Math 341: Probability: Fall 2009 Comments on Hw Problems

A key part of any math course is doing the homework. This ranges from reading the material in the book so that you can do the problems to thinking about the problem statement, how you might go about solving it, and why some approaches work and others don’t. Another important part, which is often forgotten, is how the problem fits into math. Is this a cookbook problem with made up numbers and functions to test whether or not you’ve mastered the basic material, or does it have important applications throughout math and industry? Below I’ll try and provide some comments to place the problems and their solutions in context. 1. HW #1 The first assignment was: Due Thursday, 9/17 (but as this is the first assignment, no late penalty if you put it in my mailbox by 10am on Friday the 18th): Section 1.3: #2, #3, #5; Combinatorics: (1) There are 2n people who enter as n pairs of two. The people are then randomly matched in pairs. What is the probability everyone is matched with their initial partner? There are two ways to interpret this problem; either is fine so long as you state which interpretation. In one interpretation, say there are n people from Williams and n from Amherst, matched in n pairs with each pair having someone from Williams and someone from Amherst. In the new matching, you must match someone from Williams with someone from Amherst. In the other interpretation, anyone can be matched with anyone. You may solve either problem, just clearly state which one you are doing (not surprisingly, the answers differ). (2) Consider n people ordered 1, 2, ..., n. We randomly assign another ordering to these people – what is the probability at least one person is assigned the same number twice? Section 1.4: #2, #4. Section 1.8: #2, #4, #6, #12. Section 1.3: Problem 2: This problem on Murphy’s law is quite important, and will be used later for the elementary analysis of the symmetric random walk (also known as the Gambler’s ruin). If we consider a sequence of k tosses, then the probability it is observed when we toss a fair coin k times is just p = 1/2, or the probability it does not happen is 1− p = 1− 1/2. If we toss a fair coin kN times, then the probability it does not happen in one of these blocks is (1− p) , which tends to 0 as N → ∞. Note that the sequence could still occur even if it doesn’t occur entirely in one block. For example, say our sequence is TTHT. Imagine we toss the coin 20 times, and get THHT THTH THHT TTTH HHTH. Date: December 8, 2009. 1 2 STEVEN J. MILLER ([email protected]) Note none of the blocks of four have the sequence TTHT, but it does occur in the sequence of 20 (part in the first and part in the second blocks). Section 1.3: Problem 3: For this problem, the trick is to just enumerate so that you cover all possibilities. This problem is more to test understanding of the material instead of applications for later. I find it is easiest to give each cup a label (so red cup 1, red cup 2). We might as well place the six saucers in order: red saucer 1, red saucer 2, ..., star saucer 2. There are 6! ways to arrange the 6 cups on the saucers (we ARE distinguishing between which red cup is place on a given saucer). To count how many ways to place the cups so that nothing is placed on the same color, there are three possibilities: the two reds are placed on the two whites, the two reds are placed on the two stars, or one red on a white and one red on a star. The problem is completed by counting all configurations like this. Section 1.3: Problem 5: This problem can be interpreted as saying that if we have a countable collection of events and each event happens with probability 1, then their intersection happens with probability 1. The simplest way to prove this is by induction. If X and Y happen with probability one, then P(X ∩Y ) = P(X) +P(Y )−P(X ∪Y ). Note every probability on the right hand side equals 1 (no event can have probability greater than 1, and X ⊂ X ∪ Y so P(A ∪ Y ) = 1). This implies P(X ∩ Y ) = 1. Proceed by induction, setting X = ∪r=1Ar and Y = An+1 to get P(∩ r=1Ar) = 1 for all n. The proof is completed by invoking Lemma 5 on page 7. We could have argued slightly differently above. The key is proving P(X ∩ Y ) = 1; another approach is to use partitions, and observe P(X) = P(X ∩ Y ) +P(X ∩ Y ). As P(Y ) = 1, P(Y ) = 0 and thus P(X ∩Y ) = 0 (as X ∩Y c ⊂ Y ). Thus P(X) = P(X ∩Y ), and as P(X) = 1 we finally deduce P(X ∩ Y ) = 1. Note how important in this problem the n = 2 case is in the inductive proof. Frequently in induction proofs we just need to use the result with n to prove n+ 1; however, a sizable number of times the general proof basically just reduces to understanding the n = 2 case. Combinatorics Problem (1): If anyone can be matched with anyone, there are (2n − 1)!! ways to do this, where the double factorial means we take the product of every other term (6!! = 6 ⋅ 4 ⋅ 2 and 5!! = 5 ⋅ 3 ⋅ 1). One way to see this is to note this is just ( 2n 2 )( 2n− 2 2 ) ⋅ ⋅ ⋅ ( 4 2 )( 2 2 ) ⋅ 1 n! ; we divide by n! as we have attached labels to each pair of people, and there aren’t supposed to be labels. We could also proceed by induction. The first person must be matched with someone; there are 2n − 1 ways to do this. We now pair off the remaining 2n − 2 people, which by induction happens (2n − 3)!! ways, so there are (2n − 1) ⋅ (2n − 3)!! = (2n − 1)!! ways. If you must be matched with someone from the opposite side, there are only n! ways. Combinatorics Problem (2): We solve this by inclusion-exclusion. Let Ai be the event that i is in the ith place, Aij be that i and j are in their respective places (with i ∕= j), and so on. Note P(A134) = P(A459) and so on. Then the probability that at least one MATH 341: PROBABILITY: FALL 2009 COMMENTS ON HW PROBLEMS 3 person is in the right spot is