Sol II - ENGM2032 Applied Probability and Statistics Winter 2010 Assignment#2 Solutions 1 a n = 10 8 b n = 10 ¢ 9 ¢ 8 ¢ ¡¡ ¢ 3 = 10 P 8 = 1

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Unformatted text preview: ENGM2032, Applied Probability and Statistics Winter 2010 Assignment #2: Solutions 1) a) n = 10 8 b) n = 10 ¢ 9 ¢ 8 ¢ ¡¡¡ ¢ 3 = 10 P 8 = 1 , 814 , 400 c) n = (10 ¢ 9 ¢ 8 ¢ 7) ¢ (26 ¢ 25 ¢ 24 ¢ 23) = 10 P 4 ¢ 26 P 4 = 1 , 808 , 352 , 000 d) If we know the first, second, fourth, and seventh character of the code, then we don’t know 4 characters, of which 2 are digits and 2 are roman letters. We also only have 8 digits left to choose from and 24 roman letters. The remaining number of possible codes are n = (8 ¢ 7) ¢ (24 ¢ 23) = 8 P 2 ¢ 24 P 2 = 30 , 912 So our chance of correctly guessing the correct code in the next 1000 guesses is 1000 / 30912 1 / 31. Notice that knowing half of the code increases the probability of guessing the code by a factor of almost 60,000. 2) a) n = 4 ¢ 2 ¢ 3 = 24 possible routes from A to D. b) This problem is best addressed by enumerating the possible routes, adding up the average driving times for the route, and then counting those routes with average total driving times below 7.8 hours. Let T ijk be the total average driving time when route i is taken between cities A and B, route j is taken between B and C, and route k is taken between C and D. Then T 111 = 2 . 3 + 3 . 2 + 1 . 8 = 7.3 T 121 = 2 . 3 + 2 . 8 + 1 . 8 = 6.9 T 211 = 2 . 6 + 3 . 2 + 1 . 8 = 7.6 T 221 = 2 . 6 + 2 . 8 + 1 . 8 = 7.2 T 311 = 2 . 1 + 3 . 2 + 1 . 8 = 7.1 T 321 = 2 . 1 + 2 . 8 + 1 . 8 = 6.7 T 411 = 3 . 2 + 3 . 2 + 1 . 8 = 8 . 2 T 421 = 3 . 2 + 2 . 8 + 1 . 8 = 7 . 8 T 112 = 2 . 3 + 3 . 2 + 2 . 4 = 7 . 9 T 122 = 2 . 3 + 2 . 8 + 2 . 4 = 7.5 T 212 = 2 . 6 + 3 . 2 + 2 . 4 = 8 . 2 T 222 = 2 . 6 + 2 . 8 + 2 . 4 = 7 . 8 T 312 = 2 . 1 + 3 . 2 + 2 . 4 = 7.7 T 322 = 2 . 1 + 2 . 8 + 2 . 4 = 7.3 T 412 = 3 . 2 + 3 . 2 + 2 . 4 = 8 . 8 T 422 = 3 . 2 + 2 . 8 + 2 . 4 = 8 . 1 T 113 = 2 . 3 + 3 . 2 + 2 . 1 = 7.6 T 123 = 2 . 3 + 2 . 8 + 2 . 1 = 7.2 T 213 = 2 . 6 + 3 . 2 + 2 . 1 = 7 . 9 T 223 = 2 . 6 + 2 . 8 + 2 . 1 = 7.5 T 313 = 2 . 1 + 3 . 2 + 2 . 1 = 7.4 T 323 = 2 . 1 + 2 . 8 + 2 . 1 = 7.0 T 413 = 3 . 2 + 3 . 2 + 2 . 1 = 8 . 5 T 423 = 3 . 2 + 2 . 8 + 2 . 1 = 8 . 1 and we see that there are 14 different routes having average driving time less than 7.8and we see that there are 14 different routes having average driving time less than 7....
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This note was uploaded on 02/24/2010 for the course ENGM 2032 taught by Professor Yao during the Spring '10 term at Dalhousie.

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Sol II - ENGM2032 Applied Probability and Statistics Winter 2010 Assignment#2 Solutions 1 a n = 10 8 b n = 10 ¢ 9 ¢ 8 ¢ ¡¡ ¢ 3 = 10 P 8 = 1

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