hw2sol - Math 361 X1 Homework 2 Solutions Spring 2003...

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Unformatted text preview: Math 361 X1 Homework 2 Solutions Spring 2003 Graded problems: 1(ii); 2(c); 3(b); 6; each worth 3 pts., maximal score is 12 pts. Problem 1. Variation on the birthday theme. This problem is motivated by the following letter from a reader published in Marilyn vos Savants Parade Magazine column (see also the Problem Sampler): I started counting from the first person who came into the office. I counted until I found a matching birthday in the group. Then I started a new survey with the next person. In eight surveys, the smallest number of people it took before I found a matching birthday was 12. The largest number was only 54. This suggests some natural questions, such as: What is the probability that it takes exactly 12 people to find a matching birthday? What is the probability that it takes less than (or more than) 12? What are the corresponding probabilities with 54 in place of 12? More generally, let k be a given integer between 2 and 365, and solve the following two problems. (i) What is the probability that it takes more than k people to find a matching birthday? (ii) What is the probability that it takes exactly k people to find a matching birthday (i.e., that the k th person walking into the office is the first one at which a matching birthday occurs). Remarks: Part (i) is easy, if you approach the problem the right way; in particular, do not try to solve this using the result of part (ii). In both cases, give a general formula (involving k ) for the probability in question. Numerical answers are not required. It would be easy to obtain numerical data, using a programmable calculator or software such as Mathematica; if you are curious, do that and plot (or present in table form) the probability computed in (ii) as a function of k . Solution. (i) This is just a disguised form of the standard birthday type problem. The key here is to realize that it takes more than k people to find a matching birthday, if and only if, the first k people have all distinct birthdays . Thus the probability in question is the same as the probability that there are no matching birthdays in a group of k people, which is equal to 365 364 (365- k +1) / 365 k , by the argument used for the usual birthday problem. (Take = { ( b 1 ,...,b k ) : b i = 1 , 2 ,..., 365 } , A = { ( b 1 ,...,b k ) : b i = 1 , 2 ,..., 365 , b i distinct } , so that #() = 365 k , #( A ) = 365 364 (365- k +1) and P ( A ) = #( A ) / #() = 365 364 (365- k + 1) / 365 k . (ii) There are two ways to solve this problem. Method 1: We work with an explicit outcome space , and identify the event in question as a subset A of , similar to the standard birthday problem: We take, as usual in these situations, = { ( b 1 ,...,b k ) : b i = 1 , 2 ,..., 365 } , representing all possible k-tuples of birthdays of a group of k people, and the event A = { ( b 1 ,...,b k ) : b i = 1 , 2 ,..., 365; b i distinct for i = 1 , 2 ,...,k- 1 , b k = b i for some i = 1 , 2 ,...,k,....
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hw2sol - Math 361 X1 Homework 2 Solutions Spring 2003...

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