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1. There are 2 possible outcomes on each fip: heads or tails. Since the coin is fipped three times, thereare 2 × 2 × 2 = 8 total possibilities: HHH, HHT, HTH, HTT, TTT, TTH, THT, THH.O these 8 possibilities, ho man! in"ol"e e#actl! to heads$ %e can simpl! count these up: HHT, HTH, THH. %e see that there are & outcomes that in"ol"e e#actl! to heads. Thus, the correct anser is &'8.(lternati"el!, e can dra an ana)ram table to calculate the number o outcomes that in"ol"e e#actl!2 heads.

( * +

H H T

The top ro o the ana)ram table represents the & coin fips: (, *, and +. The bottom ro o theana)ram table represents one possible a! to achie"e the desired outcome o e#actl! to heads. Thetop ro o the ana)ram !ields &, hich must be di"ided b! 2 since the bottom ro o the ana)ramtable contains 2 repetitions o the letter H. There are &'2 = & di-erent outcomes that contain e#actl!2 heads. The probabilit! o the coin landin) on heads e#actl! tice is / o to0head results total / ooutcomes = &'8. The correct anser is *.

2.3et us sa! that there are n 4uestions on the e#am. 3et us also sa! that p1 is the probabilit! that 5att!ill )et the 6rst problem ri)ht, and p2 is the probabilit! that 5att! ill )et the second problem ri)ht, andso on until pn , hich is the probabilit! o )ettin) the last problem ri)ht. Then the probabilit! that 5att!

ill )et all the 4uestions ri)ht is 7ust p1 × p2 × × pn. %e are bein) as9ed hether p1 × p2 × × pn is)reater than ;<.1 >S?@@+A>T: This tells us that or each 4uestion, 5att! has a B;< probabilit! o anserin)correctl!. Hoe"er, ithout 9noin) the number o 4uestions, e cannot determine the probabilit! that5att! ill )et all the 4uestions correct.2 >S?@@+A>T: This )i"es us some inormation about the number o 4uestions on the e#am but noinormation about the probabilit! that 5att! ill anser an! one 4uestion correctl!.1 (>C 2 >S?@@+A>T: Ta9en to)ether, the statements still do not pro"ide a de6niti"e D!esD or DnoDanser to the 4uestion. @or e#ample, i there are onl! 2 4uestions on the e#am, 5att!Es probabilit! oanserin) all the 4uestions correctl! is e4ual to .B; × .B; = .81 = 81<. On the other hand i there are F4uestions on the e#am, 5att!Es probabilit! o anserin) all the 4uestions correctl! is e4ual to .B; × .B; × .B; × .B; × .B; × .B; × .B; G 8<. %e cannot determine hether 5att!Es chance o )ettin) aperect score on the e#am is )reater than ;<.

The correct anser is A

&.n order to sol"e this problem, e ha"e to consider to di-erent scenarios. n the 6rst scenario, aoman is pic9ed rom room ( and a oman is pic9ed rom room *. n the second scenario, a man ispic9ed rom room ( and a oman is pic9ed rom room *. The probabilit! that a oman is pic9ed rom room ( is 1;'1&. that oman is then added to room *,this means that there are omen and men in room * Ori)inall! there ere & omen and men.So, the probabilit! that a oman is pic9ed rom room * is 'B.*ecause e are calculatin) the probabilit! o pic9in) a oman rom room ( (>C then rom room *, eneed to multipl! these to probabilities:1;'1& # 'B = ;'11F The probabilit! that a man is pic9ed rom room ( is &'1&. that man is then added to room *, thismeans that there are & omen and I men in room *. So, the probabilit! that a oman is pic9ed romroom * is &'B.()ain, e multipl! thse to probabilities:&'1& # &'B = B'11F To 6nd the total probabilit! that a oman ill be pic9ed rom room *, e need to ta9e both scenariosinto account. n other ords, e need to consider the probabilit! o pic9in) a oman and a oman OJa man and a oman. n probabilities, OJ means addition. e add the to probabilities, e )et:;'11F K B'11F = B'11F The correct anser is *.

4. The period rom Lul! to Lul! 8, inclusi"e, contains 8 M K 1 = da!s, so e can rephrase the 4uestion

as N%hat is the probabilit! o ha"in) e#actl! & rain! da!s out o $Since there are 2 possible outcomes or each da! J = rain or S = shine and da!s total, there are 2 #2 # 2 # 2 # 2 = &2 possible scenarios or the da! period JJJSS, JSJSS, SSJJJ, etc To 6nd theprobabilit! o ha"in) e#actl! three rain! da!s out o 6"e, e must 6nd the total number o scenarioscontainin) e#actl! & JPs and 2 SPs, that is the number o possible JJJSS ana)rams:= ' 2& = # '2 # 1 = 1;

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The probabilit! then o ha"in) e#actl! & rain! da!s out o 6"e is 1;'&2 or '1I.>ote that e ere able to calculate the probabilit! this a! because the probabilit! that an! )i"enscenario ould occur as the same. This stemmed rom the act that the probabilit! o rain = shine =;<. (nother a! to sol"e this 4uestion ould be to 6nd the probabilit! that one o the a"orablescenarios ould occur and to multipl! that b! the number o a"orable scenarios. n this case, theprobabilit! that JJJSS 1st three da!s rain, last to shine ould occur is 1'21'21'21'21'2 =1'&2. There are 1; such scenarios di-erent ana)rams o JJJSS so the o"erall probabilit! o e#actl! &rain! da!s out o is a)ain 1;'&2. This latter method or9s e"en hen the li9elihood o rain does note4ual the li9elihood o shine.

The correct anser is +.

5. There are our possible a!s to pic9 e#actl! one deecti"e car hen pic9in) our cars: C@@@, @C@@,@@C@, @@@C C = deecti"e, @ = unctional. To 6nd the total probabilit! e must 6nd the probabilit! o each one o these scenarios and add themto)ether e add because the total probabilit! is the 6rst scenario OJ the second OJ. The probabilit! o the 6rst scenario is the probabilit! o pic9in) a deecti"e car 6rst &'2; (>C then aunctional car 1F'1B (>C then another unctional car 1I'18 (>C then another unctional car 1'1F.

The probabilit! o this 6rst scenario is the product o these our probabilities:&'2; # 1F'1B # 1I'18 # 1'1F = 2'1B The probabilit! o each o the other three scenarios ould also be 2'1B since the chance o )ettin) the

C 6rst is the same as )ettin) it second, third or ourth. The total probabilit! o )ettin) e#actl! one deecti"e car out o our = 2'1B K 2'1B K 2'1B K 2'1B =8'1B.

6. The simplest a! to sol"e the problem is to reco)niQe that the total number o )ems in the ba) must bea multiple o &, since e ha"e 2'& diamonds and 1'& rubies. e had a total number that as notdi"isible b! &, e ould not be able to di"ide the stones into thirds. Ri"en this act, e can test somemultiples o & to see hether an! 6t the description in the 4uestion. The smallest number o )ems e could ha"e is I: diamonds and 2 rubies since e need at least 2rubies. s the probabilit! o selectin) to o these diamonds e4ual to '12$'I × &' = 12'&; = 2'. Since this does not e4ual '12, this cannot be the total number o )ems. The ne#t multiple o & is B, hich !ields I diamonds and & rubies:I'B × '8 = &;'F2 = '12. Since this matches the probabilit! in the 4uestion, e 9no e ha"e Idiamonds and & rubies. >o e can 6)ure out the probabilit! o selectin) to rubies:&'B × 2'8 = I'F2 = 1'12 The correct anser is +.

F.@or probabilit!, e ala!s ant to 6nd the number o a!s the re4uested e"ent could happen anddi"ide it b! the total number o a!s that an! e"ent could happen.@or this complicated problem, it is easiest to use combinatorics to 6nd our to "alues. @irst, e 6nd thetotal number o outcomes or the triathlon. There are B competitors three ill in medals and si# illnot. %e can use the +ombinatorics Rrid, a countin) method that allos us to determine the number ocombinations ithout ritin) out e"er! possible combination.

( * + C A @ R H

> > > > > >

Out o our B total places, the 6rst three, (, *, and +, in medals, so e label these ith a D.D The 6nalsi# places C, A, @, R, H, and do not in medals, so e label these ith an D>.D %e translate this intomath: B ' &I = 8. So our total possible number o combinations is 8. Jemember that meansactorial or e#ample, I = I × × × & × 2 × 1.>ote that althou)h the problem seemed to ma9e a point o di-erentiatin) the 6rst, second, and thirdplaces, our 4uestion as9s onl! hether the brothers ill medal, not hich place the! ill in. This ish! e donEt need to orr! about labelin) 6rst, second, and third place distinctl!.>o, e need to determine the number o instances hen at least to brothers in a medal. 5racticall!spea9in), this means e ant to add the number o instances to brothers in to the number oinstances three brothers in.3etEs start ith all three brothers innin) medals, here * represents a brother.

( * + C A @ R H

* * * > > > > > >

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Since all the brothers in medals, e can i)nore the part o the countin) )rid that includes those hodonEt in medals. %e ha"e & ' & = 1. That is, there is onl! one instance hen all three brothers inmedals.>e#t, letEs calculate the instances hen e#actl! to brothers in medals.

( * + C A @ R H

* * *E > > > > >

Since brothers both in and donEt in medals in this scenario, e need to consider both sides o the)rid i.e. the (*+ side and the CA@RH side. @irst, or the three ho in medals, e ha"e & ' 2 = &.@or the si# ho donEt in medals, e ha"e I ' = I. %e multipl! these to numbers to )et our totalnumber: & × I = 18.(nother a! to consider the instances o at least to brothers medalin) ould be to thin9 o simplecombinations ith restrictions. !ou are choosin) & people out o B to be inners, ho man! di-erent a!s are there to chose aspeci6c set o & rom the B i.e. all the brothers$ Lust one. Thereore, there is onl! one scenario o allthree brothers medalin). !ou are choosin) & people out o B to be inners, i 2 speci6c people o the B ha"e to be a member o the innin) )roup, ho man! possible )roups are there$ t is best to thin9 o this as a problem ochoosin) 1 out o F 2 must be chosen. +hoosin) 1 out o F can be represented as F ' 1I = F.Hoe"er, i 1 o the remainin) F can not be a member o this )roup in this case the &rd brother thereare actuall! onl! I such scenarios. Since there are & di-erent sets o e#actl! to brothers *1*2, *1*&,

*2*&, e ould ha"e to multipl! this I b! & to )et 18 scenarios o onl! to brothers medalin). The brothers in at least to medals in 18 K 1 = 1B circumstances. Our total number o circumstancesis 8, so our probabilit! is 1B ' 8. The correct anser is *.

8. set S is the set o all prime inte)ers beteen ; and 2; then:S = U2, &, , F, 11, 1&, 1F, 1BV3etPs start b! 6ndin) the probabilit! that the product o the three numbers chosenis a number less than&1. To 9eep the product less than &1, the three numbers must be 2, & and . So, hat is the probabilit!that the three numbers chosen ill be some combination o 2, &, and $HerePs the list all possible combinations o 2, &, and :case (: 2, &,

case *: 2, , &case +: &, 2, case C: &, , 2case A: , 2, &case @: , &, 2 This ma9es it eas! to see that hen 2 is chosen 6rst, there are to possible combinations. The same istrue hen & and are chosen 6rst. The probabilit! o drain) a 2, (>C a &, (>C a in case ( iscalculated as ollos remember, hen calculatin) probabilities, (>C means multipl!:case (: 1'8 # 1'F # 1'I = 1'&&I The same holds or the rest o the cases.case *: 1'8 # 1'F # 1'I = 1'&&Icase +: 1'8 # 1'F # 1'I = 1'&&Icase C: 1'8 # 1'F # 1'I = 1'&&I

case A: 1'8 # 1'F # 1'I = 1'&&Icase @: 1'8 # 1'F # 1'I = 1'&&ISo, a 2, &, and could be chosen accordin) to case (, OJ case *, OJ, case +, etc. The total probabilit!o )ettin) a 2, &, and , in an! order, can be calculated as ollos remember, hen calculatin)probabilities, OJ means add:1'&&I K 1'&&I K 1'&&I K 1'&&I K 1'&&I K 1'&&I = I'&&I>o, letPs calculate the probabilit! that the sum o the three numbers is odd. n order to )et an oddsum in this case, 2 must >OT be one o the numbers chosen. ?sin) the rules o odds and e"ens, e cansee that ha"in) a 2 ould )i"e the olloin) scenario:e"en K odd K odd = e"enSo, hat is the probabilit! that the three numbers chosen are all odd$ %e ould need an odd (>Canother odd, (>C another odd:F'8 # I'F # 'I = 21;'&&I The positi"e di-erence beteen the to probabilities is:21;'&&I M I'&&I = 2;'&&I = 1F'28 The correct answer is C.

9.

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To 6nd the probabilit! o ormin) a code ith to ad7acent Ps, e must 6nd the total number o suchcodes and di"ide b! the total number o possible 1;0letter codes. The total number o possible 1;0letter codes is e4ual to the total number o ana)rams that can beormed usin) the letters (*+CA@RH, that is 1;'2 e di"ide b! 2 to account or repetition o the Es. To 6nd the total number o 1; letter codes ith to ad7acent Ps, e can consider the to Ps as O>A3ATTAJ. The reason or this is that or an! )i"en code ith ad7acent Ps, here"er one is positioned,the other one must be positioned immediatel! ne#t to it. @or all intents and purposes, e can thin9 othe 1; letter codes as ha"in) B letters 0 is one. There are B a!s to position B letters.5robabilit! = / o ad7acent codes ' / o total possible codes

= B 1; ' 2 = B2 ' 1; = B2 ' 1;B = 1' The correct anser is +.

10. e actor the ri)ht side o the e4uation, e can come up ith a more meanin)ul relationshipbeteen p and q: p2 M 1& p K ; = q so p M 8 p M = q. %e 9no that p is an inte)er beteen 1 and1;, inclusi"e, so there are ten possible "alues or p. %e see rom the actored e4uation that the si)n oq ill depend on the "alue o p. One a! to sol"e this problem ould be to chec9 each possible "alue o p to see hether it !ields a positi"e or ne)ati"e q.Hoe"er, e can also use some lo)ic here. @or q to be ne)ati"e, the e#pressions p M 8 and p M must ha"e opposite si)ns. %hich inte)ers on the number line ill !ield opposite si)ns or thee#pressions p M 8 and p M $ Those inte)ers in the ran)e W p W 8 notice and 8 are not includedbecause the! ould both !ield a "alue o Qero and Qero is a nonne)ati"e inte)er. That means that

there are onl! to inte)er "alues or p, I and F, that ould !ield a ne)ati"e q. %ith a total o 1;possible p "alues, onl! 2 !ield a ne)ati"e q, so the probabilit! is 2'1; or 1'. The correct anser is *.

11.

3etEs consider the di-erent scenarios:

Xate ins all 6"e fips, she ends up ith Y1. Xate ins our fips, and Cann! ins one fip, Xate is let ith Y1&. Xate ins three fips, and Cann! ins to fips, Xate is let ith Y11. Xate ins to fips, and Cann! ins three fips, Xate is let ith YB. Xate ins one fip, and Cann! ins our fips, Xate is let ith YF. Xate loses all 6"e fips, she ends up ith Y.

The 4uestion as9s or the probabilit! that Xate ill end up ith more than Y1; but less than Y1. nother ords, e need to determine the probabilit! that Xate is let ith Y11 or Y1& since there is noa! Xate can end up ith Y12 or Y1.

The probability that Kate ends up with $11 after the !e "ips#

Since there are 2 possible outcomes on each fip, and there are fips, the total number o possible

outcomes is . Thus, the 6"e fips o the coin !ield &2 di-erent outcomes.

To determine the probabilit! that Xate ill end up ith Y11, e need to determine ho man! o these

&2 outcomes include a combination o e#actl! three innin) fips or Xate.

%e can create a s!stematic list o combinations that include three ins or Xate and to ins orCann!: CXXXC, CXXCX, CXCXX, CCXXX, XCXXC, XCXCX, XCCXX, XXCXC, XXCCX, XXXCC = 1; a!s.

(lternati"el!, e can consider each o the 6"e fips as 6"e spots. There are potential spots or XateEs6rst in. There are potential spots or XateEs second in because one spot has alread! been ta9en b!

XateEs 6rst in. There are & potential spots or XateEs third in. Thus, there are a!s orXateEs three "ictories to be ordered.

Hoe"er, since e are interested onl! in uni4ue innin) combinations, this number must be reduceddue to o"ercountin). +onsider the innin) combination XXXCC: This one innin) combination hasactuall! been counted I times this is & or three actorial because there are I di-erent orderin)s othis one combination:

This o"ercountin) b! I is true or all o XateEs three0"ictor! combinations. Thereore, there are

onl! a!s or Xate to ha"e three ins and end up ith Y11 as e had disco"ered earlierrom our s!stematic list.

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The probability that Kate ends up with $1 after the !e "ips#

To determine the probabilit! that Xate ill end up ith Y1&, e need to determine ho man! o the &2total possible outcomes include a combination o e#actl! our innin) fips or Xate.

()ain, e can create a s!stematic list o combinations that include our ins or Xate and one in orCann!: XXXXC, XXXCX, XXCXX, XCXXX, CXXXX = a!s.

(lternati"el!, usin) the same reasonin) as abo"e, e can determine that there area!s or XateEs our "ictories to be ordered. Then, reduce this b! our actorial or 2 due to

o"ercountin). Thus, there are a!s or Xate to ha"e our ins and end up ith Y1& thesame anser e ound usin) the s!stematic list.

The total probability that Kate ends up with either $11 or $1 after the !e "ips#

There are 1; a!s that Xate is let ith Y11. There are a!s that Xate is let ith Y1&.

Thereore, there are 1 a!s that Xate is let ith more than Y1; but less than Y1.

Since there are &2 possible outcomes, the correct anser is , anser choice %.

1&.

There is a stron) temptation to sol"e this problem b! simpl! 6ndin) the probabilit! that it ill snoB;< and the probabilit! that schools ill be closed 8;< and multipl!in) these to probabilties. Thisapproach ould !ield the incorrect anser F2<, choice C.Hoe"er, it is onl! possible to multipl! probabilities o separate e"ents i !ou 9no that the! areindependent rom each other. This act is not pro"ided in the problem. n act, e ould assume thatschool bein) closed and sno are, at least to some e#tent, dependent on each other. Hoe"er, the! arenot entirel! dependent on each other it is possible or either one to happen ithout the other.

Thereore, there is an un9non de)ree o dependence hence there is a ran)e o possible probabilities,dependin) on to hat e#tent the e"ents are dependent on each other.Set up a matri# as shon belo. @ill in the probabilit! that schools ill not be closed and the probabilit!that there ill be no sno.

Schoolsclosed

Schools notclosed

TOT(3

Sno

>o sno 10

TOT(3 &0 100

Then use subtraction to 6ll in the probabilit! that schools ill

be closed and the probabilit! that there ill be sno.

Schoolsclosed

Schools notclosed

TOT(3

Sno 90

>o sno 10

TOT(3 80 &0 100

To 6nd the )reatest possible probabilit! that schools ill be closed and it ill sno, 6ll in the remainin)cells ith the lar)est possible number in the upper let cell.

Schoolsclosed

Schools notclosed

TOT(3

Sno 80 1; 90

>o sno ; 1; 10

TOT(3 80 &0 100

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The )reatest possible probabilit! that schools ill be closed andit ill sno is 8;<. The correct anser is A.

1.

The easiest a! to attac9 this problem is to pic9 some real, eas! numbers as "alues or y and n. 3etEsassume there are & tra"elers (, *, + and 2 di-erent destinations 1, 2. %e can chart out the

possibilities as ollos:Cestination 1 Cestination 2

(*+

(* +

(+ *

*+ (

(*+

+ (*

* (+

( *+

Thus there are 8 possibilities and in 2 o them all tra"elers end up at the same destination. Thus theprobabilit! is 2'8 or 1'. *! plu))in) in y = & and n = 2 into each anser choice, e see that onl!anser choice C !ields a probabilit! o 1'.(lternati"el!, consider that each tra"eler can end up at an! one o n destinations. Thus, or each

tra"eler there are n possibilities. Thereore, or y tra"elers, there are possible outcomes.(dditionall!, the Dinnin)D outcomes are those here all tra"elers end up at the same destination.Since there are n destinations there are n Dinnin)D outcomes.

Thus, the probabilit! = .

The anser is C.

14.

There are our scenarios in hich the plane ill crash. Cetermine the probabilit! o each o thesescenarios indi"iduall!:

+(SA O>A: An)ine 1 ails, An)ine 2 ails, An)ine & or9s =

+(SA T%O: An)ine 1 ails, An)ine 2 or9s, An)ine & ails =

+(SA THJAA: An)ine 1 or9s, An)ine 2 ails, An)ine & ails =

+(SA @O?J: An)ine 1 ails, An)ine 2 ails, An)ine & ails =

To determine the probabilit! that an! one o these scenarios ill occur, sum the our probabilities:

The correct anser is C. There is a F'2 chance that the plane ill crash in an! )i"en fi)ht.

15.

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The 4uestion re4uires us to determine hether Zi9eEs odds o innin) are better i he attempts & shotsinstead o 1. @or that to be true, his odds o ma9in) 2 out o & must be better than his odds o ma9in) 1out o 1. There are to a!s or Zi9e to at least 2 shots: Aither he hits 2 and misses 1, or he hits all &:

'dds of hittin( & and )issin( 1

* of ways to hit & and )iss 1& HHZ, HZH, ZHH

Total+robability

'dds of hittin( all

* of ways to hit all 1 HHH

Zi9eEs probabilit! ohittin) at least 2 out o & ree thros =

>o, e can rephrase the 4uestion as the olloin) ine4ualit!:

(re Zi9eEs odds o hittin) at least 2 o & )reater than his odds o hittin) 1 o 1$ This can be simpli6ed as ollos:

n order or this ine4ualit! to be true, p must be )reater than . but less than 1 since this is the onl!a! to ensure that the let side o the e4uation is ne)ati"e. *ut e alread! 9no that p is less than 1since Zi9e occasionall! misses some shots. Thereore, e need to 9no hether p is )reater than .. it is, then the ine4ualit! ill be true, hich means that Zi9e ill ha"e a better chance o innin) i he

ta9es & shots.Statement 1 tells us that p W .F. This does not help us to determine hether p [ ., so statement 1 isnot su\cient.Statement 2 tells us that p [ .I. This means that p must be )reater than .. This is su\cient to anserthe 4uestion. The correct anser is *: Statement 2 alone is su\cient, but statement 1 alone is not su\cient.

16.

(lthou)h this ma! be counter0intuiti"e at 6rst, the probabilit! that any card in the dec9 ill be a heartbeore an! cards are seen is 1&'2 or 1'.

One a! to understand this is to sol"e the problem anal!ticall! or an! card b! buildin) a probabilit!

DtreeD and summin) the probabilit! o all o its Dbranches.D

@or e#ample, letEs 6nd the probabilit! that the 2nd card dealt rom the dec9 is a heart. There are tomutuall! e#clusi"e a!s this can happen: 1 both the 6rst and second cards are hearts or 2 onl! thesecond card is a heart.

+(SA 1: ?sin) the multiplication rule, the probabilit! that the 6rst card is a heart (>C the second cardis a heart is e4ual to the probabilit! o pic9in) a heart on the 6rst card or 1&'2, hich is the number o hearts in a ull dec9 di"ided b! the number o cards times the probabilit! o pic9in) a heart on thesecond card or 12'1, hich is the number o hearts remainin) in the dec9 di"ided b! the number ocards remainin) in the dec9. 1&'2 # 12'1 = 12'2;

+(SA 2: Similarl!, the probabilit! that the 6rst card is a non0heart (>C the second card is a heart ise4ual to the probabilit! that the 6rst card is >OT a heart or &B'2 times the probabilit! osubse4uentl! pic9in) a heart on the 2nd card or 1&'1.

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&B'2 # 1&'1 = &B'2;

Since these to cases are mutuall! e#clusi"e, e can add them to)ether to )et the total probabilit! o)ettin) a heart as the second card: 12'2; K &B'2; = 1'2; = 1'.

%e can do a similar anal!sis or an! card in the dec9, and, althou)h the probabilit! tree )ets morecomplicated as the card number )ets hi)her, the total probabilit! that the nth card dealt ill be a heartill ala!s end up simpli!in) to 1'.

The correct anser is (.

1,.

Since the 6rst to di)its o the license plate are 9non and there are 1; possibilities or each o theremainin) to di)its each can be an! di)it rom ; to B, the total number o combinations or di)its onthe license plate ill e4ual 1; ×1; = 1;;.*ecause there are onl! & letters that can be used or )o"ernment license plates (, *, or +, there are atotal o nine to0letter combinations that could be on the license plate & possibilities or 6rst letter × &possibilities or the second letter.

Ri"en that e ha"e 1;; possible di)it combinations and B possible letter combinations, the total

number o "ehicles to be inspected ill e4ual 1;; × B = B;;.Since it ta9es 1; minutes to inspect one "ehicle, the police ill ha"e time to inspect 18 "ehicles in threehours & hours = 18; minutes. Thus, the probabilit! o locatin) the transmitter ithin the allotted timeis 18'B;; = 1';. The correct anser is C.

18.

3etEs consider the di-erent scenarios:

Harriet ins all 6"e fips, she ends up ith Y1. Harriet ins our fips, and Tran ins one fip, Harriet is let ith Y1&. Harriet ins three fips, and Tran ins to fips, Harriet is let ith Y11. Harriet ins to fips, and Tran ins three fips, Harriet is let ith YB. Harriet ins one fip, and Tran ins our fips, Harriet is let ith YF. Harriet loses all 6"e fips, she ends up ith Y.

The 4uestion as9s or the probabilit! that Harriet ill end up ith more than Y1; but less than Y1. nother ords, e need to determine the probabilit! that Harriet is let ith Y11 or Y1& since there is noa! Harriet can end up ith Y12 or Y1.

The probability that -arriet ends up with $11 after the !e "ips#

Since there are 2 possible outcomes on each fip, and there are fips, the total number o possibleoutcomes is 2 # 2 # 2 # 2 # 2 = &2. Thus, the 6"e fips o the coin !ield &2 di-erent outcomes.

To determine the probabilit! that Harriet ill end up ith Y11, e need to determine ho man! o these&2 outcomes include a combination o e#actl! three innin) fips or Harriet and e#actl! to innin)fips or Tran. This is e4ui"alent to 6)urin) out the possible rearran)ements o THJAA HPs and T%O TPs in a @]A letterord.%e can create a s!stematic list o combinations that include three ins or Harriet and to ins or Tran: THHHT, THHTH, THTHH, TTHHH, HTHHT, HTHTH, HTTHH, HHTHT, HHTTH, HHHTT = 1; a!s.

(lternati"el!, e can count the combinations b! appl!in) the ana)ram method:

( * + C A

H H H T T

%e ta9e the actorial o the top and di"ide b! the actorial o each repeated letter on the bottom. Sincethere are to repeated letters, e )et ' & ^ 2 = 1; combinations.

Thus the probabilit! that Harriet ends up ith e#actl! Y11 ater fips is 1;'&2.

The probability that -arriet ends up with $1 after the !e "ips#

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To determine the probabilit! that Harriet ill end up ith Y1&, e need to determine ho man! o the&2 total possible outcomes include a combination o e#actl! our innin) fips or Harriet.

()ain, e can create a s!stematic list o combinations that include our ins or Harriet and one in or Tran: HHHHT, HHHTH, HHTHH, HTHHH, THHHH = a!s.

(lternati"el!, usin) the same reasonin) as abo"e, e can rite

( * + C A

H H H H T

The ormula !ields ' = combinations.

Thus the probabilit! that Harriet ends up ith e#actl! Y1& ater fips is '&2.

The total probability that -arriet ends up with either $11 or $1 after the !e "ips#

There are 1; a!s that Harriet is let ith Y11. There are a!s that Harriet is let ith Y1&.

Thereore, there are 1 a!s that Harriet is let ith more than Y1; but less than Y1.

Since there are &2 possible outcomes, the correct anser is 1'&2.

(lternati"el!, e can obser"e that the to possible a!s to Nsucceed accordin) to the terms o theproblem are connected b! a lo)ical OJ: Harriet can end up ith Y11 OJ Y1&. %hen e ha"e toa"enues to success that are connected b! a lo)ical OJ, e add the probabilities:1;'&2 K '&2 = 1'&2. The correct anser is C.

19.

@or an o"erlappin) set problem e can use a double0set matri# to or)aniQe our inormation andsol"e. The "alues here are percents, and no actual number o students is )i"en or re4uested. Thereore,e can assi)n a "alue o 1;; to the total number o students at +olle)e _. @rom the )i"en inormationin the 4uestion e ha"e:

*lue A!es >ot *lue A!es Total

*ron Hair ;

>ot *ron Hair I;

Total F; &; 1;;

The 4uestion as9s or the di-erence beteen ma#imum "alue and the minimum "alue o the centrals4uare, that is, the percent o students ho ha"e neither bron hair nor blue e!es. The ma#imum "alueis &;, as shon belo:


*ron Hair ; ; ;

>ot *ron Hair &; &; I;

Total F; &; 1;;

Thereore the ma#imum probabilit! o pic9in) such a person is ;.&. 3i9eise, the minimum "alue o the central s4uare is Qero, as shon belo:


*ron Hair 1; &; ;

>ot *ron Hair I; ; I;

Total F; &; 1;;

Thereore the minimum probabilit! o pic9in) such a person is ;, and the di-erence beteen thema#imum and the minimum probabilit! is ;.&.

2;. The chance o )ettin) (T 3A(ST one pair o cards ith the same "alue out o dealt cards should becomputed usin) the 10 x techni4ue. That is, !ou should 6)ure out the probabilit! o )ettin) >O 5(JS inthose cards an easier probabilit! to compute, and then subtract that probabilit! rom 1.

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@irst card: The probabilit! o )ettin) >O pairs so ar is 1 since onl! one card has been dealt.Second card: There is 1 card let in the dec9 ith the same "alue as the 6rst card. Thus, there are 1;cards that ill >OT orm a pair ith the 6rst card. %e ha"e 11 cards let in the dec9.5robabilit! o >O pairs so ar = 1;'11. Third card: Since e ha"e )otten no pairs so ar, e ha"e to cards dealt ith di-erent "alues. Thereare 2 cards in the dec9 ith the same "alues as those to cards. Thus, there are 8 cards that ill notorm a pair ith either o those to cards. %e ha"e 1; cards let in the dec9.5robabilit! o turnin) o"er a third card that does >OT orm a pair in an! a!, R]A> that e ha"e >Opairs so ar = 8'1;.

+umulati"e probabilit! o a"oidin) a pair *OTH on the second card (>C on the third card = product othe to probabilities abo"e = 1;'11 8'1; = 8'11.@ourth card: >o e ha"e three cards dealt ith di-erent "alues. There are & cards in the dec9 iththe same "alues thus, there are I cards in the dec9 that ill not orm a pair ith an! o the three dealtcards. %e ha"e B cards let in the dec9.5robabilit! o turnin) o"er a ourth card that does >OT orm a pair in an! a!, R]A> that e ha"e >Opairs so ar = I'B.+umulati"e probabilit! o a"oidin) a pair on the second card (>C on the third card (>C on the ourthcard = cumulati"e product = 1;'11 8'1; I'B = 1I'&&. Thus, the probabilit! o )ettin) (T 3A(ST O>A pair in the our cards is 1 0 1I'&& = 1F'&&. The correct anser is +.

1. The probabilit! that Zemphis does >OT in the competition is e4ual to 1 M p, here p is the probabilit!

that Zemphis COAS in the competition. Statement 1 states that the probabilit! that Zemphis oran! o the other cities does not in the competition is F'8. This e#plicitl! ansers the 4uestion so thisstatement alone is su\cient. Statement 2 )i"es us 1'8 as the "alue or p, the probabilit! thatZemphis COAS in the competition. %e can use this to calculate the probabilit! that Zemphis does>OT in the competition: 1 M 1'8 = F'8. This statement alone is su\cient to anser the 4uestion. The

correct anser is C

22. 41/50

23. 1/1770

24. 2/9

25. $3.6

26. 9/10

27. 13/14

28. 7/216

29. 2/25

30. ½

31. 8/25

32. 1/24

33. B

34. A

35. B

36. C

37. D

38. B

39. 4/7

40. A

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