Math135_hw1-soln

# Math135_hw1-soln - Math 135, Fall 2011: HW 1 This homework...

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Unformatted text preview: Math 135, Fall 2011: HW 1 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 (p.9 #2) . Suppose a word is picked at random from this sentence. Find: a) the chance the word has at least 4 letters; SOLUTION: All words are equally likely to be chosen. The sentence has 10 words; 7 are at least 4 letters long. We have P (at least 4 letters) = 7 10 . b) the chance that the word contains at least 2 vowels (a,e,i,o,u); SOLUTION: Assuming that the vowels need not be distinct (i.e. the three instances of “e” in “sentence” count as three vowels in the word), then exactly 4 words contain at least two vowels, so P (at least 2 vowels) = 4 10 . c) the chance that the word contains at least 4 letters and at least 2 vowels. SOLUTION: Exactly four words contain at least 4 letters and at least two vowels, so P (at least 4 letters and at least 2 vowels) = 4 10 . d) What is the distribution of the length of the word picked? SOLUTION: Let i be the length of the word. Then the values of i which occur with nonzero probability are i = 1 , 2 , 4 , 6 , 7 , and 8. We have P (1) = 1 10 , P (2) = 2 10 , P (4) = 3 10 , P (6) = 2 10 , P (7) = 1 10 , P (8) = 1 10 . e) What is the distribution of the number of vowels in the word? SOLUTION: Let j be the number of vowels in the word. Then the values of j which oc- cur with nonzero probability are j = 1 , 2 , 3. We have P (1) = 6 10 , P (2) = 2 10 , P (3) = 2 10 . Problem 2 (p.9 #7) . Suppose two 4-sided dice are rolled. Find the probabilities of the following events: a) the maximum of the two numbers rolled is less than or equal to 2; SOLUTION: The sample space is the set of ordered pairs ( i,j ), where i and j are integers between 1 and 4, and i represents the number on the upper face of the first die and j the number on the upper face of the second die. All outcomes are equally likely. There are 16 outcomes. The outcomes (1 , 1), (1 , 2), (2 , 1), and (2 , 2) are the outcomes whose maximum is less than or equal to 2. Hence P (maximum is less than or equal to 2) = 1 4 . 1 b) the maximum of the two numbers rolled is less than or equal to 3; SOLUTION: In order for the maximum of the two numbers to be less than or equal to 3, both numbers must be less than or equal to 3. There are 3 × 3 = 9 ways for this to happen, so P (maximum is less than or equal to 3) = 9 16 . c) the maximum of the two numbers rolled is equal to 3; SOLUTION: In order for the maximum to be equal to 3, at least one number must be 3, and the another number can be anything less than or equal to 3. There are two choices for which die results in a 3; once this is fixed, there are 2 choices for the outcome of second die to be strictly less than 3, yielding 2(3- 1) outcomes in which the one number is 3 and the other is strictly less than 3. Finally there is the outcome in which both numbers are 3. So thereis strictly less than 3....
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## This note was uploaded on 10/24/2011 for the course MATH 135 taught by Professor Staff during the Fall '08 term at Duke.

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Math135_hw1-soln - Math 135, Fall 2011: HW 1 This homework...

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