CS 70 Discrete Mathematics and Probability Theory Summer 2015 Chung-Wei Lin Final

19 Pages·2015·0.21 MB·English

CS 70 Discrete Mathematics and Probability Theory
Summer 2015 Chung-Wei Lin Final
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CS70,Summer2015,Final 1
SID:
Section 1: Straightforward Questions (68 points)
1. CountingandProbability(21points,3pointsforeachpart)
Forthefollowingparts,justwritedownanswers. Youdonotneedtocalculatetheexactvalueof(cid:0)a(cid:1),a!,or
b
ab. Noexplanationisrequired. Nopartialcreditwillbegiven.
(a) HowmanydifferentanagramsofDISCRETEarethere?
Answer: 8!.
2
(b) Thereare2015students. Howmanydifferentwaystochoose70studentsandletthemformateam?
Answer: (cid:0)2015(cid:1).
70
(c) Howmanydifferentsubsetsof{1,2,3,...,300}arethere?
Answer: 2300.
(d) Howmanydifferentnon-negativeintegersolutionstox +x +x +x +···+x =2015? (Notethatit
0 1 2 3 70
is0-indexed.)
Answer: (cid:0)2085(cid:1)or(cid:0)2085(cid:1).
70 2015
(e) The birthdays of three students are all in March, 1997 (there are 31 days in March). Assume their
birthdays are randomly and uniformly distributed. What is the probability that their birthdays are all
different?
Answer: 31×30×29.
31×31×31
(f) Thereare11pokercards{♠A,♠J,♠Q,♠K,♥A,♥2,♥3,♦A,♣A,♣4,♣5}. Youdrawacard
anditisanAce(A);whatistheprobabilitythatthecardistheHeart(♥)suit?
Answer: 1.
4
(g) ThenumberoftyposperboardismodelledbythePoissondistribution:
λi
Pr[X =i]= e−λ fori=0,1,2,...
i!
Ifthereis1typoperboardonaverage,whatistheprobabilitythatthereare2typosonaboard?
Answer: 1 .
2e
Pr[X =2]= 12e−1= 1 .
2! 2e
2
SID:
2. Distribution,Expectation,andVariance(9points,3pointsforeachpart)
Forthefollowingparts,justwritedownanswers. Noexplanationisrequired. Nopartialcreditwillbegiven.
(a) A special dice has 2 sides with “0”, 3 sides with “1”, and 1 side with “3”. Each side has the same
probabilityofbeingrolled. LetrandomvariableX betheoutcomeofthedice. Whatisthedistribution
ofX?
Answer: Pr[X =0]= 2,Pr[X =1]= 3,Pr[X =3]= 1.
6 6 6
(b) WhatistheexpectationofX? Calculatetheexactvalue.
Answer: 1.
E(X)=0·2+1·3+3·1 =1.
6 6 6
(c) WhatisthevarianceofX? Calculatetheexactvalue.
Answer: 1.
E(X2)=02·2+12·3+32·1 =2andVar(X)=E(X2)−E(X)2=2−1=1.
6 6 6
3
SID:
3. Markov’sInequalityandChebyshev’sInequality(9points,3pointsforeachpart)
Forthefollowingparts,justwritedownanswers. Noexplanationisrequired. Nopartialcreditwillbegiven.
(a) GivenapositivediscreterandomvariableX withitsexpectation10,useMarkov’sinequalitytoprovide
anupperboundonPr[X ≥30]?
Answer: 1.
3
Pr[X ≥30]≤ 10 = 1.
30 3
(b) Following Part (a), the variance of X is 5, use Chebyshev’s inequality to provide an upper bound on
Pr[|X−10|≥20]?
Answer: 5 .
400
Pr[|X−10|≥20]≤ 5 = 5 .
202 400
(c) Following Parts (a) and (b), provide an upper bound (this bound should be smaller than the answer in
Part(a))onPr[X ≥30]?
Answer: 5 .
400
Pr[X ≥30]=Pr[X−10≥20]≤Pr[|X−10|≥20]≤ 5 .
400
4
SID:
4. StableMarriageAlgorithm(3points)
Find a female optimal stable pairing for the following case. No explanation is required. No partial credit
willbegiven.
Men PreferenceLists Women PreferenceLists
1 C>A>B A 1>2>3
2 B>C>A B 2>1>3
3 B>A>C C 2>3>1
Answer: {(1,A),(2,B),(3,C)}.
5
SID:
5. ModularArithmeticandRSA(6points,3pointsforeachpart)
Forthefollowingparts,justwritedownanswers. Noexplanationisrequired. Nopartialcreditwillbegiven.
(a) Whatisthemultiplicativeinverseof6(mod7)?
Answer: 6.
6·6=36=1(mod7).
(b) Assume p = 3,q = 11,e = 7,d = 3 in the RSA cryptography; if the receiver receives 4, what is the
originalmessage?
Answer: 31.
N = pq=33and4d =43=64=31(mod33).
6
SID:
6. PolynomialsandErrorCorrectionCodes(6points,3pointsforeachpart)
Forthefollowingparts,justwritedownanswers. Noexplanationisrequired. Nopartialcreditwillbegiven.
(a) WorkingoverGF(7),thereisapolynomialP(x)withdegreeatmost2. GivenP(0)=3,P(1)=0,and
P(2)=0,whatisthevalueofP(3)? Calculatetheexactvalue.
Answer: 3.
(x−1)(x−2) 1
∆ (x) = = (x2−3x+2);
0
(0−1)(0−2) 2
(cid:18) (cid:19)
1 3
P(x) = 3 (x2−3x+2) = (x2−3x+2);
2 2
3
P(3) = (32−3·3+2)=3(mod7).
2
(b) Kevinwantstosendamessageof10orderedpacketstoStuart. Nomatterhowmanypacketsaresent,
there are at most 5 lost packets (erasure errors) and 2 corrupted packets (general errors). What is the
minimumnumberofpacketsKevinshouldsendsothatStuartcanrecoverthemessage?
Answer: 19.
10+5+2·2=19.
7
SID:
7. TRUEorFALSE—FirstRound(14points,1pointforeachpart)
For any discrete random variable X,Y,Z, any real numberC, and any proposition P,Q, determine whether
the following statements are true or false. Just circle the correct choice. No explanation is required. No
partialcreditwillbegiven.
(a) T F ThereexistsanEulerianwalkforthegraphabove.
(b) T F E(X+C)=E(X)+C.
(c) T F E(X+Y)=E(X)+E(Y).
(d) T F Var(X+C)=Var(X).
(e) T F Var(X+Y)=Var(X)+Var(Y).
(f) T F Ifyouflipafaircoin2ntimes,theprobabilitythatyougetexactlynHeads
increasesasnincreases.
(g) T F Theset{1,2,3}iscountablyinfinite.
(h) T F N,thesetofallnaturalnumbers,iscountablyinfinite.
(i) T F Z,thesetofallintegers,iscountablyinfinite.
(j) T F Q,thesetofallrationalnumbers,iscountablyinfinite.
(k) T F Thesetofall(n,q)wheren∈Nandq∈Q,iscountablyinfinite.
(l) T F R,thesetofallrealnumbers,iscountablyinfinite.
(m) T F (P=⇒Q)=⇒(Q=⇒P).
(n) T F ∀y∃xP=⇒∃x∀yP.
Answer: T,T,T,T,F,F,F,T,T,T,T,F,F,F.
8
SID:
Section 2: 8-Point Questions (16 points)
8. MultinomialDistribution(8points)
Thereare3coins:
• Coin1hasHeadprobability 1.
3
• Coin2hasHeadprobability 1.
2
• Coin3hasHeadprobability 3.
4
Yourepeatthefollowingrandomexperiment100times:
• YouflipCoin1first. IfyougetHead,youflipCoin2;otherwise,youflipCoin3.
What is the expectation of the number of Heads? Note that you flip coins 200 times in total. Calculate the
exactvalue.
Answer: 100.
H isarandomvariablethatcountsthetotalnumberofHeads. H isthenumberofHeadsontheitrial.
i
1 1 1 1 2 3 2 1
E(H)=2· · +1· · +1· · +0· · =1;
i
3 2 3 2 3 4 3 4
n
E(H)= ∑E(H)=100.
i
i=1
9
SID:
9. GeometricDistribution(8points)
Cocoistryingtocollectfourcoupons,C ,C ,C ,C . Shedoesnotknowwhichcoupon(s)shewillgetbefore
1 2 3 4
paying. Therearetwostoressellingcoupons:
• InStoreA,shepays3dollarstogetacoupon,andtheprobabilityofgettingeachcouponisalways 1.
4
• InStoreB,C andC arealwayspackagedtogether,andC andC arealsoalwayspackagedtogether.
1 2 3 4
She pays 8 dollars to get one package (with two coupons), and the probability of getting each pair
(either(C ,C )or(C ,C ))isalways 1.
1 2 3 4 2
Whichstoreshouldshego,oritdoesnotmatter?
Answer: StoreB.
Let A=A +A +A +A where A is the number of tries until Coco gets the i-th new coupon. Similarly,
1 2 3 4 i
B=B +B . Inthiscase,thereareonly2distinctpackages.
1 2
4 4 4 4 4 25
E(A)= ∑E(A)= + + + = ;
i
4 3 2 1 3
i=1
2 2 2
E(B)= ∑E(B)= + =3.
i
2 1
i=1
CocoshouldgotoStoreBbecause3·25 >8·3.
3
10