Math 230.01, Fall 2012: HW 2 Solutions

1 Math , Fall 2012: HW 2 SolutionsThis homework is due at the beginning of class onThursday January 26th, 2011. You are free totalk with each other and get help. However, you should write up your own Solutions and understandeverything that you 1987 World Series was tied at two games a piece before the St. Louis Cardinalswon the fifth game. According to the Associated Press, The numbers of history support the Car-dinals and the momentum they carry. Whenever the series has been tied 2-2 the team that wonthe fifth game won the series 71% of the time. If momentum is not a factor and each team has a50% chance of winning each game (independently of the previous games), what is the probabilitythat the Game 5 winner will win the series?

2 (The World Series is a best of 7 series of games playedbetween the two teams. That is, the first team to win a total of 4 games wins the series.)SOLUTION: The winner of Game 5 must win one of the last two games. This can happen inone of two ways. Either the Game 5 winner wins Game 6 (W6) or they lose Game 6 (Wc6) and winGame 7 (W7). Since the outcomes of the games (including the first 5) are assumed to be independent:P(Game 5 winner wins series) =P(W6) +P(Wc6W7)=12+12 12= 75%.So, if anything, the numbers of history suggest that the Game 5 winner is less likely to win theseries than by shear Alice and Bob play the following game.

3 Alice flips three coins while Bob flipstwo. Alice wins if she has more Heads showing than ) Find the probability that Alice : LetXbe the number of heads that Alice flips andYbe the number of headsthat Bob flips. We partition the outcome space according to the eventsY= 0,Y= 1 andY= 2, soP(Alice wins) =P(X > Y)=P(Y= 0)P(X > Y|Y= 0) +P(Y= 1)P(X > Y|Y= 1) +P(Y= 2)P(X > Y|Y= 2).Observe thatP(X > Y|Y=i) =P(X > i|Y=i), and these events are independent (theydepend on separate sets of coins), soP(X > Y|Y=i) =P(X > i) andP(Alice wins) =P(Y= 0)P(X >0) +P(Y= 1)P(X >1) +P(Y= 2)P(X >2)=1478+1212+1418=12b) What is Alice s probability of winning if she flipsn+ 1 coins and Bob flipsncoins?

4 Is thissurprising?1 SOLUTION 1 (the hard way): Again, letXbe the number of heads that Alice flips andYbe the number that Bob flips. We can do a similar (but trickier) calculation to the oneabove,P(Alice wins) =P(X > Y)=n i=0P(Y=i)P(X > i)=n i=0(ni)(12)nn+1 j=i+1(n+ 1j)(12)n+1=(12)2n+1n i=0(ni)n+1 j=i+1(n+ 1j).At this point we observe that:n i=0(ni)n+1 j=i+1(n+ 1j)=n i=0(nn i)n+1 j=i+1(n+ 1n+ 1 j)=n k=0(nk)k `=0(n+ 1`)where in the second line we substitutedk=n iand`=n+ 1 jand reversed the order ofsummation (for instance,j=i+ 1 corresponds to`=n+ 1 (i+ 1) =n i=k, so the firstsummand in the second sum is now the last).

5 Together with the observation thatn i=0(ni)n+1 j=i+1(n+ 1j)+n k=0(nk)k `=0(n+ 1`)=n i=0(ni)n+1 j=0(n+ 1j)= 2n 2n+1= 22n+1we see thatn i=0(ni)n+1 j=i+1(n+ 1j)=1222n+1= 22nso we have our answer:P(Alice wins) =12but that was tedious! Instead, we could use more probabilistic 2 (the easy way): Suppose that Alice first flipsncoins, then flips one more fora total ofn+ 1 coin flips. LetXnbe the number of heads that Alice has on her < Ythen Alice cannot win, because even if her last coin is heads she can at best tiewith Bob. IfXn> Ythen Alice has won, regardless of the outcome of her last coin flip. IfXn=Y, then Alice wins if and only if her last coin shows heads.

6 The quantitiesXnandY2have the same distribution and are independent of one another (each is the number of headsshowing innindependent coin flips). So by symmetryP(Xn> Y) =P(Xn< Y).LettingWbe the event that Alice wins, we computeP(W) =P(Xn< Y)P(W|Xn< Y) +P(Xn> Y)P(W|Xn> Y) +P(Xn=Y)P(W|Xn=Y)=P(Xn< Y) 0 +P(Xn> Y) 1 +P(Xn=Y) 12=12[2P(Xn> Y) +P(Xn=Y)]=12[P(Xn< Y) +P(Xn> Y) +P(Xn=Y)]=12 While this may seem surprising at first, the last argument shows that the game is either deter-mined by Alice s firstnflips (and Bob s flips) in favor of either player with equal probability ,or by Alice s last flip, which is Heads with probability 1 3( #5 a, d).

7 Let be the sample space corresponding to three tosses of a coin. Givea verbal description of the following events:a)A={HHH, HHT, HT H, HT T};SOLUTION: Each of the outcomes inAhas a heads as the first toss, and these are all theoutcomes in in which the first toss is a heads. HenceA= the first toss lands heads. b)B={HHH, HHT, HT H, T HH};SOLUTION: Each of outcomes inBhas at least two heads, and each outcome in withat least two heads appears inB. ThereforeB= At least two out of the three coins land heads. c) Show :P(A) =P(B) =12, butP(A B) =386=(12)2d) Find a third event,C, so thatP(C) =12andP(A B C) =P(A)P(B)P(C).

8 So the eventsA, B, Care not independent, but they have this : There are a few answers that work here. Any event that has exactly one ofthe outcomesHHH,HHT, orHT Hand any three other outcomes will suffice. For example,C={HHT, HT T, T HT, T T T}, which is the event that The last toss lands tails , will (C) =12, and we can easily check thatP(A B C) =18=(12) 4( #11).LetA, B,andCbe three events. Use the inclusion-exclusion formula fortwo events to derive the following inclusion-exclusion formula for three events:P(A B C) =P(A) +P(B) +P(C) P(AB) P(AC) P(BC) +P(ABC).SOLUTION: We haveA B C= (A B) C.

9 We apply the inclusion-exclusion formula for twoevents to the events (A B) and C:P[(A B) C] =P(A B) +P(C) P[(A B) C]Again, by the inclusion-exclusion formula applied toA B, we haveP(A B) =P(A) +P(B) P(AB)Also, (A B) C= (A C) (B C), so we apply inclusion-exclusion to these two sets:P[(A B) C] =P[(A C) (B C)]=P(AC) +P(BC) P(ABC)(Recall that (A C) (B C) =A B C=ABC.). Combining these, we getP(A B C) =P[(A B) C]=P(A B) +P(C) P[(A B) C]=P(A) +P(B) P(AB) P[(A B) C]=P(A) +P(B) +P(C) P(AB) P(AC) P(BC) +P(ABC)Problem can 5 black and 5 white balls be put into two urns to maximize the probabilitythat a white ball is drawn when we draw from a randomly chosen urn?

10 SOLUTION: Put one white ball in the first urn and the other nine balls in the second urn. This givesa probability of (1/2) 1 + (1/2) (4/9) = 13/18 of drawing a white ball. In fact, this solution is best ifthere arenblack balls andnwhite balls, and gives probability (1/2) 1+ (1/2) n 12n 1=3n 24n 2of draw-ing a white ball. To prove that we can do no better, letW={draw white ball}, 1 ={pick urn 1}and 2 ={pick urn 2}so thatP(W) =P(1)P(W|1) +P(2)P(W|2) =12[P(W|1) +P(W|2)].Let us consider two cases. In the first case, we suppose that each urn has the same number of blackballs and white balls ( 3 black and 3 white in urn 1, and 2 black and 2 white in urn 2).