tt.e. number o( outcome.; iowllic:hall fhoe\isitsaredifferentesn be obtained as follo\.,.'S. Ootlte first \is.it anyone of 900 designs m3)' be seen. On t.lte second 
\is it there are 89-') remaining design~. On th~ third visit tllere are 898 remaining design.<:. On the fourth and fifth \is its there are 897 ar.d 81)6 remaining 
n'esigns. respecti\'cly. From the multiplication rute. the number of outcomes ~·here all designs are d ifferent is qoo•8t)q•8t)a•SQ7•8q6. Therefore. tlt.e 
probability that a design is oot seen again is 
:!·73· a) F(Af""'B) =(l70+443""60)/849'J=0.0791 
bl PIA'l = <'s-363-309 ~ m • 39l/8493 = ,rm/8493 = 0.1969 
c) P(A U 8) = (l685·•<}733""l403•2•l4+:!9+46 ... j)/849j = {>915/8493 = 0.814:! 
d) P(AU B') = (168; •1•8- ;) • 3733 ~ (363- 14)-{309- '9)- 1403 ~ (9j$ - 46) ~ 139-3))/8493 = 8399/8493 = 0.9889 
e) P{A' f""'8') = (:!8- 2 • 363-14 • j06- 29 • m- 46 + 39-3V&93= 1578,/8<$9;3 = !U858 
:!-?.;. a) P(A U8\..JC) = P(A)"" P(B)- F(C). booausethe~tsarcmnruallye:xclmi\'o:. Tl:erefore, 
P(A U8'JC) = O.:l - 0~ •04 = 0:~ 
b) P{ANI"\C)=O.becoUS<ANI"\C= 0 
c) P{AN)=O ,brea.,..AN= 0 
d) P{{A UB)I"\C) = o. bec3US<~4 UB)I"\C = { .41"\C)U{BI"\C) = 0 
c) P(.4'U8'\JC' )= 1-l P(A)• P(B).o. F(C))= l- (0.~ •0~•04)=0.1 
-1·77· a) 70/JOO = 0.70 
b) (7~ ... 86-70)/100 = 0 .95 
c) 1\o, P(Af""'8) H) 
b) 
c) 
345 +5+ 12 
370 
345 +5+8 
362 
370 
358 
=--370 370 
d) 345/370 
:l-81. s) F(unsstisfactory)= {S + 10- !1)/ljO = 13/1.30 
b) P(bGthcriteria sstisfactory) = ll7/ l30 = 0 .91), No 
:l-Sj. RA) = n~/204 = 0-54~. P{B) = 9'J/W4 = 04.510, f't.>\ f""'8) = (40• 16)j:l04 = 0 .2745 
a) fttt U8) =it'\).._ f\:8) -llA f"'\8) = 0-5490 • 04510 -0.-1745 = 0.7~ 
b) lt4f18') = {1~ • 44)/204 = O.:l745and/t-1 U8') = f(.4) • ~')- /t-1 f""'B') = 0..:;4~ • {l- 04510)- 0 .2745 = 0.82:).5 
c) 1\:.4' U8') = 1 - f\:-4. f"'\8) = 1 -0.2745 = 0 .7255 
:l-8.5. P(A) = 43:29/2:225:!= 0 .1945. P(B) = 953/:!:225:!= 0 .04:!8, P(Af""'B) = it$.1/ :!M.S.:l = 0 .0109, 
P{A() B') = {984~3103)/"'5' = 0.1837 
a) 1'(,.\ U 8) = P(A) • P(B)- P(Af""'B) = 0 .1945 • 0.0428-0.0109 = 0 .22(>4 
b) P{AU 8') = P{A) ~ P{B')- P{A() 8') = 0.1945 ~ (l-0.0428)- 0.1837 = 0.9680 
c) f'(A' U 8') = 1- P(Af""'B) = 1-0.0109 = 0 .9891 
:l.-87. a) F(A) = 86/100 
b) P{B) = 79/100 
c) P( A IB) = P(AnB) = 70/ 100 = 70 
P(B) 79 / 100 79 
d) P( BIA ) = P(AnB) = 70/ 100 = 70 
P(A) 861100 86 
:l.·Bi). let A deoote theC'X!Ot tf!.Ata leaf oomplet...s theoolor transformation and Jet 8 denote t.heeo.oe:nt IJ!At a leaf oompletl'.i the te:xt~:-raJ 
transformation. The total number of o:periments is :;oo. 
() P(B IA) - P(An B) _ 243/300 0.903 
a - P(A) - (243+26) / 300 
(b) P(A iB')= P(An B') = 26/300 = 0.591 
P(B') (18+26)/300 
:l-<)l. a) l~/lOO b) 1'J.j 'J.8 c) 34/l2:l 
:l-9$. lrtA denote tlt-ceo.oent that a d olys is is high and let 8 deootet.heeo.l!nt that pntrefactinn is high. TJt.etotal number of experiments is 100. 
(a) P(B' IA)= P(AnB') = 18/ 100 = 0.5625 
P(A) (14+18)/ 100 
(b) P( AIB)= P(An B) = 14/ 100 O. l918 
. P(B) (14+59) / 100 
(c) P(A' IB') = P(A'n B') 9/ 100 0.333 
P(B') (18+9)/ 100 
1-95. a) 20/ 100 
b) 19/99 
c) {;0/100X19/99) = 0.038 
d) tftl:.echips were replaced. tl:.e probabi!ity\~·ollld be(20/l00) = 0 .-1. 
2-97. a) P={8- l)/(3s0-l)=O.OW 
b) p = (8/:}.50) )( 1{8- l)/{:}.50 - lll = 0 .000458 
C) P = {34ljj;O) X ({341- 1)/{3;0- 1)) = 0 .9547 
P(An B) P(B) 
Nn,;f8 C A.then P(AIB)= =--=I 
P(B) P(B) 
P(An B) (40+ 16)1204 56 
= 0.6087 2-1 0 I. a) P(A!B)- -
- P(B) - (40+ 16+36)1204 92 
b) P(A' I B)= P(An B) 36/ 204 36 = 0.3913 ( 40+ 16+ 36)1204 P(B) 92 
P( A I B') = P(An B') 56/204 56 c) (12+44+56)1204 =-=0.5 P(B') 11 2 
P(An B) (40+ 16)1204 40 + 16 d) P(B! 4)= = 
. P(A) (40+ 12+ 16 + 44)/ 204 11 2 
( .) P(An B) ~-103. •l P B I A = P(A) 
( 170 + 443 +60 )8493 
(1685+ 3733+ 1403) / 8493 
Also f.l!.e pmbsbilityoffailure to r ft\ver th:an 1000 wells is 
673 
= 0.0987 
682 1 
= 0.5 
, P(B n A'l (2+ 14+29 +46+ 3)18493 
P(B !A )- - .,..--->-----'---:--
- P(B') - (28 + 363+309+933+39) / 8493 ~=0.0562 1672 
Lct Cdeoote t.J!.ee\'f:Ot th.st fC\~·er thansoo wells a re present. 
P(B!C) = P(An C) 
P(C) 
(2+ 14+29+46+3)18493 
( 28 + 363 + 309 + 39)18493 
:!·105. a) pt.>\(') 8) = P(.1 8)P{B)"' {04 XO,s) = o.~o 
b) P(A'("' 8) = 1'(.<18)1'(8) = (0.6Xo.;) = OJO 
~=0.0650 
739 
1·107. l.et F d.enote theewmt tJ!.at .:~connector fa ilsand" let W d.eootetlt.-e:C\'tnt theta connector is ,,·et. 
P(F) = P(f~V)P(IV)• P(+ V')P(IV') 
= (o .osXO.lO )+{O.O!X0.91)) = O.i'Jl4 
~H09. let R deootetlte~ tJ!.sta product exhibits surface rouglmess. Let N,A.and W deoote ft,t:f!\"ents that tt.e. blades are new,a\"l'f"ilge.aOO 
worn. re.specthoely. Tt~D. 
P(R) = P(R(N)P(N)" P(R(A)P{A) • P(R(W)P(\V) 
= {O.OlX0.2_;) • (0 .03)(6.60) + (o .o.;XI'US) 
=CJ.M.S 
~-m. a) (0.8SXO.:l7 ) = 0.~6 
b) {0 .1:lXO.lj+0.,;2) = 0.0.078 
~-llj. tet A and 8 deootethcf:\'t'llt t~.a t the first andseonod part selected P.asexoessh'l'shrinkage. respecth'Ciy. 
a) 1'(8) = P( 4• )P(A)+ P( 4• 11'(.<') 
= {4/~Xs/2.5) ~ {.;/~)(20/2,5) = 0 .21) 
b) {.et Cdeootetf!.ef:\'Cnt t~.s t the third paM selected has l".l(OCSSi\'es.hrinkage. 
P(C)= P(CIAn B)P(AnB)+P(CIAnB')P(An B') 
+P<CIA'n B)P(A'n B)+ P(CIA'n B')P(A'n B') 
3 [4][5) 4[20][5) 4[5][20]5[19](20] 
= 23 24 25 + 23 24 25 + 23 24 25 + 23 24 25 
= 0.20 
2·115. 
0 
success 
large stone 192 
small stone 81 
overnJI summary 273 
PN 
success 
large stone 55 
small stone 234 
overall summary 289 
fajlure 
71 
6 
77 
fDilure 
25 
36 
61 
st~mp&e size 
263 
87 
350 
sampfe size 
so 
270 
350 
St~mple percentnge, 
75% 
25% 
100% 
Silmple pe.rcenta~ 
23% 
77% 
100% 
conditiont~J suceess rnt& 
73% 
93% 
78% 
conditional succes.s rate 
69% 
83% 
83% 
Tt.e.overaJI suocess rntedependson tJt.e snooess rates for each stone size group. but also the probabilityoftf!.egroups. {tis tf!.e weig.htedaverageoft!u: 
groupsuocess rt~tewcighted bythegroupsizeas follov.'S 
P{O\wall snoctSS) = P(s"llOCeSSI largestone)~large stone)) .a. P(s-uooe:ssj small stone)P(s:mall stone). 
For open surgery, the dominant group (large stone) has a smaller suoor.ss rate while for PN, the dominant group{small stone) h:asa larger suoor.ss rate. 
:H17. P(A) = 43:2:~/2-i:!.S2 = 0.1945. P(8) = 953/2-i:!.S2 = 0.0428 
a ) P(Afl 8) = P{,.-\ j 8)P(8) = (l42/953X9.;j./21-2.52) = 0 .0109 
b) P(AU 8) = P(A) ~ P(8) - P(Afl 8) = 0 .1945 ~ 0.0428-0.0109 = 0.2264 
c) F(A'U 8' ) = 1- P(Afl 8) = 1-0.0109 = 0.9891 
d) P(..-\) = P(A I B)P(B) ~ F(A I B')P(8') = {242/953X953/2:22-S2) - (408]/1l29'JX2l:!'!)9/22'2.;2) = 0.1945 
:1.-u9. P(A) = (168.; • 373;1-140.3)/8493 = o.SO~p. P(8) = h70 • 2 ~ «l - 14- 29 ~ 60 • 46- 3)/8493 = 0.0903 
a ) P(Ar'l 8) = 1'(81 A)f'(A) = (67j/6821X68;>/8493l = 0.0,92 
b) P(AU 8) = P{,.-\) ~ P{8) - P(Afl 8) = 0 .80,'p ~ 0 .0903- 0 .07'1)2 = 0.8142 
c) P(A'U 8') = 1- P(:\ fl 8) = 1-0.0792 = 0:0208 
dl P!Al = P!A I 8JP(8l • P!A 18')1'(8') = (673/767X767/B493l • (6>48/77,6Xm6/8493J = 0.803> 
2-121. l.et R deoote red oolor and Fdenote that tlte font sit.(' is oot the smallest. Tlti!n P(R) = t/4. P(F) = 4/.;. Becatse tr.e Websitesa.regtt'IE:Nited 
randomJy tltese t\l!llts are i odepeodent. 
Therefore, P(R() F) = P(R)P(f) = b/4)(4/Sl = 0 .2 
2·1~. P{,.-\' ) = l-P(A) = 0 .7 and I'( A" 8) = 1- F(A 18) = O.i 
2-12..5. a) ~IA) =4/499and 
/VI) = ~JM). ~··)It,') 
= 14 / 4')9)(; I soo)•(s I 499X49S I soo) 
=5/500 
Therefore. A a 00 8 are ne! i~odent. 
b) .4aod8areindqx:nd.cnt. 
:!·12?. a) P(Afl 8 )= 22/100, P(A) = j0/100, P(8) = 77/100, n.-en P(Afl 8 ) $ P(A)P(B), tr..ttt:fore.Aand Bare oot independent. 
b) P(8)A) = P(A() 8)1P(A) = (:>'I>OO )If30I>OO ) = 0 .733 
:H:l9. {tis useful to work oncoftheseo:erciscswitheare to illustratetf!.e l a\~"'Sof probabiJity. {.et Hi deoote tl:-ee\'ent that t~.e ith:s.amp!e 
oontaim l>~gh Je-.~sof ~nta~inati~n. 
. a) P(H1 () H20 ~ () H4 () H5 ) = P(H1 )P(fl, )P(~ )P(H4 )P(H5 ) 
by lrxiepeodeoce.Also,JV/i) = 0:9. Thetdore. ti'IA'!answcr LS0.9 = 0-59 
. . . . 
, b) A, =(H,r'lfla()~()H4r'lH;l 
A; =(H1f'\H,r'lH3nH4nH5,) 
Aj = (H11"\ H21"\ Hjr'l ~ () H5.) 
A4 =<H,nH,n~nH4nH5 l 
A.;= (H1 fl H2 fl ~(I H4 fl H5 ) 
Thclcqucsttd probability is tr.e probabilit;'oft.Jt.e u.nionA1U A2U ~U A4U A.; aDd fl..ese~'elltsare mnttallyo:clush'f:.Aiso. by-independence P(.-\j) = 
0.9 (0.1) = 0 .0656. n.erefore. tf!.earuv.·er 1s .;(o.o6.;6) = Oj:28. 
c) U:t 8 deoote tf.eel.'ent tt-.at nosam_p!eoontains high !e-.'clsof contamination. TJ!..e l'«!.l+'('Sted _probability is P(8') = 1 • P(8). From p.art (a), 
P(B') = 1-0-59 = 041. 
~Hjl. (a) l£t I and(; deootean infestedaOOgood sample. There are 3 ways to obtain font oonsealtivt .s.amplessh.o\rillg.t!u:s.ig,nsohft.e 
infesta~on:J lliGG. GJ lllG. GGill l . T"'-~ore. the probability is 
jx{0.2 0 .8 ) = 0 .00:}0?2. 
{b) Tlt.cre are 10 h·aystoobtainthreeout of four oonst<:llti\ttsamp!esshov,.;ng tlt.esig,nsofinfesta tion. The probability is 10x(0.2 3 • o.s3) = 
0 .0409(>. 
:Hjj. (a )T~.e -proba hi !ity that one teclmician obtains equivsJcoce ~t 100 ml. is o .1. 
So tlte probability-that both technicia~obtainf:<!.nivsJmceat 100 mLisO.l = 0.01. 
(b) The probabilitytl!.at ooeteclmicianobtains oqui\'aleoct b.."t\\o'OOJ,Saod l04 mLis 0.71 
So tl.e probabilitytltat both technicians obtain cquiW~Jeoce betv.·een 98 and 104 mLis0.7 = 049. 
' (c) The probability-that the a\ 'el''llgc\'Oiume at equh-a!eooe from tr.e technician is 100 mLis9(0.l ) = 0.09. 
:l·l$.5. l.et Ad-eoote tit~~ lft.ata sample is ptoduoed ineavityoneoftr..e mold. 
s) 8yb><!cpcndenoe. P(..,n • .,nA3r'lA4r'lA5) = [! r = 0.0000$. 
b) l.et Sj be tlt-c~oent that a U fh'f:sa.mplcsarcprodooed inc,nity i. lkca.use tlte s ·sare m~tusl!y('XclllSh'f:. P{81U 8:1U ... U 8g) = P(81) ~ 
1'(8,)• ... • P(8g). 
From pan a .. 1'(8; )=[! r. Tlrer<fore. tllcanswcds8[! r = 0.00024. 
c) 8yiodcpcodence. P(A I n A2 n A3 n A, n A~)= [! r [~ ln .. oumhcrnh«t"""""' in wnich fouroutof6-. .. mplesar<from 
"''ityoneiss. Trmtbre, tllcanswcr is s[~ n ~ l = 0.00 I 07 
2·139· P(A) = ~~9/:1:2252 = O.t()tt.;. P{B) = 953/:1:22.;2 = 0.0428. P(Afl 8) = :142/2::!.2.;2 = 0 .0109 
Because P(A)•P(8) = {0.1945X0.0428) = 0.0083 '* 0.0109 = P(Afl 8),Aand Bare oot iodepcndmt. 
l ·>4>. P!Al = f3's"3"s l/14TSTSl = 0.2;, 1'(8) = I4T4"3"5li(4"$'5'3"Sl = o.S, 
P!An 8J = f3"4"3"sll(<"3"s"3"sl = o .2 
Bo::au.se P{.-\)•P(8) = {0.2..;X0.8) = 0 .2 = P(.:\ fl 8).Aaoci Bareinckpendent. 
section 2-1 
P(AiB )P(B) P(BiA ) = P( AiB )P(B) 
P(A) P(AIB )P(B) + P(AIB' )P(B') 
= _----=.o·:..:.4.:...:x ..:.co ·c:.s __ = 0 .s9 
0.4x0.8 + 0.2x0.2 
2-145. (a) P = {o~pX0.9]8) ~ (0.27)(0.981) ... {0.21X0.96.;)- {0.13X0:992) ... {o.o8X0.959) = 0.9Jf)j8 
(b) P = (0.21)(0.965) 0.207552 
0.97638 
i · l47· tet 0 deootea product that reoci\'eda good tt'\i ev.·.l.ct H .. M.a:nd Pd.eoote products tl!..<t t wtre high. mod.."'NNte.a:nd poor perfonners, 
respecth'dy. 
a) /{G)= i{G!H )JVI)-/{GIMll~,\0- IWI~Pl 
.. 0 ·9.5{0 40 }• 0 .60(0 .3.5 )"-0 .10(0 ·'1.5) 
.: (1.-61.5 
b) UsingtlteNSult from parta., 
P( HiG ) = P( GIH )P(H) 0.95(0.40) = 0_618 
P(G) 0.615 
C) P(HIG' ) = P( G'I H)P(H) = 0.05(0.40) 0.052 
P(G) 1- 0.615 
~-149. lkootus follows: S =signal. 0 =organic pollntsots. V = \-olati!esohmts. C = chlorim tedoompou.Dtti. 
a) P{S) = P{S!O)P{O)- P{S!V)P{V) + P{S!C)P{C) = 0 .997{0.M)- o .m;(O.'J) • 0.8')7(0.1:\) = 0 .9847 
b) P<CjS) = P{SjC)P{C)fP{S) = ( O.il97X0.1J)/0.9847 = O.llS., 
!H.51. tet Ldeoote thet'\l!rlt tr..a t a per:son is !.WBS. 
l.d A deooteihe t'\'eot t~a~ t a person visit; Hospital 1. 
l.et 8&mote t.lt.eevent that a per.soo\isits Hospital i. 
tet Cdenote tf!.et!\'ent that a person,isits Hospitil3. 
l.et 0 denote t.lt.eeo.oent th:ata person,isits Hospital4. 
P(DiL) - P(LiD)P(D) 
- P(Li A)P(A)+ P(LI B)P(B)+ P(Li C)P(C)+ P(LI D)P(D) 
=~~~~~~~~(0_.0~55~9)~~~- ~~~~5~)~~--~~~~ 
co.o368)(0.2378J + co.oJ 86 xo.J 1 42J + (0.0436)(0.2535> + co.o559)(o. 1 945) 
= 0.2540 
'2-·l.)j. Oeooteas follov.--s: A= affi liatesite.S =search s ite, 8 =blue. G =green. 
p SiB = P(BiS)P(S) 
( ) P(BIS)P(S)+P(BIA)P(A) 
Suetplemefltal ExertGeS 
= -:-:--:-:-:"( 0=-'.4"')(.::0.:'-'7 ):-:-:--=-
(0.4)(0. 7) + (0.8)(0.3) 
= 0.5 
2-·1.5.5· U:t Bdeootet.he~tth.ata glass breaks. 
l.et l.dmotet.Jt.e event that large p.!lckaging is used. 
P{B) = P(Bji.)P{l.) + P{Bji:)P{t:) 
.. 0 .0!(0.60) ... 0.02{040) = 0.014 
!H57. letA = o:cel!cnt su.rfilttfinish; 8 = o:cel!cnt length. 
a) P(A) = 8i/lOO = O.B.i 
b) P(B)= 90/100 = 0 .90 
c) P(A")=l -0.82=o.l8 
d) P(A() B)= 80/100 = 0.80 
i!) P(A v 8) = ~.9::!: 
0 P(A.UB)=0.9Jl 
.i·l.S9· I f A,B,Care mutv.ailyo:clu.si'-'!. t.Jt.en P{ AV 8 V C) = P(A) ... P{B)- P(C) = Oj + !14 - !1..; = l .i. wh.ichgreater t.hanl. Therefore. P{A). 
P(B).and P(C)ca.nnot «tnaJ tlt.egi\l!OvaJ!les. 
'2-·l6l. {a) P(th.~ first one selected is not ionized)= 2.0/lOO = 0 .::!: 
(b) P(tt.eseoond is oot ioniz.edghl!n the first one\\-as. ionized)= .iOf<» = 0 .2.0:! 
(c) P(bothareionit.ed) 
= P(th.~ first one selected is ionized) x P(th.o:sooond is iooiudg.h'f!fl the first ooe.was ionized) 
= (80 /100))( {79/99) .. ().(~8 
{d) l f ssmpless.elected "'·ere replacOO prior to t lw: nv:t selection, 
P(theseooOO i.s not ioniud gh 'i:ntlte first one was ioniud) = 'i0/ 100 = 0.2. 
n.e C\oent of tft..i! first selection and ibe t'\'ttlt oftl'.e .secood selection a re indepmdent. 
'2-·l6,l. a) P(..-\) = 0.!13 
b) P(A.) = 0.97 
c) P{BIA) =040 
d) P(B!A.) = i).O_; 
e) I'{ A () B)= I'{ B!A )P(A) = (040XO.OJ) = 0.012 
0 I'{ A() B.)= I'{ B' lA )P{A) = (0.60XO.OJ) = 0.018 
g ) P(B) = P( B!A )P(A) • P( BfA .)P(A.) = (040X0.03) +(O.O;X0.97) = 0.0(>05 
1PO 
2.-16..;. {a) P= 1--{1-0.002.) = 0 .18143 
l 2 (b) P=0 (0.998 )0.002 = 0.005976 
100 10 
{c) P= 1-l(l-c).i)O:l) I = il.86494 
, , ~H67. l.et Ai denote thee-vent that t.J!f ith readback issltOO!'.SSful. Byiode-pend..."'llle, 
P(Al 1""'1A:!1"1Aj) = P~ltl )P(A2)PfA_:J) = (0.02) = 0.000008. 
2.-169. (a) .4• 1"18 =so 
(b)B·=37 
(c) .4UB =93 
~H7t. l£t Dj &note the~t that the prima ryfaiJu.re mode i.s type i and let A &oote tf!.et'\..-:nt that a board passes the tes.t. The sample space isS 
= M.,A.· 0 1 • .4.' D2.if D3,A' D4 .• 4' D5) . 
i ·173· -a) P(A) =-l9/l00 = 0 .19 
b) P(Af"' 8) =.15/100 = 0 .15 
c) P(AV8) ={19 .... 9-5-l5)/UlU=U.'!}-) 
d ) P(A' 1"1 8) = SO/JOO = 0.80 
e) P(AfB) = P(A() 8)/P{B) = 0 .158 
::!.·175· a) No. P<E1(') F-2(') Ej,) :\ o. 
b) No,E.1' (') E.::!.'isooHJ. 
c) P(f>1'U E,' U E:)'l = P{f>J')" P(E, ') " P!E:)'l- P(f>!' () E, ') • P(E1' () 8:) ~ • P(E,' () E:)'l" P(E1' () E,' () E:)'l = 40/ :140 
d) P(E1r'\ E,r'\ f~) = 200/ :140 
e) P(F.1U Ejl = P(E1) - P(F-3)- P(E1r'\ f-3) = :134/ :140 
0 P(E1U E,U E,jl = 1- P(E1' () E,'() ~') = 1-0 = 1 
::!.·177. tet Ai denote tf:.CE'\l:llt t".at tlt~ ith: bolt selected is oot torqnOO. to t.Jt.e proper limit. 
a) TJt.en, 
P(A 1 n A2n A3 nA4) = P(A4 !A1 nA 2 IIA 3)P(A 1 IIA2 11 A) 
= P(A4 !A1 nA2 nA3)P(A3 ~< 1 nA2 )P(A2 !A1 )P(A 1) 
= [: ~ ][: ~ ][:: ][ ~~ l = 0.282. 
b) tet 8 denote the~t that at least oneoftltesclt\":ted bolts are oot properlytof"<ll+'ed. Tft.liS, 8' is t.Jt.eE'\'E'nt that aJI bolts are properly 
torqued. Then, 
[ 15)[14][13][12] , P(B) = I - P(B') = I- 20 (9 lS )7 = 0. d 8. 
~!-!79. A1 = bytelephooe,A2 =website: P(Al) = 0.9::!. P(A2) = 0.9.;: 
Byindt'prndenct P(AlU A::!_)=- P(Al)'"' P(..•\a)- P(.o\l(')A::!_) = 0.9::!. - 0.9.; - 0 .92{0.9.;) = 0 .9!)6. 
::!. -181. tet 0 denote the C\~t t.Jt.a t a container is inoorrectlyfilled and let H deootethcE'\'t'llt t~.a ta oontaioer is fi lled undcr l;ig.h-spood operotion. 
Then, 
a) P(O) = P(DIH )P(H)" P(D)H")P(ff) = O.O!(OJO), 0.00!(0.70) = 0.0037 
b) P(H !D ) = P(D I H )P(H ) = 0.0 1(0.30) = 0_8 1 OS. 
P(D) 0.0037 
a) [ 2 ](73][72) [73][ 2 ][72] [73][72][ 2 ] P(D= I)= -- - + -- - + -- -= 0.0778 75 74 73 75 74 73 75 74 73 
b) P(D=2)= [2][-1 ][
7
3]+ [2][
7
3][-'] + [
7
3][2][-1 ]= 0.00108 75 74 73 75 74 73 75 74 73 
c) l.et A represent l.ttf!\'E'trt that tlt.e two items NOT inspected are oot def«:ti\'c. n.en. 
P(A) = b:J/7.;X7i/ 74) = 0.947. 
,.,ss. a) {O.:)XO .~~X0.98Sl " {0.7X0.98XO.W) = 0.~64 
. 
1 
P(Eiroutei)P(routel) 0.02485(0.30) 
b) P(rourel E) = = = 0.3 159 
P(E) 1- 0.9764 
::!·187. l.et /l j denote t.t'..eeYent tr.at the ith:wss.her stlectedis thicker tr.an target. 
a) [ 30)[ 29][28] = 0.207 50 49 48 
b) 30/48 .. 0.625 
c) Tl:e req1:ested probability can bewriuen in termsofwl:rt~.eror oot the first and s.eonod wasJ!.erselecteda:re t.Jt.Jcker tlt.an the target. That 
is. 
[ 30)[ 29][ 28] + [30 ][ 20][ 29] +[ 20)[ 30][ 29] + [ 20 ][.!2.][ 30 l = 0.60. 50 49 48 50 49 48 50 49 48 50 49 48 
2-189. a) 
b) 
c) 
d ) 
e) 
t) 
P(A u B)= I 12+68+246 = 0.453 
940 
246 P( An B )=-= 0.262 
940 
P( A' UB ) = 514+ 68+ 246 = 0.88 1 
940 
P(lfn B')= 514 = 0.547 
940 
P(A !B)= P(A u B) = 246/ 940 = 0.783 
P(B) 314/ 940 
P(B!A) = P(Bn A) = 246/ 940 = 0.687 
P(A) 358 / 940 
::!.-191. l.e1.4i deootetJt.e e\~nt that the itn tO\\· operates. Then. 
P(A,) = 0 .98.P!Ao) = {0.99X0.99) = 0 .9801,P(A3l = 0.9801, 1'(. .... ) = 0.98. 
Tr.e probability the cim:.it does not operate is 
~t'1 lit'<> )Jt4~ )Jt'\t) = {o.g;Xo.<ll99X0.0199XO.OO) 
: l,s8XU) 
-8 (b) P = {1- 0.9.3X1- 0 .91)l1- 0 .97Xl-0.9Q)(l- o.91)X1- 0 .9:}) = a.(>46x 10 
{c) P = 1- (l - 0.91)(1 -0.97)(1- 0 .90) = O:!>~m3 
7 
::!.·195· {a) .}6 
6 {b) Number of p.."'tlnutatiomofsix letters is ::!.-6 • Nuty,ber of w-ays to select one number = 10. Number of positions among tr.esi>: letters to 
platt tft.eooe number= 7. Number of passwords= ::!.6 x 10 x 7. 
5 • (c) ::!6 10 
~-197. (a) letAdeooteth.at a partoontOrms tospecificationund let Bdeootea simplcoomponent. 
For supplier 1: 
P{A) = !<}8.~/1000 = M94 
For supplier~ : 
I'<Al = 19!)0/ zooo = o .99s 
{b) 
For supplier 1: P(AlB.) = 9!)0/!000 = 0 .99 
For supplier~ : P(..;\!8.) = 394/400 = 0 .91).:; 
{c) 
For supplier 1: P(Af8) = ~/1000 = 0 .998 
For supplier~: P(AIB) = ts(}&/1600 = 0 .997.5 
{d) TlteunUSt!AI result is that for both a simp!eoomponentand for a oomplex assembly. supplier 1 h:asa greater probabilityth:ata part 
oonforms to specifications. HO\\'tl.'er.supplier 1 has a lower probability of oonfonnanoeCM!taU. TJ!.eo\Willl oonforming probabilityd'epeods 
on bothtt.e:oonfonning probability of eaclt pa.rt type and also tft.e probabiJityof each part type. Supplier 1 prodllCES moreoftt.e:oomplo: 
partssot.hato~ll oonfonnanoefromsupplier 1 is lo,<~er. 
!H99. l.et ndeoote the number of washers selected. 
n 
a) The probabiJitythat none arc thicker. that is.all are less thantlte target is04 byiod.cp..~""nce. The fOllowing result$ aN obtained: 
ll 0 .4" 
0.4 
2 0. 16 
3 0.064 
T'*:tefore. n = 3 
b) Tft.e requ-ESted probability is theoomplcmcnt oft.r.e proOObilityrequ.csted in part a ). Therefor-e. n = 3 
~-::l:Ol . tet Edeootethee>.l!nt that noneoftlte bolts are identified as inoorrectlytorquOO. 
tet X denote t!bc number of bolts in the s.:~mplc that are incorrect. Tlte N:q•testcd pmb:tbilityis P(E"). Then. 
P{E) = P{E!X = o)P{X = 0 ) • P{E!X = 1) P{X = 1) • P{E!X = 1) P{X = 1) • P{E!X = 3) P{X = ;)) • P{E!X = 4) P{X = 4) 
and F(X = o) = hs/ :!O Xl4/ l9XJ3/t8)(J:!/l7) = 0 .:!817. 
The remainin& probability for X can be determined from tr.eoounting metitods. Then 
5!15!4!16! = 0.4696 
4 !3!12!201 
P{X.= 4) = {.;/ 10X4/:19X:J!>BX1/ 17) = 0.0010. P{f!j X= 0) = 'ct;1E I X=!)= 0 .05. 4 _6 P(E j X =~) =O.OS .::O.OO:!.S,P(E!X=J)=O.OS =l.::l:,;Xt0-4 .P(f. I X=-4)=0.0~ =6.25x10 .Th.en. 
run= l (0 .~8l7)+0.0.!j (046!)6)'"'0 .00:!.5{0 .:!l67)"'la.5xlO {O.Oj09)'"-6.!l..:;Xl0 {0.00!0) 
.:: OJ06 
sndi'{E.)=O.fi94 
~-:!0$. Tltetotal sam_plesizeiska ... a ... kb • b = {k • J)a • {k +l)b. Therefore 
F(A = k(a + b) F(B = ka + a 
) (k + 1)a + (k + 1)b' ) (k + 1)a + (k + l)b 
and 
ka P(AnB) = ka 
(k + 1)a + (k + 1)b (k + 1X(I + b) 
Tl>en. 
P(A)P(B) = k(a + bXka + a) 
[(k + 1)a + (k + J)bf 
k(a + bXk + !)a = ka P(An B) 
(k + J)2 (a + b)2 (k + JXa + b) 
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