Math135_hw2-soln

Math135_hw2-soln - Math 135, Fall 2011: HW 2 This homework...

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Unformatted text preview: Math 135, Fall 2011: HW 2 This homework is due at the beginning of class on Wednesday September 7th, 2011 . You are free to talk with each other and get help. However, you should write up your own solutions and understand everything that you write. Problem 1. The 1987 World Series was tied at two games a piece before the St. Louis Cardinals won 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 won the fifth game won the series 71% of the time. If momentum is not a factor and each team has a 50% chance of winning each game (independently of the previous games), what is the probability that the Game 5 winner will win the series? (The World Series is a best of 7 series of games played between 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 in one of two ways. Either the Game 5 winner wins Game 6 ( W 6 ) or they lose Game 6 ( W c 6 ) and win Game 7 ( W 7 ). Since the outcomes of the games (including the first 5) are assumed to be independent: P (Game 5 winner wins series) = P ( W 6 ) + P ( W c 6 W 7 ) = 1 2 + 1 2 1 2 = 75% . So, if anything, the numbers of history suggest that the Game 5 winner is less likely to win the series than by shear chance. Problem 2. Suppose Alice and Bob play the following game. Alice flips three coins while Bob flips two. Alice wins if she has more Heads showing than Bob. a) Find the probability that Alice wins. SOLUTION: Let X be the number of heads that Alice flips and Y be the number of heads that Bob flips. We partition the outcome space according to the events Y = 0, Y = 1 and Y = 2, so P (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 that P ( X > Y | Y = i ) = P ( X > i | Y = i ), and these events are independent (they depend on separate sets of coins), so P ( X > Y | Y = i ) = P ( X > i ) and P (Alice wins) = P ( Y = 0) P ( X > 0) + P ( Y = 1) P ( X > 1) + P ( Y = 2) P ( X > 2) = 1 4 7 8 + 1 2 1 2 + 1 4 1 8 = 1 2 b) What is Alices probability of winning if she flips n + 1 coins and Bob flips n coins? Is this surprising? 1 SOLUTION 1 (the hard way): Again, let X be the number of heads that Alice flips and Y be the number that Bob flips. We can do a similar (but trickier) calculation to the one above, P (Alice wins) = P ( X > Y ) = n X i =0 P ( Y = i ) P ( X > i ) = n X i =0 n i 1 2 n n +1 X j = i +1 n + 1 j 1 2 n +1 = 1 2 2 n +1 n X i =0 n i n +1 X j = i +1 n + 1 j . At this point we observe that: n X i =0 n i n +1 X j = i +1 n + 1 j = n X i =0 n n- i n +1 X j = i +1 n + 1 n + 1- j = n X k =0 n k k X =0 n + 1 where in the second line we substituted k = n- i and = n + 1- j and reversed the order of summation (for instance, j = i + 1 corresponds to = n + 1- (...
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Math135_hw2-soln - Math 135, Fall 2011: HW 2 This homework...

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