This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ECEN 303: Assignment 4 Problems: 1. De M´ er´ e’s puzzle. A sixsided die is rolled three times independently. Which is more likely: a sum of 11 or a sum of 12? (This question was posed by the French nobleman de M´ er´ e to his friend Pascal in the 17th century.) A sum of 11 is obtained with the following 6 combinations: (6 , 4 , 1) (6 , 3 , 2) (5 , 5 , 1) (5 , 4 , 2) (5 , 3 , 3) (4 , 4 , 3) . A sum of 12 is obtained with the following 6 combinations: (6 , 5 , 1) (6 , 4 , 2) (6 , 3 , 3) (5 , 5 , 2) (5 , 4 , 3) (4 , 4 , 4) . Each combination of 3 distinct numbers corresponds to 6 permutations, while each combi nation of 3 numbers, two of which are equal, corresponds to 3 permutations. Counting the number of permutations in the 6 combinations corresponding to a sum of 11, we obtain 6 + 6 + 3 + 6 + 3 + 3 = 27 permutations. Counting the number of permutations in the 6 combinations corresponding to a sum of 12, we obtain 6 + 6 + 3 + 3 + 6 + 1 = 25 per mutations. Since all permutations are equally likely, a sum of 11 is more likely than a sum of 12. Note also that the sample space has 6 3 = 216 elements, so we have Pr(11) = 27 / 216, Pr(12) = 25 / 216. 2. The birthday problem. Consider n people who are attending a party. We assume that every person has an equal probability of being born on any day during the year, indpendently of everyone else, and ignore the additional compication presented by leap years (i.e., nobody is born on February 29). What is the probability that each person has a distinct birthday? The sample space consists of all possible choices for the birthday of each person. Since there are n persons, and each has 365 choices for their birthday, the sample space has 365 n elements. Let us now consider those choices of birthdays for which no two persons have the same birthday. Assuming that n ≤ 365, there are 365 choices for the first person, 364 for the second, etc., for a total of 365 · 364 ··· (365 − n + 1). Thus, Pr(no two birthdays coincide) = 365 · 364 ··· (365 − n + 1) 365 n . It is interesting to note that for n as small as 23, the probability that there are two persons with the same birthday is larger than 1/2. 3. An urn contains m red and n white balls. (a) We draw two balls randomly and simultaneously. Describe the sample space and calcu late the probability that the selected balls are of different color, by using two approaches: 1 a counting approach based on the discrete uniform law, and a sequential approach based on the multiplication rule. We number the red balls from 1 to m , and the white balls from m + 1 to m + n . One possible sample space consists of all pairs of integers ( i,j ) with 1 ≤ i,j ≤ m + n and i negationslash = j . The total number of possible outcomes is ( m + n )( m + n − 1). The number of outcomes corresponding to redwhite selection, (i.e., i ∈ { 1 ,...,m } and j ∈ {...
View
Full
Document
This note was uploaded on 02/11/2010 for the course ECEN 303 taught by Professor Chamberlain during the Fall '07 term at Texas A&M.
 Fall '07
 Chamberlain

Click to edit the document details