part6 - EE 5375/7375 Random Processes Homework#1 Solutions...

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EE 5375/7375 Random Processes September 2, 2003 Homework #1 Solutions Problem 1. textbook problem 2.37 In how many ways can 10 students occupy 10 desks? 12 desks? Imagine the sequence of students selecting desks. When there are 10 desks, the first student has 10 choices, the second student 9 choices, and so on. Therefore 10 students can fill 10 desks in 10! = 3 , 628 , 800 ways. If there are 12 desks, the first student has 12 choices, the second student has 11 choices, and so on. Therefore 10 students can fill 12 desks in 12 × 11 × · · · × 3 = 239 , 500 , 800 ways. Problem 2. textbook problem 2.43 You win a lottery if you correctly predict the numbers of 6 balls drawn from an urn containing balls numbered 1, 2,..., 49, without replacement and without regard to ordering. What is the probability of winning if you buy one ticket? There are ( 49 6 ) = 13 , 983 , 816 ways of choosing 6 balls randomly out of 49 balls. Each of these ways is equally likely, so the chance of winning the lottery with one ticket is 1 / 13 , 983 , 816 = 7 . 2 × 10 - 8 . Problem 3. textbook problem 2.53 In each lot of 100 items, 2 items are tested, and the lot is rejected if either of the tested items is found defective. (a) Find the probability of accepting a lot with 5 defective items. Repeat for 10 defective items. (b) Recompute the probabilities in part (a) if 3 items are tested, and a lot is accepted when at most 1 of the 3 tested items is found defective. (a) Define A 1 as the event that the first item is non-defective and A 2 as the event that the second item is non-defective. P (lot accepted) = P ( A 1 A 2 ) = P ( A 2 | A 1 ) P ( A 1 ) With 5 defective items and 95 non-defective items in the lot, P ( A 1 ) = 95 100 Given that the first item is non-defective, there are only 94 non-defective items left in the lot for the second test. Hence P ( A 2 | A 1 ) = 94 99 Combining, P (lot accepted) = (95)(94) (100)(99) = 0 . 902 If there are 10 defective and 90 non-defective items in the lot, similar reasoning leads to P (lot accepted) = P ( A 1 A 2 ) = P ( A 2 | A 1 ) P ( A 1 ) = 89 99 90 100 = 0 . 809 (b) Define the additional event A 3 as the event that the third item is non-defective. If a lot is accepted when at most 1 of the 3 tested items is defective, then the lot is accepted in any of 4 events: ( A 1 A 2 A 3 ) or ( A c 1 A 2 A 3 ) or ( A 1 A c 2 A 3 ) or ( A 1 A 2 A c 3 ). P (lot accepted) = P ( A 1 A 2 A 3 ) + P ( A c 1 A 2 A 3 ) + P ( A 1 A c 2 A 3 ) + P ( A 1 A 2 A c 3 ) 1
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We can use conditional probabilities to write P ( A 1 A 2 A 3 ) = P ( A 3 | A 1 A 2 ) P ( A 2 | A 1 ) P ( A 1 ) Similarly, P ( A c 1 A 2 A 3 ) = P ( A 3 | A c 1 A 2 ) P ( A 2 | A c 1 ) P ( A c 1 ) P ( A 1 A c 2 A 3 ) = P ( A 3 | A 1 A c 2 ) P ( A c 2 | A 1 ) P ( A 1 ) P ( A 1 A 2 A c 3 ) = P ( A c 3 | A 1 A 2 ) P ( A 2 | A 1 ) P ( A 1 ) With 5 defective items and 95 non-defective items in the lot, P ( A 1 ) = 95 100 P ( A 2 | A 1 ) = 94 99 P ( A 3 | A 1 A 2 ) = 93 98 Filling in the other probabilities in a similar way, P (lot accepted) = 95 100 94 99 93 98 + 5 100 95 99 94 98 + 95 100 5 99 94 98 + 95 100 94 99 5 98 = (93)(94)(95) (98)(99)(100) + 3 (5)(95)(94) (100)(99)(98) If there are 10 defective and 90 non-defective items in the lot, P (lot accepted) = (88)(89)(90) (98)(99)(100) + 3 (10)(90)(89) (100)(99)(98) Problem 4. textbook problem 2.57 A nonsymmetric binary communications channel is shown in Fig. P2.2. Assume the inputs are equiprobable.
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