p chapter 2 solutions 3rd fall 2007

p chapter 2 solutions 3rd fall 2007 - Probability, Third...

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Sheet1 Page 1 Probability, Third Edition By David J Carr & Michael A Gauger Published by BPP Professional Education Solution 2.1 (i) There are n1 =4 ways to choose the color of the cap, n2 =6 ways to choose the color of the shirt, and n3 =2 ways to choose the color of the pants. So the number of different color combinations for the uniform is: 462 = nnn Err:501 48 123 (ii) There are n1 =4 ways to choose the color of the cap, n2 =4 ways to choose the color of the shirt, and n3 =1 way to choose the color of the pants. So the number of different color combinations for the uniform is: 441 nnn Err:501 16 123 (iii) First, note that the set of pants colors is a subset of the set of cap colors, which in turn is a subset of the shirt colors. If we first select the pant colors, there are n1 =2 choices. Once the color of the pants is determined, there are n2 =-=41 3 ways to choose a different color for the cap. Once the color of both the pants and the cap have been chosen, there are n3 =62-=4 ways to choose a different color for the shirt. So the number of different color combinations for the uniform is: 234 = nnn Err:501 24 123 Solution 2.2 (i) The number of ordered samples of 3 objects from a set of 10 objects with replacement is: 103 =1,000 Solutions to practice questions h Chapter 2
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Sheet1 Page 2 (ii) The number of ordered samples of 3 objects from a set of 10 objects without replacement is: P=10 = 9 8 720 10 3 1 Probability (iii) This part of the problem is quite difficult. Let nn, , and n be the number of possible choices for the 12 3 first, second and third balls chosen. Then notice that the values of n2 and n3 depend on whether the previous balls were odd or even. The following table gives a breakdown of all the possible outcomes: Ball 1 n1 Ball 2 n2 n3 12 3nn n Odd Odd Even Even 5 5 5 5 Odd Even Odd Even 4 5 5 5 8 9 9 10 160 225 225 250 Hence the total number of possible outcomes if the odd balls are not replaced is: 160 225 225 250 = 860 Solution 2.3 Solutions to practice questions h Chapter 2
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Sheet1 Page 3 Any runner can win at most one medal. So we are choosing an ordered sample of 3 objects from a set of 30 objects without replacement. The number of possible outcomes is: 30 P3 = 30 29 28 = 24360 Solution 2.4 The number of heads could be 3, 4 or 5. Since the order of the outcomes does not matter, the number of possible outcomes that include at three or more heads is: C + C + C = 10 5 1 = 16 53 5 4 55 Solution 2.5 (i) The seven distinct letters in WYOMING can be rearranged in 7! =5,040 ways. (ii) 7! Err:520 2520 2! (iii) There are 13 letters in the word MASSACHUSETTS. The letter A occurs twice, the letter T occurs twice, and the letter S occurs four times. Each of the other five letters occurs just once. So the number of ordered rearrangements is: 13! Err:520
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This note was uploaded on 09/08/2008 for the course STT 490 taught by Professor Dudziak during the Spring '08 term at Michigan State University.

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p chapter 2 solutions 3rd fall 2007 - Probability, Third...

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